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KooM1QkVr1JhZisA3F1kluga4m4 | maths | sequences-and-series | summation-of-series | The sum of the infinite series <br/>$$1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ......$$ is equal to : | [{"identifier": "A", "content": "$${9 \\over 4}$$"}, {"identifier": "B", "content": "$${13 \\over 4}$$"}, {"identifier": "C", "content": "$${15 \\over 4}$$"}, {"identifier": "D", "content": "$${11 \\over 4}$$"}] | ["B"] | null | $$S = 1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + ....$$<br><br>$${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {7 \over {{3^3}}} + {{12} \over {{3^4}}} + ....$$<br><br>$$2S = 1 + {1 \over 3} + {5 \over {{3^2}}} + {5 \over {{3^3}}} + {5 \over {{3^4}}} + ....$$ + up to infinite terms<br><br>$${{2S} \over 3} = {4 \over 3} + {5 \over 3}\left\{ {{{1/3} \over {1 - {1 \over 3}}}} \right\} = {5 \over 6} + {4 \over 3} = {{13} \over 6}$$
<br><br>$$ \Rightarrow $$ S = $${13 \over 4}$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
ZBMT4NvKr5BSRHqBGT1kluxngbh | maths | sequences-and-series | summation-of-series | The sum of the series <br/><br/>$$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $$ is equal to : | [{"identifier": "A", "content": "$${{41} \\over 8}e + {{19} \\over 8}{e^{ - 1}} - 10$$"}, {"identifier": "B", "content": "$${{41} \\over 8}e - {{19} \\over 8}{e^{ - 1}} - 10$$"}, {"identifier": "C", "content": "$${{41} \\over 8}e + {{19} \\over 8}{e^{ - 1}} + 10$$"}, {"identifier": "D", "content": "$$ - {{41} \\over 8}e + {{19} \\over 8}{e^{ - 1}} - 10$$"}] | ["B"] | null | $$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $$<br><br>Put 2n + 1 = r, where r = 3, 5, 7, .......<br><br>$$ \Rightarrow n = {{r - 1} \over 2}$$<br><br>$${{{n^2} + 6n + 10} \over {(2n + 1)!}} = {{{{\left( {{{r - 1} \over 2}} \right)}^2} + 3r - 3 + 10} \over {r!}} $$
<br><br>$$= {{{r^2} + 10r + 29} \over {4r!}}$$ = $${{{r(r - 1) + 11r + 29} \over {4r!}}} $$<br><br>Now, $$\sum\limits_{r = 3,5,7} {{{r(r - 1) + 11r + 29} \over {4r!}}} $$
<br><br>=$${1 \over 4}\sum\limits_{r = 3,5,7,......} {\left( {{1 \over {(r - 2)!}} + {{11} \over {(r - 1)!}} + {{29} \over {r!}}} \right)} $$<br><br>$$ = {1 \over 4}\left\{ {\left( {{1 \over {1!}} + {1 \over {3!}} + {1 \over {5!}} + ......} \right) + 11\left( {{1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + ......} \right) + 29\left( {{1 \over {3!}} + {1 \over {5!}} + {1 \over {7!}} + ......} \right)} \right\}$$<br><br>$$ = {1 \over 4}\left\{ {{{e - {1 \over e}} \over 2} + 11\left( {{{e + {1 \over e} - 2} \over 2}} \right) + 29\left( {{{e - {1 \over e} - 2} \over 2}} \right)} \right\}$$<br><br>$$ = {1 \over 8}\left\{ {e - {1 \over e} + 11e + {{11} \over e} - 22 + 29e - {{29} \over e} - 58} \right\}$$<br><br><br>$$ = {1 \over 8}\left\{ {41e - {{19} \over e} - 80} \right\}$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
zd4RTaPsITd0LFIGzU1kmiz8sd6 | maths | sequences-and-series | summation-of-series | S<sub>n</sub>(x) = log<sub>a<sup>1/2</sup></sub>x + log<sub>a<sup>1/3</sup></sub>x + log<sub>a<sup>1/6</sup></sub>x + log<sub>a<sup>1/11</sup></sub>x + log<sub>a<sup>1/18</sup></sub>x + log<sub>a<sup>1/27</sup></sub>x + ...... up to n-terms, where a > 1. If S<sub>24</sub>(x) = 1093 and S<sub>12</sub>(2x) = 265, then value of a is equal to ____________. | [] | null | 16 | $${S_n}(x) = {\log _a}{x^2} + {\log _a}{x^3} + {\log _a}{x^6} + {\log _a}{x^{11}}$$<br><br>$${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$$<br><br>$${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$$<br><br>$${S_r} = 2 + 3 + 6 + 11$$
<br><br>$$ \therefore $$ T<sub>n</sub> = 2 + (1 + 3 + 5 +......+ (n - 1))
<br><br>= 2 + $${{n - 1} \over 2}\left[ {2.1 + \left( {n - 2} \right)2} \right]$$
<br><br>= 2 + $$\left( {n - 1} \right)\left[ {1 + \left( {n - 2} \right)} \right]$$
<br><br>= n<sup>2</sup> - 2n + 3
<br><br>General term $${T_r} = {r^2} - 2r + 3$$<br><br>$${S_n}(x) = \sum\limits_{r = 1}^n {{{\log }_a}x({r^2} - 2r + 3)} $$<br><br>$${S_{24}}(x) = \sum\limits_{r = 1}^{24} {{{\log }_a}x({r^2} - 2r + 3)} $$<br><br>$${S_{24}}(x) = {\log _a}x\sum\limits_{r = 1}^{24} {({r^2} - 2r + 3)} $$<br><br>$$1093 = 4372{\log _a}x$$<br><br>$${\log _a}x = {1 \over 4}$$<br><br>$$x = {a^{1/4}}$$ .....(i)<br><br>$${S_{12}}(2x) = {\log _a}(2x)\sum\limits_{r = 1}^{12} {({r^2} - 2r + 3)} $$<br><br>$$265 = 530{\log _a}(2x)$$<br><br>$${\log _a}(2x) = {1 \over 2}$$<br><br>$$2x = {a^{1/2}}$$ ....(ii)<br><br>From (i) and (ii), we get<br><br>$$2{a^{{1 \over 4}}} = {a^{{1 \over 2}}}$$
<br><br>$$ \Rightarrow $$ $${\left( {2{a^{{1 \over 4}}}} \right)^4} = {\left( {{a^{{1 \over 2}}}} \right)^4}$$
<br><br>$$ \Rightarrow $$ 16$$a$$ = $$a$$<sup>2</sup>
<br><br>$$ \Rightarrow $$ $$a = 16$$ | integer | jee-main-2021-online-16th-march-evening-shift |
qu0cXEoBuHthtOcLsa1kmlisv2k | maths | sequences-and-series | summation-of-series | If $$\alpha$$, $$\beta$$ are natural numbers such that <br/>100<sup>$$\alpha$$</sup> $$-$$ 199$$\beta$$ = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through ($$\alpha$$, $$\beta$$) and origin is : | [{"identifier": "A", "content": "540"}, {"identifier": "B", "content": "550"}, {"identifier": "C", "content": "530"}, {"identifier": "D", "content": "510"}] | ["B"] | null | RHS = $$\sum\limits_{r = 0}^{99} {(100 - r)(100 + r)} $$<br><br>$$ = {(100)^3} - {{99 \times 100 \times 199} \over 6} = {(100)^3} - (1650)199$$<br><br>LHS = (100)<sup>$$\alpha$$</sup> $$-$$ (199)<sup>$$\beta$$</sup><br><br>So, $$\alpha$$ = 3, $$\beta$$ = 1650<br><br>Slope = tan$$\theta$$ = $${\beta \over \alpha }$$<br><br>$$ \Rightarrow $$ tan$$\theta$$ = 550 | mcq | jee-main-2021-online-18th-march-morning-shift |
XHol4qKySWK9HAFu5P1kmlix67q | maths | sequences-and-series | summation-of-series | $${1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}}$$ is equal to | [{"identifier": "A", "content": "$${{101} \\over {404}}$$"}, {"identifier": "B", "content": "$${{25} \\over {101}}$$"}, {"identifier": "C", "content": "$${{101} \\over {408}}$$"}, {"identifier": "D", "content": "$${{99} \\over {400}}$$"}] | ["B"] | null | $$S = \sum\limits_{r = 1}^{100} {{1 \over {{{(2n + 1)}^2} - 1}}} $$<br><br>$$ = \sum\limits_{r = 1}^{100} {{1 \over {(2n + 1 + 1)(2n + 1 - 1)}}} $$<br><br>$$ = \sum\limits_{r = 1}^{100} {{1 \over {2n(2n + 2)}}} $$<br><br>$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {{1 \over {n(n + 1)}}} $$<br><br>$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {{{(n + 1) - n} \over {n(n + 1)}}} $$<br><br>$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $$<br><br>$$S = {1 \over 4}\left( {\left( {1 - {1 \over 2}} \right) + \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right) + ...... + \left( {{1 \over {100}} - {1 \over {101}}} \right)} \right)$$<br><br>$$ \therefore $$ $$S = {1 \over 4}\left[ {{{100} \over {101}}} \right] = {{25} \over {101}}$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
1krruwb6f | maths | sequences-and-series | summation-of-series | If sum of the first 21 terms of the series $${\log _{{9^{1/2}}}}x + {\log _{{9^{1/3}}}}x + {\log _{{9^{1/4}}}}x + .......$$, where x > 0 is 504, then x is equal to | [{"identifier": "A", "content": "243"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "81"}] | ["D"] | null | $$s = 2{\log _9}x + 3{\log _9}x + ....... + 22{\log _9}x$$<br><br>$$s = {\log _9}x(2 + 3 + ..... + 22)$$<br><br>$$s = {\log _9}x\left\{ {{{21} \over 2}(2 + 22)} \right\}$$<br><br>Given, $$252{\log _9}x = 504$$<br><br>$$ \Rightarrow {\log _9}x = 2 \Rightarrow x = 81$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1krrx00sq | maths | sequences-and-series | summation-of-series | For k $$\in$$ N, let $${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $$, where $$\alpha > 0$$. Then the value of $$100{\left( {{{{A_{14}} + {A_{15}}} \over {{A_{13}}}}} \right)^2}$$ is equal to _____________. | [] | null | 9 | $${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $$<br><br>$${A_{14}} = {1 \over {( - 14)( - 13)......( - 1)(1).......(6)}} = {1 \over {14!.6!}}$$<br><br>$${A_{15}} = {1 \over {( - 15)( - 14)......( - 1)(1).......(5)}} = {1 \over {15!.5!}}$$<br><br>$${A_{13}} = {1 \over {( - 13)......( - 1)(1).......(7)}} = {1 \over {13!.7!}}$$<br><br>$${{{A_{14}}} \over {{A_{13}}}} = {1 \over {14!.6!}} \times - 13! \times 7! = {{ - 7} \over {14}} = - {1 \over 2}$$<br><br>$${{{A_{15}}} \over {{A_{13}}}} = {1 \over {15! \times 5!}} \times - 13! \times 7! = {{42} \over {15 \times 14}} = {1 \over 5}$$<br><br>$$100{\left( {{{{A_{14}}} \over {{A_{13}}}} + {{{A_{15}}} \over {{A_{13}}}}} \right)^2} = 100{\left( { - {1 \over 2} + {1 \over 5}} \right)^2} = 9$$ | integer | jee-main-2021-online-20th-july-evening-shift |
1krrx9mvg | maths | sequences-and-series | summation-of-series | Let $$\left\{ {{a_n}} \right\}_{n = 1}^\infty $$ be a sequence such that a<sub>1</sub> = 1, a<sub>2</sub> = 1 and $${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$$ for all n $$\ge$$ 1. Then the value of $$47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}} $$ is equal to ______________. | [] | null | 7 | $${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$$, let $$\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} = P$$<br><br>Divide by 8<sup>n</sup> we get<br><br>$${{{a_{n + 2}}} \over {{8^n}}} = {{2{a_{n + 1}}} \over {{8^n}}} + {{{a_n}} \over {{8^n}}}$$<br><br>$$ \Rightarrow 64{{{a_{n + 2}}} \over {{8^{n + 2}}}} = {{16{a_{n + 1}}} \over {{8^{n + 1}}}} + {{{a_n}} \over {{8^n}}}$$<br><br>$$64\sum\limits_{n = 1}^\infty {{{{a_{n + 2}}} \over {{8^{n + 2}}}}} = 16\sum\limits_{n = 1}^\infty {{{{a_{n + 1}}} \over {{8^{n + 1}}}}} + \sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} $$<br><br>$$64\left( {P - {{{a_1}} \over 8} - {{{a_2}} \over {{8^2}}}} \right) = 16\left( {P - {{{a_1}} \over 8}} \right) + P$$<br><br>$$ \Rightarrow 64\left( {P - {1 \over 8} - {1 \over {64}}} \right) = 16\left( {P - {1 \over 8}} \right) + P$$<br><br>$$64P - 8 - 1 = 16P - 2 + P$$<br><br>$$47P = 7$$ | integer | jee-main-2021-online-20th-july-evening-shift |
1krw23dgt | maths | sequences-and-series | summation-of-series | If the value of<br/><br/> $${\left( {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....upto\,\infty } \right)^{{{\log }_{(0.25)}}\left( {{1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....upto\,\infty } \right)}}$$<br/><br/> is $$l$$, then $$l$$<sup>2</sup> is equal to _______________. | [] | null | 3 | $$l = {\left( {\underbrace {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}}}_S + ....} \right)^{{{\log }_{0.25}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ...} \right)}}$$<br><br>$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....$$<br><br>$${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + .....$$$${{2x} \over 3} = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + ....$$<br><br>$${{2S} \over 3} = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + .....$$<br><br>$$S = {3 \over 2}\left( {{{4/3} \over {1 - 1/3}}} \right) = 3$$<br><br>Now, $$l = {\left( 3 \right)^{{{\log }_{0.25}}\left( {{{1/3} \over {1 - 1/3}}} \right)}}$$<br><br>$$l = {3^{{{\log }_{\left( {(1/4)} \right)}}\left( {{1 \over 2}} \right)}} = {3^{1/2}} = \sqrt 3 $$<br><br>$$\Rightarrow$$ l<sup>2</sup> = 3 | integer | jee-main-2021-online-25th-july-morning-shift |
1ktbcep0a | maths | sequences-and-series | summation-of-series | The sum of the series <br/><br/>$${1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$ when x = 2 is : | [{"identifier": "A", "content": "$$1 + {{{2^{101}}} \\over {{4^{101}} - 1}}$$"}, {"identifier": "B", "content": "$$1 + {{{2^{100}}} \\over {{4^{101}} - 1}}$$"}, {"identifier": "C", "content": "$$1 - {{{2^{100}}} \\over {{4^{100}} - 1}}$$"}, {"identifier": "D", "content": "$$1 - {{{2^{101}}} \\over {{2^{400}} - 1}}$$"}] | ["D"] | null | $$S = {1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$<br><br>$$S + {1 \over {1 - x}} = {1 \over {1 - x}} + {1 \over {x + 1}} + ...... = {2 \over {1 - {x^2}}} + {2 \over {1 + {x^2}}} + ....$$<br><br>$$S + {1 \over {1 - x}} = {{{2^{101}}} \over {1 - {x^{400}}}}$$<br><br>$$S = 1 - {{{2^{101}}} \over {{2^{400}} - 1}}$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktehkmgu | maths | sequences-and-series | summation-of-series | If 0 < x < 1, then $${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$$, is equal to : | [{"identifier": "A", "content": "$$x\\left( {{{1 + x} \\over {1 - x}}} \\right) + {\\log _e}(1 - x)$$"}, {"identifier": "B", "content": "$$x\\left( {{{1 - x} \\over {1 + x}}} \\right) + {\\log _e}(1 - x)$$"}, {"identifier": "C", "content": "$${{1 - x} \\over {1 + x}} + {\\log _e}(1 - x)$$"}, {"identifier": "D", "content": "$${{1 + x} \\over {1 - x}} + {\\log _e}(1 - x)$$"}] | ["A"] | null | Let $$t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty $$<br><br>$$ = \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty $$<br><br>$$ = 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + .....\infty } \right)$$<br><br>$$ = {{2{x^2}} \over {1 - x}} - (\ln (1 - x) - x)$$<br><br>$$ \Rightarrow t = {{2{x^2}} \over {1 - x}} + x - \ln (1 - x)$$<br><br>$$ \Rightarrow t = {{x(1 + x)} \over {1 - x}} - \ln (1 - x)$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktehsn1f | maths | sequences-and-series | summation-of-series | If for x, y $$\in$$ R, x > 0, y = log<sub>10</sub>x + log<sub>10</sub>x<sup>1/3</sup> + log<sub>10</sub>x<sup>1/9</sup> + ...... upto $$\infty$$ terms <br/><br/>and $${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$$, then the ordered pair (x, y) is equal to : | [{"identifier": "A", "content": "(10<sup>6</sup>, 6)"}, {"identifier": "B", "content": "(10<sup>4</sup>, 6)"}, {"identifier": "C", "content": "(10<sup>2</sup>, 3)"}, {"identifier": "D", "content": "(10<sup>6</sup>, 9)"}] | ["D"] | null | $${{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}$$<br><br>$$ \Rightarrow {\log _{10}}x = 6 \Rightarrow x = {10^6}$$<br><br>Now, <br><br>$$y = ({\log _{10}}x) + \left( {{{\log }_{10}}{x^{{1 \over 3}}}} \right) + \left( {{{\log }_{10}}{x^{{1 \over 9}}}} \right) + ....\infty $$<br><br>$$ = \left( {1 + {1 \over 3} + {1 \over 9} + ....\infty } \right){\log _{10}}x$$<br><br>$$ = \left( {{1 \over {1 - {1 \over 3}}}} \right){\log _{10}}x = 9$$<br><br>So, (x, y) = (10<sup>6</sup>, 9) | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktg3am8n | maths | sequences-and-series | summation-of-series | If 0 < x < 1 and $$y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$$, then the value of e<sup>1 + y</sup> at $$x = {1 \over 2}$$ is : | [{"identifier": "A", "content": "$${1 \\over 2}{e^2}$$"}, {"identifier": "B", "content": "2e"}, {"identifier": "C", "content": "$${1 \\over 2}\\sqrt e $$"}, {"identifier": "D", "content": "2e<sup>2</sup>"}] | ["A"] | null | $$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$$<br><br>$$ = ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)$$<br><br>$$ = {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ....} \right)$$<br><br>$$ = {x \over {1 - x}} + \ln (1 - x)$$<br><br>$$x = {1 \over 2} \Rightarrow y = 1 - \ln 2$$<br><br>$${e^{1 + y}} = {e^{1 + 1 - \ln 2}}$$<br><br>$$ = {e^{2 - \ln 2}} = {{{e^2}} \over 2}$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktioa9ro | maths | sequences-and-series | summation-of-series | The sum of 10 terms of the series
<br/><br/>$${3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....$$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${{120} \\over {121}}$$"}, {"identifier": "C", "content": "$${{99} \\over {100}}$$"}, {"identifier": "D", "content": "$${{143} \\over {144}}$$"}] | ["B"] | null | $$S = {{{2^2} - {1^2}} \over {{1^2} \times {2^2}}} + {{{3^2} - {2^2}} \over {{2^2} \times {3^2}}} + {{{4^2} - {3^2}} \over {{3^2} \times {4^2}}} + ...$$<br><br>$$ = \left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right] + \left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] + \left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right] + .... + \left[ {{1 \over {{{10}^2}}} - {1 \over {{{11}^2}}}} \right]$$<br><br>$$ = 1 - {1 \over {121}}$$<br><br>$$ = {{120} \over {121}}$$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktkeexuz | maths | sequences-and-series | summation-of-series | If $$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$$, then 160 S is equal to ________. | [] | null | 305 | $$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$$<br><br>$${1 \over 5}S = {7 \over 5} + {9 \over {{5^3}}} + {{13} \over {{5^4}}} + ....$$<br><br>On subtracting<br><br>$${4 \over 5}S = {7 \over 5} + {2 \over {{5^2}}} + {4 \over {{5^3}}} + {6 \over {{5^4}}} + ....$$<br><br>$$S = {7 \over {14}} + {1 \over {10}}\left( {1 + {2 \over 5} + {3 \over {{5^2}}} + ...} \right)$$<br><br>$$S = {7 \over 4} + {1 \over {10}}{\left( {1 - {1 \over 5}} \right)^{ - 2}}$$<br><br>$$ = {7 \over 4} + {1 \over {10}} \times {{25} \over {16}} = {{61} \over {32}}$$<br><br>$$\Rightarrow$$ 160S = 5 $$\times$$ 61 = 305 | integer | jee-main-2021-online-31st-august-evening-shift |
1kto7yln2 | maths | sequences-and-series | summation-of-series | Let S<sub>n</sub> = 1 . (n $$-$$ 1) + 2 . (n $$-$$ 2) + 3 . (n $$-$$ 3) + ..... + (n $$-$$ 1) . 1, n $$\ge$$ 4.<br/><br/>The sum $$\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $$ is equal to : | [{"identifier": "A", "content": "$${{e - 1} \\over 3}$$"}, {"identifier": "B", "content": "$${{e - 2} \\over 6}$$"}, {"identifier": "C", "content": "$${e \\over 3}$$"}, {"identifier": "D", "content": "$${e \\over 6}$$"}] | ["A"] | null | Let T<sub>r</sub> = r(n $$-$$ r)<br><br>T<sub>r</sub> = nr $$-$$ r<sup>2</sup><br><br>$$ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {{T_r} = \sum\limits_{r = 1}^n {(nr - {r^2})} } $$<br><br>$${S_n} = {{n\,.\,(n)(n + 1)} \over 2} - {{n(n + 1)(2n + 1)} \over 6}$$<br><br>$${S_n} = {{n(n - 1)(n + 1)} \over 6}$$<br><br>Now, $$\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $$<br><br>$$ = \sum\limits_{r = 4}^\infty {\left( {2.{{n(n - 1)(n + 1)} \over {6\,.\,n(n - 1)(n - 2)!}} - {1 \over {(n - 2)!}}} \right)} $$<br><br>$$ = \sum\limits_{r = 4}^\infty {\left( {{1 \over 3}\left( {{{n - 2 + 3} \over {(n - 2)!}}} \right) - {1 \over {(n - 2)!}}} \right)} $$<br><br>$$ = \sum\limits_{r = 4}^\infty {{1 \over 3}.{1 \over {(n - 3)!}} = {1 \over 3}(e - 1)} $$<br><br>Option (a) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l545a1mq | maths | sequences-and-series | summation-of-series | <p>Let $$\{ {a_n}\} _{n = 0}^\infty $$ be a sequence such that $${a_0} = {a_1} = 0$$ and $${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$$ for all n $$\ge$$ 0. Then, $$\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}}} $$ is equal to:</p> | [{"identifier": "A", "content": "$${6 \\over {343}}$$"}, {"identifier": "B", "content": "$${7 \\over {216}}$$"}, {"identifier": "C", "content": "$${8 \\over {343}}$$"}, {"identifier": "D", "content": "$${{49} \\over {216}}$$"}] | ["B"] | null | <p>$${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$$ & $${a_0} = {a_1} = 0$$</p>
<p>$${a_2} = 2{a_1} - {a_0} + 1 = 1$$</p>
<p>$${a_3} = 2{a_2} - {a_1} + 1 = 3$$</p>
<p>$${a_4} = 2{a_3} - {a_2} + 1 = 6$$</p>
<p>$${a_5} = 2{a_4} - {a_3} + 1 = 10$$</p>
<p>$$\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}} = {{{a_2}} \over {{7^2}}} + {{{a_3}} \over {{7^3}}} + {{{a_4}} \over {{7^4}}} + \,\,...} $$</p>
<p>$$s = {1 \over {{7^2}}} + {3 \over {{7^3}}} + {6 \over {{7^4}}} + {{10} \over {{7^5}}} + \,\,...$$</p>
<p>$${{{1 \over 7}s = {1 \over {{7^3}}} + {3 \over {{7^4}}} + {6 \over {{7^5}}} + \,\,...} \over {{{6s} \over 7} = {1 \over {{7^2}}} + {2 \over {{7^3}}} + {3 \over {{7^4}}} + \,\,...}}$$</p>
<p>$${{{{6s} \over {49}} = \,\,\,\,\,\,\,\,\,{1 \over {{7^3}}} + {2 \over {{7^4}}} + \,\,...} \over {{{36s} \over {49}} = {1 \over {{7^2}}} + {1 \over {{7^3}}} + {1 \over {{7^4}}} + \,\,...}}$$</p>
<p>$${{36s} \over {49}} = {{{1 \over {{7^2}}}} \over {1 - {1 \over 7}}}$$</p>
<p>$${{36s} \over {49}} = {7 \over {49 \times 6}}$$</p>
<p>$$s = {7 \over {216}}$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54apl39 | maths | sequences-and-series | summation-of-series | <p>The sum of the infinite series $$1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + {{35} \over {{6^4}}} + {{51} \over {{6^5}}} + {{70} \over {{6^6}}} + \,\,.....$$ is equal to :</p> | [{"identifier": "A", "content": "$${{425} \\over {216}}$$"}, {"identifier": "B", "content": "$${{429} \\over {216}}$$"}, {"identifier": "C", "content": "$${{288} \\over {125}}$$"}, {"identifier": "D", "content": "$${{280} \\over {125}}$$"}] | ["C"] | null | <p>$$S = 1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + $$ ...... (1)</p>
<p>$${1 \over 6}S = {1 \over 6} + {5 \over {{6^2}}} + {{12} \over {{6^3}}} + $$ ...... (2)</p>
<p>$$S - {1 \over 6}S = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $$ ........</p>
<p>$$ \Rightarrow {{5S} \over 6} = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $$ ....... (3)</p>
<p>Now, multiplying both sides by $${1 \over 6}$$, we get</p>
<p>$$ \Rightarrow {{5S} \over {36}} = {1 \over 6} + {4 \over {{6^2}}} + {7 \over {{6^3}}} + {{10} \over {{6^4}}} + $$ ...... (4)</p>
<p>Subtract equation (4) from equation (3), we get</p>
<p>$${{25} \over {36}}S = 1 + {3 \over 6} + {3 \over {{6^2}}} + {3 \over {{6^3}}} + $$ ......</p>
<p>$$ \Rightarrow {{25S} \over {36}} = 1 + {{{3 \over 5}} \over {1 - {1 \over 6}}}$$</p>
<p>$$ = 1 + {3 \over 6} \times {6 \over 5}$$</p>
<p>$$ = 1 + {3 \over 5} = {8 \over 5}$$</p>
<p>$$ \Rightarrow S = {8 \over 5} \times {{36} \over {25}} = {{288} \over {125}}$$</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l56q6r0z | maths | sequences-and-series | summation-of-series | <p>Let $$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + \,.....$$. Then 4S is equal to</p> | [{"identifier": "A", "content": "$${\\left( {{7 \\over 3}} \\right)^2}$$"}, {"identifier": "B", "content": "$${{{7^3}} \\over {{3^2}}}$$"}, {"identifier": "C", "content": "$${\\left( {{7 \\over 3}} \\right)^3}$$"}, {"identifier": "D", "content": "$${{{7^2}} \\over {{3^3}}}$$"}] | ["C"] | null | <p>$$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + $$ ..... ...... (i)</p>
<p>$${1 \over 7}S = {2 \over 7} + {6 \over {{7^2}}} + {{12} \over {{7^3}}} + {{20} \over {{7^4}}} + $$ .... ....... (ii)</p>
<p>(i) - (ii)</p>
<p>$${6 \over 7}S = 2 + {4 \over 7} + {6 \over {{7^2}}} + {8 \over {{7^3}}} + $$ ...... ....... (iii)</p>
<p>$${6 \over {{7^2}}}S = {2 \over 7} + {4 \over {{7^2}}} + {6 \over {{7^3}}} + $$ ..... ......... (iv)</p>
<p>(iii) - (iv)</p>
<p>$${\left( {{6 \over 7}} \right)^2}S = 2 + {2 \over 7} + {2 \over {{7^2}}} + {2 \over {{7^3}}} + $$ ......</p>
<p>$$ = 2\left[ {{1 \over {1 - {1 \over 7}}}} \right] = 2\left( {{7 \over 6}} \right)$$</p>
<p>$$\therefore$$ $$4S = 8{\left( {{7 \over 6}} \right)^3} = {\left( {{7 \over 3}} \right)^3}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l57p5mqd | maths | sequences-and-series | summation-of-series | <p>If the sum of the first ten terms of the series</p>
<p>$${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$$</p>
<p>is $${m \over n}$$, where m and n are co-prime numbers, then m + n is equal to ______________.</p> | [] | null | 276 | <p>$${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$$</p>
<p>$$ = {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$$</p>
<p>$$ = {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$$</p>
<p>$${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $$</p>
<p>$$ = {1 \over 4}\left[ {1 - {1 \over 5} + {1 \over 5} - {1 \over {13}} + \,\,....\,\, + \,\,{1 \over {181}} - {1 \over {221}}} \right]$$</p>
<p>$$ \Rightarrow {S_{10}} = {1 \over 4}\,.\,{{220} \over {221}} = {{55} \over {221}} = {m \over n}$$</p>
<p>$$\therefore$$ $$m + n = 276$$</p> | integer | jee-main-2022-online-27th-june-morning-shift |
1l58f35j3 | maths | sequences-and-series | summation-of-series | <p>If $$A = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$ and $$B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$, then $${A \over B}$$ is equal to :</p> | [{"identifier": "A", "content": "$${{11} \\over 9}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$$${{11} \\over 9}$$"}, {"identifier": "D", "content": "$$-$$$${{11} \\over 3}$$"}] | ["C"] | null | <p>$$A = \sum\limits_{n = 1}^\infty {{1 \over {{{(3 + {{( - 1)}^n})}^n}}}} $$ and $$B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{(3 + {{( - 1)}^n})}^n}}}} $$</p>
<p>$$A = {1 \over 2} + {1 \over {{4^2}}} + {1 \over {{2^3}}} + {1 \over {{4^4}}} + $$ ........</p>
<p>$$B = {{ - 1} \over 2} + {1 \over {{4^2}}} - {1 \over {{2^3}}} + {1 \over {{4^4}}} + $$ ......</p>
<p>$$A = {{{1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$$, $$B = {{ - {1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$$</p>
<p>$$A = {{11} \over {15}}$$, $$B = {{ - 9} \over {15}}$$</p>
<p>$$\therefore$$ $${A \over B} = {{ - 11} \over 9}$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59jzej6 | maths | sequences-and-series | summation-of-series | <p>The sum 1 + 2 . 3 + 3 . 3<sup>2</sup> + ......... + 10 . 3<sup>9</sup> is equal to :</p> | [{"identifier": "A", "content": "$${{2\\,.\\,{3^{12}} + 10} \\over 4}$$"}, {"identifier": "B", "content": "$${{19\\,.\\,{3^{10}} + 1} \\over 4}$$"}, {"identifier": "C", "content": "$$5\\,.\\,{3^{10}} - 2$$"}, {"identifier": "D", "content": "$${{9\\,.\\,{3^{10}} + 1} \\over 2}$$"}] | ["B"] | null | <p>Let $$S = 1\,.\,{3^0} + 2\,.\,{3^1} + 3\,.\,{3^2} + \,\,......\,\, + \,\,10\,.\,{3^9}$$</p>
<p>$$3S = 1\,.\,{3^1} + 2\,.\,{3^2} + \,\,..........\,\, + \,\,10\,.\,{3^{10}}$$</p>
<p>___________________________________________________________</p>
<p>$$ - 2S = (1\,.\,{3^0} + 1\,.\,{3^1} + 1\,.\,{3^2} + \,\,........\,\, + \,\,1\,.\,{3^9}) - 10\,.\,{3^{10}}$$</p>
<p>$$ \Rightarrow S = {1 \over 2}\left[ {10\,.\,{3^{10}} - {{{3^{10}} - 1} \over { - 3 - 1}}} \right]$$</p>
<p>$$ \Rightarrow S = {{19\,.\,{3^{10}} + 1} \over 4}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5ajtlod | maths | sequences-and-series | summation-of-series | <p>For a natural number n, let $${\alpha _n} = {19^n} - {12^n}$$. Then, the value of $${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}}$$ is ___________.</p> | [] | null | 4 | <p>$${\alpha _n} = {19^n} - {12^n}$$</p>
<p>Let equation of roots 12 & 19 i.e.</p>
<p>$${x^2} - 31x + 228 = 0$$</p>
<p>$$ \Rightarrow (31 - x) = {{228} \over x}$$ (where x can be 19 or 12)</p>
<p>$$\therefore$$ $${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}} = {{31({{19}^9} - {{12}^9}) - ({{19}^{10}} - {{12}^{10}})} \over {57({{19}^8} - {{12}^8})}}$$</p>
<p>$$ = {{{{19}^9}(31 - 19) - {{12}^9}(31 - 12)} \over {57({{19}^8} - {{12}^8})}}$$</p>
<p>$$ = {{228({{19}^8} - {{12}^8})} \over {57({{19}^8} - {{12}^8})}} = 4$$.</p> | integer | jee-main-2022-online-25th-june-morning-shift |
1l5ajy4fx | maths | sequences-and-series | summation-of-series | <p>The greatest integer less than or equal to the sum of first 100 terms of the sequence $${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$$ ...... is equal to ___________.</p> | [] | null | 98 | <p>$$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, + $$ ....</p>
<p>$$ = \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)} $$</p>
<p>$$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$$</p>
<p>$$ = 98 + 2{\left( {{2 \over 3}} \right)^{100}}$$</p>
<p>$$\therefore$$ $$[S] = 98$$</p> | integer | jee-main-2022-online-25th-june-morning-shift |
1l5vzhxr8 | maths | sequences-and-series | summation-of-series | <p>The value of $$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}} + \,\,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,.....\,\, + \,\,11}}$$ is equal to:</p> | [{"identifier": "A", "content": "$${{20} \\over {11}}$$"}, {"identifier": "B", "content": "$${{11} \\over {6}}$$"}, {"identifier": "C", "content": "$${{241} \\over {132}}$$"}, {"identifier": "D", "content": "$${{21} \\over {11}}$$"}] | ["B"] | null | <p>Given,</p>
<p>$$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}}\, + \,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,....\,\, + \,11}}$$</p>
<p>General term,</p>
<p>$${T_n} = {1 \over {1 + 2 + 3\, + \,\,....\,\, + \,\,n}}$$</p>
<p>$$ = {1 \over {{{n(n + 1)} \over 2}}}$$</p>
<p>$$ = {2 \over {n(n + 1)}}$$</p>
<p>$$ = 2\left[ {{{n + 1 - n} \over {n(n + 1)}}} \right]$$</p>
<p>$$ = 2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$$</p>
<p>$${t_1} = 2\left[ {{1 \over 1} - {1 \over 2}} \right]$$</p>
<p>$${t_2} = 2\left[ {{1 \over 2} - {1 \over 3}} \right]$$</p>
<p>$${t_3} = 2\left[ {{1 \over 3} - {1 \over 4}} \right]$$</p>
<p>$$ \vdots $$</p>
<p>$${t_n} = 2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$$</p>
<p>$$\therefore$$ $${S_n} = {t_1} + {t_2} + {t_3}\, + \,\,....\,\, + \,\,{t_n}$$</p>
<p>$$ = 2\left[ {{1 \over 1} - {1 \over 2} + {1 \over 2} - {1 \over 3} + {1 \over 3} - {1 \over 4}\, + \,\,...\,\, + \,\,{1 \over n} - {1 \over {n + 1}}} \right]$$</p>
<p>$$ = 2\left[ {1 - {1 \over {n + 1}}} \right]$$</p>
<p>$$ = 2\left[ {{{n + 1 - 1} \over {n + 1}}} \right]$$</p>
<p>$$ = {{2n} \over {n + 1}}$$</p>
<p>$$\therefore$$ $${S_{11}} = {{2 \times 11} \over {11 + 1}} = {{22} \over {12}} = {{11} \over 6}$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6dx9fyy | maths | sequences-and-series | summation-of-series | <p>Let $$a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$$ and $$b_{n}=a_{n}+b_{n-1}$$ for every <br/><br/>natural number $$n \geqslant 2$$. Then $$\sum\limits_{n = 1}^{15} {{a_n}.{b_n}} $$ is equal to ___________.</p> | [] | null | 27560 | <p>Given,</p>
<p>$${a_n} = {a_{n - 1}} + 2$$</p>
<p>$$ \Rightarrow {a_n} - {a_{n - 1}} = 2$$</p>
<p>$$\therefore$$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.</p>
<p>Also given $${a_1} = 1$$</p>
<p>$$\therefore$$ Series is = 1, 3, 5, 7 ......</p>
<p>$$\therefore$$ $${a_n} = 1 + (n - 1)2 = 2n - 1$$</p>
<p>Also $${b_n} = {a_n} + {b_{n - 1}}$$</p>
<p>When $$n = 2$$ then</p>
<p>$${b_2} - {b_1} = {a_2} = 3$$</p>
<p>$$ \Rightarrow {b_2} - 1 = 3$$ [Given $${b_1} = 1$$]</p>
<p>$$ \Rightarrow {b_2} = 4$$</p>
<p>When $$n = 3$$ then</p>
<p>$${b_3} - {b_2} = {a_3}$$</p>
<p>$$ \Rightarrow {b_3} - 4 = 5$$</p>
<p>$$ \Rightarrow {b_3} = 9$$</p>
<p>$$\therefore$$ Series is = 1, 4, 9 ......</p>
<p>= 1<sup>2</sup>, 2<sup>2</sup>, 3<sup>2</sup> ....... n<sup>2</sup></p>
<p>$$\therefore$$ $${b_n} = {n^2}$$</p>
<p>Now, $$\sum\limits_{n = 1}^{15} {\left( {{a_n}\,.\,{b_n}} \right)} $$</p>
<p>$$ = \sum\limits_{n = 1}^{15} {\left[ {(2n - 1){n^2}} \right]} $$</p>
<p>$$ = \sum\limits_{n = 1}^{15} {2{n^3} - \sum\limits_{n = 1}^{15} {{n^2}} } $$</p>
<p>$$ = 2\left( {{1^3} + {2^3} + \,\,...\,\,{{15}^3}} \right) - \left( {{1^2} + {2^2} + \,\,...\,\,{{15}^2}} \right)$$</p>
<p>$$ = 2 \times {\left( {{{15 \times 16} \over 2}} \right)^2} - \left( {{{15(16) \times 31} \over 6}} \right)$$</p>
<p>$$ = 27560$$</p> | integer | jee-main-2022-online-25th-july-morning-shift |
1l6f0wdoi | maths | sequences-and-series | summation-of-series | <p>The sum $$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}} $$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{7}{87}$$"}, {"identifier": "B", "content": "$$\\frac{7}{29}$$"}, {"identifier": "C", "content": "$$\\frac{14}{87}$$"}, {"identifier": "D", "content": "$$\\frac{21}{29}$$"}] | ["B"] | null | <p>$$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}} = {3 \over 4}\sum\limits_{n = 1}^{21} {{1 \over {4n - 1}} - {1 \over {4n + 3}}} } $$</p>
<p>$$ = {3 \over 4}\left[ {\left( {{1 \over 3} - {1 \over 7}} \right) + \left( {{1 \over 7} - {1 \over {11}}} \right) + \,\,....\,\, + \,\,\left( {{1 \over {83}} - {1 \over {87}}} \right)} \right]$$</p>
<p>$$ = {3 \over 4}\left[ {{1 \over 3} - {1 \over {87}}} \right] = {3 \over 4}{{84} \over {3.87}} = {7 \over {29}}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6gjc5oa | maths | sequences-and-series | summation-of-series | <p>The series of positive multiples of 3 is divided into sets : $$\{3\},\{6,9,12\},\{15,18,21,24,27\}, \ldots$$ Then the sum of the elements in the $$11^{\text {th }}$$ set is equal to ____________.</p> | [] | null | 6993 | <p>Given series</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbx3j2/7964fa43-5af9-4fc1-96c1-0a4102aef97d/c56543e0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbx3j3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbx3j2/7964fa43-5af9-4fc1-96c1-0a4102aef97d/c56543e0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbx3j3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Morning Shift Mathematics - Sequences and Series Question 87 English Explanation"> </p>
<p>$$\therefore$$ 11<sup>th</sup> set will have $$1 + (10)2 = 21$$ term</p>
<p>Also upto 10<sup>th</sup> set total $$3 \times k$$ type terms will be $$1 + 3 + 5\, + \,......\, + \,19 = 100 - $$ term</p>
<p>$$\therefore$$ Set $$11 = \{ 3 \times 101,\,3 \times 102,\,......\,3 \times 121\} $$</p>
<p>$$\therefore$$ Sum of elements $$ = 3 \times (101 + 102\, + \,...\, + \,121)$$</p>
<p>$$ = {{3 \times 222 \times 21} \over 2} = 6993$$</p> | integer | jee-main-2022-online-26th-july-morning-shift |
1l6hzs0vf | maths | sequences-and-series | summation-of-series | <p>If $$\sum\limits_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$$, where m and n are co-prime, then $$m+n$$ is equal to _____________.</p> | [] | null | 166 | <p>$$\sum\limits_{k = 1}^{10} {{k \over {{k^4} + {k^2} + 1}}} $$</p>
<p>$$ = {1 \over 2}\left[ {\sum\limits_{k = 1}^{10} {\left( {{1 \over {{k^2} - k + 1}} - {1 \over {{k^2} + k + 1}}} \right)} } \right.$$</p>
<p>$$ = {1 \over 2}\left[ {1 - {1 \over 3} + {1 \over 3} - {1 \over 7} + {1 \over 7} - {1 \over {13}}\, + \,...\, + \,{1 \over {91}} - {1 \over {111}}} \right]$$</p>
<p>$$ = {1 \over 2}\left[ {1 - {1 \over {111}}} \right] = {{110} \over {2\,.\,111}} = {{55} \over {111}} = {m \over n}$$</p>
<p>$$\therefore$$ $$m + n = 55 + 111 = 166$$</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6klic2q | maths | sequences-and-series | summation-of-series | <p>$$
\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\cdots+
\frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$$ is equal to _____________.</p> | [] | null | 120 | <p>$${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{2n(2{n^2} + 3n + 1) - 3{n^2} - 3n + n} \over {n(4n + 3)}}$$</p>
<p>$$ = {{{n^2}(4n + 3)} \over {n(4n + 3)}} = n$$</p>
<p>$$\therefore$$ $${T_n} = n$$</p>
<p>$${S_n} = \sum\limits_{n = 1}^{15} {{T_n} = {{15 \times 16} \over 2} = 120} $$</p> | integer | jee-main-2022-online-27th-july-evening-shift |
1l6m6bitt | maths | sequences-and-series | summation-of-series | <p>Consider the sequence $$a_{1}, a_{2}, a_{3}, \ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$$ for $$\mathrm{n}=1,2,3, \ldots .$$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\frac{\mathrm{a}_{3}+\frac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\frac{\mathrm{a}_{30}+\frac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)$$, then $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$30"}, {"identifier": "B", "content": "$$-$$31"}, {"identifier": "C", "content": "$$-$$60"}, {"identifier": "D", "content": "$$-$$61"}] | ["C"] | null | <p>$${a_{n + 2}} = {2 \over {{a_{n + 1}}}} + {a_n}$$</p>
<p>$$ \Rightarrow {a_n}{a_{n + 1}} + 1 = {a_{n + 1}}{a_{n + 2}} - 1$$</p>
<p>$$ \Rightarrow {a_{n + 2}}{a_{n + 1}} - {a_n}\,.\,{a_{n + 1}} = 2$$</p>
<p>For</p>
<p>$$\matrix{
{n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \cr
{n = 2} & {{a_4}{a_3} - {a_3}{a_2} = 2} \cr
{n = 3} & {{a_5}{a_4} - {a_4}{a_3} = 2} \cr
{} & {\matrix{
. \cr
. \cr
. \cr
. \cr
} } \cr
{n = n} & {{{{a_{n + 2}}{a_{n + 1}} - {a_n}{a_{n + 1}} = 2} \over {{a_{n + 2}}{a_{n + 1}} = 2n + {a_1}{a_2}}}} \cr
} $$</p>
<p>Now,</p>
<p>$${{({a_1}{a_2} + 1)} \over {{a_2}{a_3}}}\,.\,{{({a_2}{a_3} + 1)} \over {{a_3}{a_4}}}\,.\,{{({a_3}{a_4} + 1)} \over {{a_4}{a_5}}}\,.\,.....\,.\,{{({a_{30}}{a_{31}} + 1)} \over {{a_{31}}{a_{32}}}}$$</p>
<p>$$ = {3 \over 4}\,.\,{5 \over 6}\,.\,{7 \over 8}\,.\,....\,.\,{{61} \over {62}}$$</p>
<p>$$ = {2^{ - 60}}\left( {{}^{61}{C_{31}}} \right)$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6npcsi3 | maths | sequences-and-series | summation-of-series | $${6 \over {{3^{12}}}} + {{10} \over {{3^{11}}}} + {{20} \over {{3^{10}}}} + {{40} \over {{3^9}}} + \,\,...\,\, + \,\,{{10240} \over 3} = {2^n}\,.\,m$$, where m is odd, then m . n is equal to ____________. | [] | null | 12 | <p>$${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right) = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {1 \over {{3^{12}}}} + {5 \over 5}\left( {{1 \over {{3^{12}}}}\,.\,{2^{12}}\,.\,{3^{12}} - {1 \over {{3^{12}}}}} \right) = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {1 \over {{3^{12}}}} + {2^{12}} - {1 \over {{3^{12}}}} = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {2^n}\,.\,m = {2^{12}}$$</p>
<p>$$ \Rightarrow m = 1$$ and $$n = 12$$</p>
<p>$$m\,.\,n = 12$$</p> | integer | jee-main-2022-online-28th-july-evening-shift |
1l6p3pptf | maths | sequences-and-series | summation-of-series | <p>If $$\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102}=\frac{\mathrm{k}}{101}$$, then 34 k is equal to _________.</p> | [] | null | 286 | <p>$$S = {1 \over {2 \times 3 \times 4}} + {1 \over {3 \times 4 \times 5}} + {1 \over {4 \times 5 \times 6}}\, + \,....\, + \,{1 \over {100 \times 101 \times 102}}$$</p>
<p>$$ = {1 \over {(3 - 1)\,.\,1}}\left[ {{1 \over {2 \times 3}} - {1 \over {101 \times 102}}} \right]$$</p>
<p>$$ = {1 \over 2}\left( {{1 \over 6} - {1 \over {101 \times 102}}} \right)$$</p>
<p>$$ = {{143} \over {102 \times 101}} = {k \over {101}}$$</p>
<p>$$\therefore$$ $$34k = 286$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
1l6re3sg0 | maths | sequences-and-series | summation-of-series | <p>$$
\begin{aligned}
&\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\
&a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 .
\end{aligned}
$$</p>
<p>Then $$a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$$ is equal to</p> | [{"identifier": "A", "content": "483"}, {"identifier": "B", "content": "528"}, {"identifier": "C", "content": "575"}, {"identifier": "D", "content": "624"}] | ["B"] | null | <p>Given,</p>
<p>$${a_0} = {a_1} = 0$$</p>
<p>and $${a_{n + 2}} = 3{a_{n + 1}} - 2{a_n} + 1$$</p>
<p>For $$n = 0,\,{a_2} = 3{a_1} - 2{a_0} + 1$$</p>
<p>$$ = 3\,.\,0 - 2\,.\,0 + 1$$</p>
<p>$$ = 1$$</p>
<p>For $$n = 1,\,{a_3} = 3{a_2} - 2{a_1} + 1$$</p>
<p>$$ = 3\,.\,1 - 2\,.\,0 + 1$$</p>
<p>$$ = 4$$</p>
<p>For $$n = 2,\,{a_4} = 3{a_3} - 2{a_2} + 1$$</p>
<p>$$ = 3\,.\,4 - 2\,.\,1 + 1$$</p>
<p>$$ = 11$$</p>
<p>For $$n = 3,\,{a_5} = 3{a_4} - 2{a_3} + 1$$</p>
<p>$$ = 3\,.\,11 - 2\,.\,4 + 1$$</p>
<p>$$ = 26$$</p>
<p>For $$n = 4,\,{a_6} = 3{a_5} - 2{a_4} + 1$$</p>
<p>$$ = 3\,.\,26 - 2\,.\,11 + 1$$</p>
<p>$$ = 57$$</p>
<p>$$\therefore$$ $${S_n} = 1 + 4 + 11 + 26 + 57\, + \,....\, + \,{t_n}$$</p>
<p>$${S_n} = 1 + 4 + 11 + 26\, + \,....\, + \,{t_{n - 1}} + {t_n}$$</p>
<p>$$0 = 1 + 3 + 7 + 15 + 31\, + \,.....\, - {t_n}$$</p>
<p>$$ \Rightarrow {t_n} = 1 + 3 + 7 + 15 + 31\, + \,....$$</p>
<p>Now, find the sum of the series,</p>
<p>$${t_n} = 1 + 3 + 7 + 15 + 31\, + \,.....\, + \,{x_{n - 1}} + {x_n}$$ .....(1)</p>
<p>$${t_n} = $$ $$1 + 3 + 7 + 15\, + \,.....\, + \,{x_{n - 1}} + {x_n}$$ ......(2)</p>
<p>Subtracting (2) from (1), we get</p>
<p>-------------------------------------------------------------------------</p>
<p>$$0 = 1 + 2 + 4 + 8 + 16\, + \,....\, + \,{x_n}$$</p>
<p>$$ \Rightarrow {x_n} = 1 + 2 + 4 + 8 + 16\, + \,.....\, + \,$$ n terms</p>
<p>$$ = {{1({2^n} - 1)} \over {2 - 1}}$$</p>
<p>$$ = {2^n} - 1$$</p>
<p>$$\therefore$$ $${t_n} = \sum\limits_{n = 1}^n {{x_n}} $$</p>
<p>$$ = \sum\limits_{n = 1}^n {({2^n} - 1)} $$</p>
<p>$$ = \sum\limits_{n = 1}^n {{2^n} - \sum\limits_{n = 1}^n 1 } $$</p>
<p>$$ = {{2({2^n} - 1)} \over {2 - 1}} - n$$</p>
<p>$$ = {2^{n + 1}} - 2 - n$$</p>
<p>$${t_1} = {2^2} - 2 - 1 = 1 = {a_2}$$</p>
<p>$${t_2} = {2^3} - 2 - 2 = 4 = {a_3}$$</p>
<p>$${t_3} = {2^4} - 2 - 3 = 11 = {a_4}$$</p>
<p>$$\therefore$$ $${a_{22}} = {t_{21}} = {2^{22}} - 2 - 21 = {2^{22}} - 23$$</p>
<p>$${a_{23}} = {t_{22}} = {2^{23}} - 2 - 22 = {2^{23}} - 24$$</p>
<p>$${a_{24}} = {t_{23}} = {2^{24}} - 2 - 23 = {2^{24}} - 25$$</p>
<p>$${a_{25}} = {t_{24}} = {2^{25}} - 2 - 24 = {2^{25}} - 26$$</p>
<p>Now,</p>
<p>$${a_{25}}{a_{23}} - 2{a_{25}}{a_{22}} - 2{a_{23}}{a_{24}} + 4{a_{22}} {a_{24}}$$</p>
<p>$$ = {a_{25}}({a_{23}} - 2{a_{22}}) - 2{a_{24}}({a_{23}} - 2{a_{22}})$$</p>
<p>$$ = ({a_{23}} - 2{a_{22}})({a_{25}} - 2{a_{24}})$$</p>
<p>$$ = [({2^{23}} - 24) - 2({2^{22}} - 23)][({2^{25}} - 26) - 2({2^{24}} - 25)]$$</p>
<p>$$ = [({2^{23}} - 24 - {2^{23}} + 46)][({2^{25}} - 26 - {2^{25}} + 50)]$$</p>
<p>$$ = (22)(24)$$</p>
<p>$$ = 528$$</p> | mcq | jee-main-2022-online-29th-july-evening-shift |
1ldo67p4f | maths | sequences-and-series | summation-of-series | <p>The sum $$\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}} $$ is equal to :</p> | [{"identifier": "A", "content": "$${{11e} \\over 2} + {7 \\over {2e}}$$"}, {"identifier": "B", "content": "$${{13e} \\over 4} + {5 \\over {4e}} - 4$$"}, {"identifier": "C", "content": "$${{11e} \\over 2} + {7 \\over {2e}} - 4$$"}, {"identifier": "D", "content": "$${{13e} \\over 4} + {5 \\over {4e}}$$"}] | ["B"] | null | $\begin{aligned} & \sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !} \\\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{2 n(2 n-1)+8 n+8}{(2 n) !} \\\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\\\ & e=1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots . \\\\ & e^{-1}=1-1+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots . \\\\ & \left(\mathrm{e}+\frac{1}{\mathrm{e}}\right)=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots . .\right) \\\\ & e-\frac{1}{e}=\left(1+\frac{1}{3 !}+\frac{1}{5 !}+\ldots . .\right)\end{aligned}$
<br/><br/>Now
<br/><br/>$$
\begin{aligned}
& \frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}\right)+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\\\
& =\frac{1}{2}\left[\frac{e+\frac{1}{\mathrm{e}}}{2}\right]+2\left[\frac{\mathrm{e}-\frac{1}{\mathrm{e}}}{2}\right]+4\left[\frac{\mathrm{e}+\frac{1}{\mathrm{e}}-2}{2}\right] \\\\
& =\frac{\left(\mathrm{e}+\frac{1}{\mathrm{e}}\right)}{4}+e-\frac{1}{\mathrm{e}}+2 \mathrm{e}+\frac{2}{\mathrm{e}}-4 \\\\
& =\frac{13}{4} e+\frac{5}{4 e}-4
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-evening-shift |
ldoawens | maths | sequences-and-series | summation-of-series | The sum $1^{2}-2 \cdot 3^{2}+3 \cdot 5^{2}-4 \cdot 7^{2}+5 \cdot 9^{2}-\ldots+15 \cdot 29^{2}$ is _________. | [] | null | 6952 | $S=1^{2}-2.3^{2}+3.5^{2}-4.7^{2}+\ldots \ldots+15.29^{2}$
<br/><br/>Separating odd placed and even placed terms we get
<br/><br/>$$
\begin{aligned}
& \mathrm{S}=\left(1.1^2+3.5^2+\ldots .15 .(29)^2\right)-\left(2.3^2+4.7^2\right. \\
& +\ldots .+14 .(27)^2
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\sum_{r=1}^{8}(2 r-1)(4 r-3)^{2}-\sum_{r=1}^{7} 2 r(4 r-1)^{2} \\\\
& =\sum_{r=1}^{8} (32 r^{3}-64 r^{2}+42 r-9)-2\sum_{r=1}^{7} 16 r^{3}-8 r^{2}+r \\\\
& =32 \times 36^{2}-64 \times 204+1512-72 \\\\
& -2\left(16 \times 28^{2}-1120+28\right) \\\\
& =6592
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-evening-shift |
1ldom86ia | maths | sequences-and-series | summation-of-series | <p>The sum of 10 terms of the series</p>
<p>$${1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....$$ is</p> | [{"identifier": "A", "content": "$${{58} \\over {111}}$$"}, {"identifier": "B", "content": "$${{56} \\over {111}}$$"}, {"identifier": "C", "content": "$${{55} \\over {111}}$$"}, {"identifier": "D", "content": "$${{59} \\over {111}}$$"}] | ["C"] | null | $$
\begin{aligned}
& T_n=\frac{n}{1+n^2+n^4} \\\\
& =\frac{n}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)} \\\\
& =\frac{1}{2}\left[\frac{\left(\mathrm{n}^2+\mathrm{n}+1\right)-\left(\mathrm{n}^2-\mathrm{n}+1\right)}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\\\
& \Rightarrow \mathrm{T}_{\mathrm{n}}=\frac{1}{2}\left[\frac{1}{\left(\mathrm{n}^2-\mathrm{n}+1\right)}-\frac{1}{\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\\\
& \mathrm{S}_{\mathrm{n}}=\sum_{n=1}^{10} T_n \\\\
& =\frac{1}{2} \sum\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right) \\\\
& =\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)\right. \\\\
& \left.\dots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\left(\frac{1}{91}-\frac{1}{111}\right)\right]
\end{aligned}
$$
<br/><br/>$$ \therefore $$ $ S=\frac{1}{2}\left(1-\frac{1}{111}\right)=\frac{55}{111} $ | mcq | jee-main-2023-online-1st-february-morning-shift |
ldqzs9si | maths | sequences-and-series | summation-of-series | The $8^{\text {th }}$ common term of the series
<br/><br/>$$
\begin{aligned}
& S_1=3+7+11+15+19+\ldots . . \\\\
& S_2=1+6+11+16+21+\ldots . .
\end{aligned}
$$
<br/><br/>is : | [] | null | 151 | <p>First common term is 11</p>
<p>Common difference of series of common terms is LCM (4, 5) = 20</p>
<p>$$a_8=a+7d$$</p>
<p>$$=11+7\times20=151$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldr5km4h | maths | sequences-and-series | summation-of-series | <p>If $${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$$, then $${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$$ is equal to :</p> | [{"identifier": "A", "content": "$${{51} \\over {144}}$$"}, {"identifier": "B", "content": "$${{49} \\over {138}}$$"}, {"identifier": "C", "content": "$${{50} \\over {141}}$$"}, {"identifier": "D", "content": "$${{52} \\over {147}}$$"}] | ["C"] | null | <p>$$\sum\limits_{i = 1}^{25} {{a_i} = \sum {{{ - 2} \over {4{n^2} - 16n + 15}} = \sum {{{ - 2} \over {(2n - 5)(2n - 3)}}} } } $$</p>
<p>$$ = \sum\limits_{i = 1}^{25} {\left( {{1 \over {2n - 3}} - {1 \over {2n - 5}}} \right)} $$</p>
<p>$$ = \left[ {\left( {{1 \over { - 1}} - {1 \over { - 3}}} \right) + \left( {{1 \over 1} - {1 \over { - 1}}} \right) + \left( {{1 \over 3} - {1 \over 1}} \right)......} \right.$$</p>
<p>$$ = {1 \over {2(25) - 3}} + {1 \over 3} = {{50} \over {141}}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldr7tnf5 | maths | sequences-and-series | summation-of-series | <p>Let $$\sum_\limits{n=0}^{\infty} \frac{\mathrm{n}^{3}((2 \mathrm{n}) !)+(2 \mathrm{n}-1)(\mathrm{n} !)}{(\mathrm{n} !)((2 \mathrm{n}) !)}=\mathrm{ae}+\frac{\mathrm{b}}{\mathrm{e}}+\mathrm{c}$$, where $$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}$$ and $$e=\sum_\limits{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n} !}$$ Then $$\mathrm{a}^{2}-\mathrm{b}+\mathrm{c}$$ is equal to ____________.</p> | [] | null | 26 | <p>$$\sum\limits_{n = 0}^\infty {{{{n^3}(2n!) + (2n - 1)(n!)} \over {n!\,.\,(2n)!}}} $$</p>
<p>$$ = \sum\limits_{n = 0}^\infty {{{{n^3}} \over {n!}} + {{2n - 1} \over {2n!}}} $$</p>
<p>$$ = \sum\limits_{n = 0}^\infty {{3 \over {(n - 2)!}} + {1 \over {(n - 3)!}} + {1 \over {(n - 1)!}} + {1 \over {(2n - 1)!}} - {1 \over {(2n)!}}} $$</p>
<p>$$ = 3e + e + e - {1 \over e}$$</p>
<p>$$ = 5e - {1 \over e}$$</p>
<p>$$\therefore$$ $$a = 5,b = - 1,c = 0$$</p>
<p>$$\therefore$$ $${a^2} - b + c = 26$$</p> | integer | jee-main-2023-online-30th-january-morning-shift |
1ldsg00ct | maths | sequences-and-series | summation-of-series | <p>Let $$a_1=b_1=1$$ and $${a_n} = {a_{n - 1}} + (n - 1),{b_n} = {b_{n - 1}} + {a_{n - 1}},\forall n \ge 2$$. If $$S = \sum\limits_{n = 1}^{10} {{{{b_n}} \over {{2^n}}}} $$ and $$T = \sum\limits_{n = 1}^8 {{n \over {{2^{n - 1}}}}} $$, then $${2^7}(2S - T)$$ is equal to ____________.</p> | [] | null | 461 | <p>$$\because$$ $${a_n} = {a_{n - 1}} + (n - 1)$$ and $${a_1} = {b_1} = 1$$</p>
<p>$${b_n} = {b_{n - 1}} + {a_{n - 1}}$$</p>
<p>$$\therefore$$ $${b_{n + 1}} = 2{b_n} - {b_{n - 1}} + n - 1$$</p>
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<th class="tg-amwm">$$n$$</th>
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<th class="tg-amwm">$$b_n-n$$</th>
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<p>$$\therefore$$ $$2S - T = \left( {\sum\limits_{n = 1}^8 {{{{b_n} - n} \over {{2^{n - 1}}}}} } \right) + {{{b_9}} \over {{2^8}}} + {{{b_{10}}} \over {{2^9}}}$$</p>
<p>$$ = {{461} \over {128}}$$</p>
<p>$$\therefore$$ $${2^7}(2S - T) = 461$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldwxkndi | maths | sequences-and-series | summation-of-series | <p>If $${{{1^3} + {2^3} + {3^3}\, + \,...\,up\,to\,n\,terms} \over {1\,.\,3 + 2\,.\,5 + 3\,.\,7\, + \,...\,up\,to\,n\,terms}} = {9 \over 5}$$, then the value of $$n$$ is</p> | [] | null | 5 | Given $\frac{1^{3}+2^{3}+3^{3}+\ldots \text { up to } n \text { terms }}{1.3+2.5+3.7+\ldots \text { up to } n \text { terms }}=\frac{9}{5}$
<br/><br/>
Now
<br/><br/>
Let $S=1.3+2.5+3.7+\ldots$
<br/><br/>
$$
\begin{aligned}
& T_{n}=n \cdot(2 n+1) \\\\
& \therefore S=\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\\\
& \Rightarrow \frac{\left(\frac{n(n+1)}{2}\right)^{2}}{n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]}=\frac{9}{5} \\\\
& \Rightarrow 5 n^{2}-19 n-30=0 \\\\
& \Rightarrow(5 n+6)(n-5)=0 \\\\
& \therefore n=5
\end{aligned}
$$ | integer | jee-main-2023-online-24th-january-evening-shift |
lgnyx1yy | maths | sequences-and-series | summation-of-series | If the sum of the series
<br/><br/>$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^{2}}-\frac{1}{2 \cdot 3}+\frac{1}{3^{2}}\right)+\left(\frac{1}{2^{3}}-\frac{1}{2^{2} \cdot 3}+\frac{1}{2 \cdot 3^{2}}-\frac{1}{3^{3}}\right)+$
<br/><br/>$\left(\frac{1}{2^{4}}-\frac{1}{2^{3} \cdot 3}+\frac{1}{2^{2} \cdot 3^{2}}-\frac{1}{2 \cdot 3^{3}}+\frac{1}{3^{4}}\right)+\ldots$
<br/><br/>is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha+3 \beta$ is equal to __________. | [] | null | 7 | We can rewrite the given series as follows :
<br/><br/>$$S = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right) + \left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right) + \left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right) + \ldots$$
<br/><br/>The first few terms of the series are :
<br/><br/>$$S = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\right) - \left(\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{18} + \frac{1}{36} + \ldots\right) - \left(\frac{1}{27} + \frac{1}{54} + \ldots\right) + \ldots$$
<br/><br/>We can now see that each group of terms forms a geometric series with a common ratio of $\frac{1}{2}$:
<br/><br/>$$S = \frac{\frac{1}{2}}{1-\frac{1}{2}} - \frac{\frac{1}{3}}{1-\frac{1}{2}} + \frac{\frac{1}{9}}{1-\frac{1}{2}} - \frac{\frac{1}{27}}{1-\frac{1}{2}} + \ldots$$
<br/><br/>The series can be rewritten as :
<br/><br/>$$S = 2 \left(\frac{1}{{2}} - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)$$
<br/><br/>Now, we can simplify and rewrite the series inside the parentheses as:
<br/><br/>$$S = 2 \left[\frac{1}{{2}} + (- \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)]$$
<br/><br/>The series inside the parentheses is an infinite geometric series with the first term $a = - \frac{1}{3}$ and the common ratio $r = -\frac{1}{3}$:
<br/><br/>$$S = 2 \left(\frac{1}{{2}} +\frac{a}{1-r}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{1+\frac{1}{3}}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{\frac{4}{3}}\right) $$
<br/><br/>= $$2\left( {{1 \over 2} - {1 \over 3} \times {3 \over 4}} \right)$$
<br/><br/>$$ = 2\left( {{1 \over 2} - {1 \over 4}} \right) = 2\left( {{1 \over 4}} \right) = {1 \over 2}$$
<br/><br/>Thus, the sum of the series is $\frac{1}{2}$, and $\alpha = 1$ and $\beta = 2$ are co-prime.
<br/><br/>Therefore, $\alpha + 3\beta = 1 + 6 = 7$ | integer | jee-main-2023-online-15th-april-morning-shift |
1lgq0vrkc | maths | sequences-and-series | summation-of-series | <p>The sum to $$20$$ terms of the series $$2 \cdot 2^{2}-3^{2}+2 \cdot 4^{2}-5^{2}+2 \cdot 6^{2}-\ldots \ldots$$ is equal to __________.</p> | [] | null | 1310 | $$
\begin{aligned}
& \sum_{r=1}^{10}\left(2 \cdot(2 r)^2-(2 r+1)^2\right) \\\\
& =\sum_{r=1}^{10}\left(8 r^2-4 r^2-4 r-1\right) \\\\
& =\sum_{r=1}^{10}\left(4 r^2-4 r-1\right) \\\\
& =\frac{4 \cdot 10 \cdot 11 \cdot 21}{6}-4 \frac{10 \cdot 11}{2}-10 \\\\
& =44 \cdot 35-220-10 \\\\
& =1540-230=1310
\end{aligned}
$$ | integer | jee-main-2023-online-13th-april-morning-shift |
1lgreevjx | maths | sequences-and-series | summation-of-series | <p>Let $$< a_{\mathrm{n}} > $$ be a sequence such that $$a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$$. If $$28 \sum_\limits{k=1}^{10} \frac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}$$, where $$\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}$$ are the first $$\mathrm{m}$$ prime numbers, then $$\mathrm{m}$$ is equal to</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["C"] | null | Given the sum of the first n terms, $S_n = \frac{n^2+3n}{(n+1)(n+2)}$, we can find the n<sup>th</sup> term $a_n$ as the difference between the sum of the first n terms and the sum of the first n-1 terms :
<br/><br/>So,
$$a_n = S_n - S_{n-1}$$
<br/><br/>Solving, we get :
<br/><br/>$$a_n = \frac{n^2+3n}{(n+1)(n+2)} - \frac{(n-1)^2+3(n-1)}{n(n+1)}$$
<br/><br/>Simplifying further, we find :
<br/><br/>$$a_n = \frac{4}{n(n+1)(n+2)}$$
<br/><br/>Then, we find the reciprocal of $a_n$:
<br/><br/>$$\frac{1}{a_n} = \frac{n(n+1)(n+2)}{4}$$
<br/><br/>Now, we sum this over the first 10 terms :
<br/><br/>$$\sum\limits_{k=1}^{10} \frac{1}{a_k} = \sum\limits_{k=1}^{10} \frac{k(k+1)(k+2)}{4}$$
<br/><br/>Evaluating the sum :
<br/><br/>$$
\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} \left[\sum\limits_{k=1}^{10} k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)\right]
$$
<br/><br/>This can be rewritten as the sum of differences :
<br/><br/>$$
\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} [(1 \cdot 2 \cdot 3 \cdot 4 - 0) + (2 \cdot 3 \cdot 4 \cdot 5 - 1 \cdot 2 \cdot 3 \cdot 4) + \cdots + (10 \cdot 11 \cdot 12 \cdot 13 - 9 \cdot 10 \cdot 11 \cdot 12)]
$$
<br/><br/>$$
\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} (10 \cdot 11 \cdot 12 \cdot 13 - 0) = \frac{1}{16} \cdot 17160
$$
<br/><br/>Now, given the condition that :
<br/><br/>$$
28 \sum\limits_{k=1}^{10} \frac{1}{a_k} = p_1p_2p_3...p_m
$$
<br/><br/>Substituting the sum we've calculated:
<br/><br/>$$
28 \cdot \frac{1}{16} \cdot 17160 = p_1p_2p_3...p_m
$$
<br/><br/>This simplifies to :
<br/><br/>$$
30030 = p_1p_2p_3...p_m
$$
<br/><br/>The prime factorization of 30030 is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$, which consists of 6 primes.
<br/><br/>Therefore, m is equal to 6.
| mcq | jee-main-2023-online-12th-april-morning-shift |
1lgsw5qty | maths | sequences-and-series | summation-of-series | <p>For $$k \in \mathbb{N}$$, if the sum of the series $$1+\frac{4}{k}+\frac{8}{k^{2}}+\frac{13}{k^{3}}+\frac{19}{k^{4}}+\ldots$$ is 10 , then the value of $$k$$ is _________.</p> | [] | null | 2 | From the given series :
<br/><br/>$$10 = 1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$$
<br/><br/>We isolate the 1 to get :
<br/><br/>$$9 = \frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$$ .......(1)
<br/><br/>Divide each term in the equation by $k$ :
<br/><br/>$$\frac{9}{k} = \frac{4}{k^2}+\frac{8}{k^3}+\frac{13}{k^4}+\ldots$$ .......(2)
<br/><br/>Subtracting the second equation from the first, we obtain :
<br/><br/>$$9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$$
<br/><br/>This is equivalent to :
<br/><br/>$$S = 9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$$ .........(3)
<br/><br/>Divide both sides by $k$ again, we get :
$$\frac{S}{k} = \frac{4}{k^2} + \frac{4}{k^3} + \frac{5}{k^4} + \ldots$$ ..........(4)
<br/><br/>Subtracting equation (4) from (3), we have :
<br/><br/>$$(1-\frac{1}{k})S = \frac{4}{k} + \frac{1}{k^3} + \frac{1}{k^4} + \ldots$$
<br/><br/>In this equation, you've treated the infinite series on the right-hand side as a geometric series with a ratio of $1/k$, so the sum of this series can be expressed as $\frac{1/k^3}{1 - 1/k}$.
<br/><br/>Therefore, we get :
<br/><br/>$$9\left(1-\frac{1}{k}\right)^2 = \frac{4}{k} + \frac{1/k^3}{1 - 1/k}.$$
<br/><br/>Now, simplifying this equation, we get :
<br/><br/>$$9(k-1)^2 = 4k^2 - 4k + 1$$
<br/><br/>$$ \Rightarrow $$$$9(k^2 - 2k + 1) = 4k^2 - 4k + 1$$
<br/><br/>$$ \Rightarrow $$$$9k^2 - 18k + 9 = 4k^2 - 4k + 1$$
<br/><br/>$$ \Rightarrow $$$$9k^2 - 4k^2 - 18k + 4k + 9 - 1 = 0$$
<br/><br/>$$ \Rightarrow $$$$5k^2 - 14k + 8 = 0$$
<br/><br/>This is a standard form quadratic equation, $ax^2 + bx + c = 0$, where $a=5$, $b=-14$, and $c=8$.
<br/><br/>The solutions can be found using the quadratic formula, $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
<br/><br/>Substituting the values for $a$, $b$, and $c$, we have :
<br/><br/>$$ \Rightarrow $$$$k = \frac{14 \pm \sqrt{(-14)^2 - 4\times5\times8}}{2\times5}$$
<br/><br/>$$ \Rightarrow $$$$k = \frac{14 \pm \sqrt{196 - 160}}{10}$$
<br/><br/>$$ \Rightarrow $$$$k = \frac{14 \pm \sqrt{36}}{10}$$
<br/><br/>$$ \Rightarrow $$$$k = \frac{14 \pm 6}{10}$$
<br/><br/>So, the solutions are $k = 2$ or $k = 0.8$. However, since the question mentions $k \in \mathbb{N}$, i.e., $k$ is a natural number, the only valid solution is $k = 2$. | integer | jee-main-2023-online-11th-april-evening-shift |
1lguwu480 | maths | sequences-and-series | summation-of-series | <p>Let $$S=109+\frac{108}{5}+\frac{107}{5^{2}}+\ldots .+\frac{2}{5^{107}}+\frac{1}{5^{108}}$$. Then the value of $$\left(16 S-(25)^{-54}\right)$$ is equal to ___________.</p> | [] | null | 2175 | We have, $S=109+\frac{108}{5}+\frac{107}{5^2}+\ldots+\frac{2}{5^{107}}+\frac{1}{5^{108}}$ ...........(i) <br/><br/>$\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2}+\frac{107}{5^3}+\ldots .+\frac{2}{5^{108}}+\frac{1}{5^{109}}$ .............(ii)
<br/><br/>On subtracting Eq. (ii) from Eq. (i), we get
<br/><br/>$$
\begin{aligned}
\frac{4 S}{5} & =109-\frac{1}{5}-\frac{1}{5^2}-\ldots .-\frac{1}{5^{108}}-\frac{1}{5^{109}} \\\\
\frac{4 S}{5} & =109-\left[\frac{1}{5}+\frac{1}{5^2}+\ldots+\frac{1}{5^{108}}+\frac{1}{5^{109}}\right]
\end{aligned}
$$
<br/><br/>This form a GP with $r=\frac{1}{5}$
<br/><br/>$$
\begin{aligned}
\frac{4 S}{5} & =109-\frac{1}{5}\left[\frac{1-\frac{1}{5^{109}}}{1-\frac{1}{5}}\right]\\\\
&=109-\frac{1}{4}\left[1-\frac{1}{5^{109}}\right] \\\\
& =109-\frac{1}{4}+\frac{1}{4} \times \frac{1}{5^{109}}
\end{aligned}
$$
<br/><br/>$$ \therefore $$ $$
4 S=\frac{5}{4}\left[435+\frac{1}{5^{109}}\right]
$$
<br/><br/>$$ \Rightarrow 16 S=2175+\frac{1}{5^{108}}
$$
<br/><br/>$$
16 S-(25)^{-54}=2175
$$ | integer | jee-main-2023-online-11th-april-morning-shift |
1lgvpd2gn | maths | sequences-and-series | summation-of-series | <p>If $$\mathrm{S}_{n}=4+11+21+34+50+\ldots$$ to $$n$$ terms, then $$\frac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_{9}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "227"}, {"identifier": "B", "content": "226"}, {"identifier": "C", "content": "220"}, {"identifier": "D", "content": "223"}] | ["D"] | null | Given that
<br/><br/>$$
\begin{aligned}
& \mathrm{S}_n=4+11+21+24+50+\ldots+\mathrm{T}_n \\\\
& \mathrm{~S}_n=~~~~~~~~4+11+21+34++\mathrm{T}_{n-1}+\mathrm{T}_n \\
& -\quad-\quad-\quad-\quad-\quad-\quad- \\
& \hline 0=4+7+10+13+16+\ldots\left(\mathrm{T}_n-\mathrm{T}_{n-2}\right)-\mathrm{T}_n
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \mathrm{T}_n=4+7+10+13+16+\ldots . \text { to } n \text { terms } \\\\
& \Rightarrow \mathrm{T}_n=\frac{n}{2}[2 \times 4+(n-1) 3] \\\\
& \mathrm{T}_n=\frac{3}{2} n^2+\frac{5}{2} n
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { So } \mathrm{S}_n=\Sigma \mathrm{T}_n=\frac{3}{2} \Sigma n^2+\frac{5}{2} \Sigma n \\\\
& \Rightarrow \mathrm{S}_n=\frac{3}{2} \times \frac{n(n+1)(2 n+1)}{6}+\frac{5}{2} \times \frac{n(n+1)}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow S_n=\frac{n(n+1)}{4}(2 n+1+5)=\frac{n(n+1)(n+3)}{2} \\\\
& \text { Hence, } \frac{1}{60}\left(S_{29}-S_9\right)=\frac{1}{60} \times \frac{1}{2}(29 \times 30 \times 32-9 \times 10 \times 12) \\\\
& =223
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxwg676 | maths | sequences-and-series | summation-of-series | <p>The sum of all those terms, of the arithmetic progression 3, 8, 13, ...., 373, which are not divisible by 3, is equal to ____________.</p> | [] | null | 9525 | The sum of all those terms, of the arithmetic progression $3,8,13 \ldots, 373$.
<br/><br/>Which are not divisible by 3 is
<br/><br/>$$
\begin{aligned}
& =(3+8+13+18+\ldots+373) -(3+18+33+\ldots+363) \\\\
& =\frac{75}{2}(3+373)-\frac{25}{2}(3+363) \\\\
& =\frac{75}{2} \times 376-\frac{25}{2} \times 366 \\\\
& =75 \times 188-25 \times 183 \\\\
& =14100-4575 \\\\
& =9525
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-morning-shift |
1lgyl2od3 | maths | sequences-and-series | summation-of-series | <p>Let $$\mathrm{a}_{\mathrm{n}}$$ be the $$\mathrm{n}^{\text {th }}$$ term of the series $$5+8+14+23+35+50+\ldots$$ and $$\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$$. Then $$\mathrm{S}_{30}-a_{40}$$ is equal to :</p> | [{"identifier": "A", "content": "11280"}, {"identifier": "B", "content": "11290"}, {"identifier": "C", "content": "11310"}, {"identifier": "D", "content": "11260"}] | ["B"] | null | Let $\mathrm{S}_n=5+8+14+23+\ldots .+a_n$
<br/><br/>and $\mathrm{S}_n=0+5+8+14+\ldots .+a_n$
<br/><br/>On subtracting, we get
<br/><br/>$$
\begin{aligned}
& 0=5+3+6 \ldots-a_n \\\\
& \Rightarrow a_n=5+3+6+9+\ldots(n-1) \text { terms } \\\\
& =5+\left[\frac{(n-1)}{2}(6+(n-2) 3)\right]
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =5+\left[\frac{(n-1)}{2}(6+3 n-6)\right] \\\\
& =5+\frac{(n-1)(3 n)}{2} \\\\
& =\frac{10+3 n^2-3 n}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { So, } a_{40}=\frac{3(40)^2-3(40)+10}{2} \\\\
& =\frac{4800-120+10}{2}=2345
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, } S_n=\sum_{k=1}^n a_k \\\\
& \Rightarrow S_{30}=\frac{3 \sum_{n=1}^{30} n^2-3 \sum_{n=1}^{30} n+10 \sum_{n=1}^{30} 1}{2} \\\\
& =\frac{3 \times(30)(30+1)(60+1)}{12}-\frac{3 \times 30 \times 31}{4}
+\frac{10 \times 30}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\frac{28365-1395+300}{2}=\frac{27270}{2} \\\\
& =13635 \\\\
& \therefore S_{30}-a_{40}=13635-2345=11290
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh200ul0 | maths | sequences-and-series | summation-of-series | <p>The sum of the first $$20$$ terms of the series $$5+11+19+29+41+\ldots$$ is :</p> | [{"identifier": "A", "content": "3420"}, {"identifier": "B", "content": "3450"}, {"identifier": "C", "content": "3250"}, {"identifier": "D", "content": "3520"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{S}_n=5+11+19+29+41+\ldots .+\mathrm{T}_n \\\\
& \mathrm{S}_n=~~~~~~~~ 5+11+19+29+\ldots .+\mathrm{T}_{n-1}+\mathrm{T}_n \\\\
& \hline 0=5+6+8+10+12+\ldots . . \mathrm{T}_n
\end{aligned}
$$
<br/><br/>$$
\begin{array}{rlrl}
&0 =5+\frac{n-1}{2}[2 \times 6+(n-2)(2)]-T_n \\\\
&\Rightarrow T_n =5+(n-1)(n+4) \\\\
&\Rightarrow T_n =5+n^2+3 n-4 \\\\
&\Rightarrow T_n =n^2+3 n+1 \\\\
&\Sigma T_n =\Sigma n^2+3 \Sigma n+\Sigma 1
\end{array}
$$
<br/><br/>$$
\Rightarrow \quad S_n=\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}+n
$$
<br/><br/>When, $n=20$
<br/><br/>Then,
<br/><br/>$$
\begin{aligned}
S_{20} & =\frac{20 \times 21 \times 41}{6}+\frac{3 \times 20 \times 21}{2}+20 \\\\
& =2870+630+20=3520
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2xwwfd | maths | sequences-and-series | summation-of-series | <p> If $$\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1$$ and $$1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$$ then $$m^{2}-n^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "220"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "240"}, {"identifier": "D", "content": "180"}] | ["C"] | null | Given $\operatorname{gcd}(m, n)=1$ and
<br/><br/>$$
\begin{aligned}
& \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\
& =1012 m^2 n \\\\
& \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\
& =1012 m^2 n \\\\
& \Rightarrow(1-2)(1+2)+(3-4)(3+4)+\ldots .(2021-2022) \\
& (2021+2022)+(2023)^2=(1012) m^2 n \\\\
& \Rightarrow(-1)(1+2)+(-1)(3+4)+\ldots .+ \\
& \quad(-1)(2021+2022)+\left(2023^2\right)=(1012) m^2 n
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \Rightarrow(-1)[1+2+3+4+\ldots+2022]+(2023)^2 =1012 m^2 n \\\\ & \Rightarrow(-1)\left[\frac{(2022) \cdot(2022+1)}{2}\right]+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(-1)\left[\frac{(2022)(2023)}{2}\right]+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(2023)[2023-1011]=(1012) m^2 n \\\\ & \Rightarrow m^2 n=2023 \\\\\end{aligned}$
<br/><br/>$\begin{array}{lc}\Rightarrow m^2 n=289 \times 7 \\\\ \Rightarrow m^2=289 \text { and } n=7 \\\\ \Rightarrow m=17 \text { and } n=7(\text { such that } \operatorname{gcd}(m, n)=1) \\\\ \therefore m^2-n^2=289-49=240\end{array}$ | mcq | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lse57f85 | maths | sequences-and-series | summation-of-series | <p>The sum of the series $$\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots$$ up to 10 -terms is</p> | [{"identifier": "A", "content": "$$\\frac{45}{109}$$\n"}, {"identifier": "B", "content": "$$-\\frac{55}{109}$$\n"}, {"identifier": "C", "content": "$$\\frac{55}{109}$$\n"}, {"identifier": "D", "content": "$$-\\frac{45}{109}$$"}] | ["B"] | null | <p>General term of the sequence,</p>
<p>$$\begin{aligned}
& \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4} \\
& \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2} \\
& \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} \\
& \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\
& \mathrm{T}_{\mathrm{r}}=\frac{\frac{1}{2}\left[\left(\mathrm{r}^2+\mathrm{r}-1\right)-\left(\mathrm{r}^2-\mathrm{r}-1\right)\right]}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\
& =\frac{1}{2}\left[\frac{1}{\mathrm{r}^2-\mathrm{r}-1}-\frac{1}{\mathrm{r}^2+\mathrm{r}-1}\right]
\end{aligned}$$</p>
<p>Sum of 10 terms,</p>
<p>$$\sum_\limits{\mathrm{r}=1}^{10} \mathrm{~T}_{\mathrm{r}}=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
1lsg5cobe | maths | sequences-and-series | summation-of-series | <p>Let $$S_n$$ be the sum to $$n$$-terms of an arithmetic progression $$3,7,11$$,
If $$40<\left(\frac{6}{n(n+1)} \sum_\limits{k=1}^n S_k\right)<42$$, then $$n$$ equals ________.</p> | [] | null | 9 | <p>$$\begin{aligned}
& \mathrm{S}_{\mathrm{n}}= 3+7+11+\ldots \ldots \mathrm{n} \text { terms } \\
&=\frac{\mathrm{n}}{2}(6+(\mathrm{n}-1) 4)=3 \mathrm{n}+2 \mathrm{n}^2-2 \mathrm{n} \\
&=2 \mathrm{n}^2+\mathrm{n} \\
& \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}=2 \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K}^2+\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K} \\
&=2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\
&=\mathrm{n}(\mathrm{n}+1)\left[\frac{2 \mathrm{n}+1}{3}+\frac{1}{2}\right] \\
&=\frac{\mathrm{n}(\mathrm{n}+1)(4 \mathrm{n}+5)}{6} \\
& \Rightarrow 40<\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{k=1}^n S_{\mathrm{k}}<42 \\
& 40<4 \mathrm{n}+5<42 \\
& 35<4 n<37 \\
& \mathrm{n}=9
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift |
1lsgcmznt | maths | sequences-and-series | summation-of-series | <p>Let $$\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots$$ upto 10 terms and $$\beta=\sum_\limits{n=1}^{10} n^4$$. If $$4 \alpha-\beta=55 k+40$$, then $$\mathrm{k}$$ is equal to __________.</p> | [] | null | 353 | <p>$$\begin{gathered}
\alpha=1^2+4^2+8^2 \ldots . \\
t_n=a^2+b n+c
\end{gathered}$$</p>
<p>$$\begin{aligned}
& 1=a+b+c \\
& 4=4 a+2 b+c \\
& 8=9 a+3 b+c
\end{aligned}$$</p>
<p>On solving we get, $$\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1$$</p>
<p>$$\begin{aligned}
& \alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2 \\
& 4 \alpha=\sum_{n=1}^{10}\left(n^2+3 n-2\right)^2, \beta=\sum_{n=1}^{10} n^4 \\
& 4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^3+5 n^2-12 n+4\right)=55(353)+40
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-morning-shift |
luxwdj8i | maths | sequences-and-series | summation-of-series | <p>If $$\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots . .+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots \ldots+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$$, then $$\alpha$$ is equal to ___________.</p> | [] | null | 1011 | <p>$$\begin{aligned}
& \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\
& \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\
& \quad \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) \\
& \quad=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right) \\
& \quad=\left(1+\frac{1}{3}+\frac{1}{5}+\ldots .+\frac{1}{2023}\right) \\
& \quad-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2024}\right) \\
& \quad=\left(1+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1012}\right)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1011}\right) \\
& \frac{-1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1012}\right) \\
& =\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023}-\frac{1}{2024} \\
& \Rightarrow \alpha+1012=2023 \\
& \Rightarrow \alpha=1011
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-evening-shift |
luy6z4j2 | maths | sequences-and-series | summation-of-series | <p>If the sum of the series $$\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}$$ is equal to 5, then $$50 \mathrm{~d}$$ is equal to :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "20"}] | ["A"] | null | <p>$$\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5$$</p>
<p>Multiply and divide by $$d$$</p>
<p>$$\begin{aligned}
& \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\
& \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\
& \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \\
& \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{10}{1+10 d}=5 \\
& 1+10 d=2 \\
& d=\frac{1}{10} \\
& 50 d=50 \times \frac{1}{10}=5
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv2erzk5 | maths | sequences-and-series | summation-of-series | <p>The value of $$\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$$ is</p> | [{"identifier": "A", "content": "$$\\frac{305}{301}$$\n"}, {"identifier": "B", "content": "$$\\frac{306}{305}$$\n"}, {"identifier": "C", "content": "$$\\frac{32}{31}$$\n"}, {"identifier": "D", "content": "$$\\frac{31}{30}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n(n+1)^2}{\sum_\limits{n=1}^{100} n^2(n+1)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n^3+2 n^2+n}{\sum_\limits{n=1}^{100} n^3+n^2} \\
& =\frac{\left(\frac{100(101)}{2}\right)^2+\frac{2 \cdot 100(101)(201)}{6}+\frac{100(101)}{2}}{\left(\frac{100(101)}{2}\right)^2+\frac{100(101)(201)}{6}} \\
& =\frac{300(101)+4(201)+6}{300(101)+2(201)}=\frac{5185}{5117}=\frac{305}{301}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv5gt51f | maths | sequences-and-series | summation-of-series | <p>Let the positive integers be written in the form :</p>
<p><img src="data:image/png;base64,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"/></p>
<p>If the $$k^{\text {th }}$$ row contains exactly $$k$$ numbers for every natural number $$k$$, then the row in which the number 5310 will be, is __________.</p> | [] | null | 103 | <p>To solve this problem, we need to determine in which row the number $5310$ appears when positive integers are arranged in rows such that the $k^\text{th}$ row contains exactly $k$ numbers.</p>
<h3><strong>Understanding the Pattern</strong></h3>
<p><p><strong>First row ($k = 1$)</strong>: Contains 1 number.</p></p>
<p><p><strong>Second row ($k = 2$)</strong>: Contains 2 numbers.</p></p>
<p><p><strong>Third row ($k = 3$)</strong>: Contains 3 numbers.</p></p>
<p><p><strong>…</strong></p></p>
<p><p><strong>$k^\text{th}$ row</strong>: Contains $k$ numbers.</p></p>
<h3><strong>Total Numbers Up to the $k^\text{th}$ Row</strong></h3>
<p>The total number of integers up to the $k^\text{th}$ row is given by the sum of the first $k$ natural numbers:</p>
<p>$ S(k) = 1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2} $</p>
<h3><strong>Finding the Row Containing 5310</strong></h3>
<p>We need to find the smallest integer $k$ such that $S(k-1) < 5310 \leq S(k)$. This means:</p>
<p>$ \frac{(k-1)k}{2} < 5310 \leq \frac{k(k+1)}{2} $</p>
<h4><strong>Step 1: Estimate $k$</strong></h4>
<p>Let's approximate $k$ by solving the inequality:</p>
<p>$ \frac{k(k+1)}{2} \geq 5310 $</p>
<p>Multiply both sides by 2:</p>
<p>$ k(k+1) \geq 10620 $</p>
<p>This is a quadratic inequality. We can approximate $k$ by taking the square root:</p>
<p>$ k \approx \sqrt{10620} \approx 103 $</p>
<h4><strong>Step 2: Calculate $S(k)$ for $k = 102$ and $k = 103$</strong></h4>
<p>For $k = 102$:</p>
<p>$ S(102) = \frac{102 \times 103}{2} = \frac{10506}{2} = 5253 $</p>
<p>For $k = 103$:</p>
<p>$ S(103) = \frac{103 \times 104}{2} = \frac{10712}{2} = 5356 $</p>
<h4><strong>Step 3: Determine the Correct Row</strong></h4>
<p>Since $S(102) = 5253 < 5310 \leq 5356 = S(103)$, the number $5310$ lies between $5254$ and $5356$, inclusive. Therefore, it is in the $103^\text{rd}$ row.</p>
<h3><strong>Conclusion</strong></h3>
<p>The number $5310$ will be in the <strong>$103^\text{rd}$ row</strong>.</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lv5gt1x2 | maths | sequences-and-series | summation-of-series | <p>Let $$\alpha=\sum_\limits{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$$ and $$\beta=\left(\sum_\limits{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$$. If $$140<\frac{2 \alpha}{\beta}<281$$, then the value of $$n$$ is _________.</p> | [] | null | 5 | <p>$$\begin{aligned}
\alpha= & \sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r \\
& =4 \sum_{r=0}^n r^2{ }^n C_r+2 \sum_{r=0}^n r \cdot{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\
& =4 n(n+1) 2^{n-2}+2 \cdot n \cdot 2^{n-1}+2^n \\
& =2^n(n(n+1)+n+1)=2^n(n+1)^2 \\
& \beta=\sum_{r=0}^n\left(\frac{{ }^n C_r}{r+1}\right)+\left(\frac{1}{n+1}\right) \\
& (1+x)^n=\sum_{r=0}^n{ }^n C_r x^r
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \int_\limits0^1(1+x)^n d x=\left.\sum_{r=0}^n \frac{{ }^n C_r x^{r+1}}{r+1}\right|_0 ^1=\sum_\limits{r=0} \frac{{ }^n C}{r+1} \\
& \left.\frac{(1+x)^{+1}}{n+1}\right|_0 ^1=\frac{2^n-1}{n+1} \\
\Rightarrow & \beta=\frac{2^{n+1}-1+1}{(n+1)}=\frac{2}{n+1} \\
\Rightarrow & \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n \quad 1)}{\left(\frac{2^{n+1}}{n+1}\right)}=(n+1)^3 \in(140,281) \\
\Rightarrow & (n+1)^3=216 \\
\Rightarrow & n+1=6 \Rightarrow n=5
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lv7v4g04 | maths | sequences-and-series | summation-of-series | <p>If $$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$$ and $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=\mathrm{n}$$, then the point $$(\mathrm{m}, \mathrm{n})$$ lies on the line</p> | [{"identifier": "A", "content": "$$11(x-1)-100 y=0$$\n"}, {"identifier": "B", "content": "$$11 x-100 y=0$$\n"}, {"identifier": "C", "content": "$$11(x-1)-100(y-2)=0$$\n"}, {"identifier": "D", "content": "$$11(x-2)-100(y-1)=0$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\
& \text { and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\
& =\frac{1}{\sqrt{1}+\sqrt{2}} \times \frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\
& \quad+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{100}-\sqrt{99} \\
& =\sqrt{100}-\sqrt{1} \\
& =10-1 \\
& \Rightarrow m=9
\end{aligned}$$</p>
<p>and $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$$</p>
<p>$$\begin{aligned}
& \frac{2-1}{1 \times 2}+\frac{3-2}{2 \times 3}+\ldots+\frac{100-99}{100 \times 99}=n \\
& \Rightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{99}-\frac{1}{100}=n \\
& \Rightarrow n=1-\frac{1}{100} \\
& \Rightarrow n=\frac{99}{100} \\
& (m, n)=\left(9, \frac{99}{100}\right)
\end{aligned}$$</p>
<p>Satisfies the line $$11 x-100 y=0$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
lvb2953r | maths | sequences-and-series | summation-of-series | <p>If $$\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x \neq 0$$, and $$(60)^2 \mathrm{~S}(60)=\mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b}$$, where $$a, b \in N$$, then $$(a+b)$$ equal to _________.</p> | [] | null | 3660 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwawri93/7135ddd1-f250-4342-8c29-b0037fcb56d9/81a82a80-146c-11ef-a0c7-b1f23fa7cdc4/file-1lwawri94.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwawri93/7135ddd1-f250-4342-8c29-b0037fcb56d9/81a82a80-146c-11ef-a0c7-b1f23fa7cdc4/file-1lwawri94.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Sequences and Series Question 2 English Explanation"></p>
<p>$$\begin{aligned}
& (-x)(S(x))=\frac{(1+x)\left[(1+x)^{60}-1\right]}{(1+x-1)}-60(1+x)^{61} \\
& (-x) S(x)=\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}-60(1+x)^{61} \\
& x S(x)=60(1+x)^{61}-\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}
\end{aligned}$$</p>
<p>Multiplying $$x$$ on both side,</p>
<p>$$x^2 S(x)=60 x(1+x)^{61}-(1+x)\left[(1+x)^{60}-1\right]$$</p>
<p>Putting $$x=60$$</p>
<p>$$\begin{aligned}
& (60)^2 S(60)=60 \times 60(61)^{61}-(61)\left[61^{60}-1\right] \\
& =60 \times 60(61)^{61}-(61) \cdot 61^{60}+61 \\
& =(61)^{61}[60 \times 60-1]+61 \\
& =(3600-1) \cdot 61^{61}+61 \\
& a=3600-1, \quad b=61 \Rightarrow a+b=3660
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-evening-shift |
WSabZw5gL3VAFWgcOz1qpqahkk8e9ppyl | maths | sets-and-relations | number-of-sets-and-relations | Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y $$ \subseteq $$ X, Z $$ \subseteq $$ X and Y $$ \cap $$ Z is empty, is : | [{"identifier": "A", "content": "3<sup>5</sup>"}, {"identifier": "B", "content": "2<sup>5</sup>"}, {"identifier": "C", "content": "5<sup>3</sup>"}, {"identifier": "D", "content": "5<sup>2</sup>"}] | ["A"] | null | For any element x<sub>i</sub> present in X, 4 cases arises while making subsets Y and Z.
<br><br><b>Case- 1 :</b> x<sub>i</sub> $$ \in $$ Y, x<sub>i</sub> $$ \in $$ Z $$ \Rightarrow $$ Y $$ \cap $$ Z $$ \ne $$ $$\phi $$
<br><br><b>Case- 2 :</b> x<sub>i</sub> $$ \in $$ Y, x<sub>i</sub> $$ \notin $$ Z $$ \Rightarrow $$ Y $$ \cap $$ Z = $$\phi $$
<br><br><b>Case- 3 :</b> x<sub>i</sub> $$ \notin $$ Y, x<sub>i</sub> $$ \in $$ Z $$ \Rightarrow $$ Y $$ \cap $$ Z = $$\phi $$
<br><br><b>Case- 4 :</b> x<sub>i</sub> $$ \notin $$ Y, x<sub>i</sub> $$ \notin $$ Z $$ \Rightarrow $$ Y $$ \cap $$ Z = $$\phi $$
<br><br>$$ \therefore $$ For every element, number of ways = 3 for which Y $$ \cap $$ Z = $$\phi $$
<br><br>$$ \Rightarrow $$ Total ways = 3 × 3 × 3 × 3 × 3 [$$ \because $$ no. of elements in set X = 5]
<br><br>= 3<sup>5</sup> | mcq | aieee-2012 |
l91h0jo8 | maths | sets-and-relations | number-of-sets-and-relations | Let A and B be two sets containing four and
two elements respectively. Then, the number
of subsets of the set A $\times$ B , each having atleast
three elements are | [{"identifier": "A", "content": "219"}, {"identifier": "B", "content": "256"}, {"identifier": "C", "content": "275"}, {"identifier": "D", "content": "510"}] | ["A"] | null | Given,<br/><br/>
$$
\begin{aligned}
&n(A)=4, n(B) =2 \\\\
&\Rightarrow n(A \times B) =8
\end{aligned}
$$<br/><br/>
Total number of subsets of set $(A \times B)=2^8$<br/><br/>
Number of subsets of set $A \times B$ having no element (i.e. $\phi)=1$<br/><br/>
Number of subsets of set $A \times B$ having one element $={ }^8 C_1$<br/><br/>
Number of subsets of set $A \times B$ having two elements $={ }^8 C_2$<br/><br/>
$\therefore$ Number of subsets having atleast three elements<br/><br/>
$$
\begin{aligned}
&=2^8-\left(1+{ }^8 C_1+{ }^8 C_2\right) \\\\
&=2^8-1-8-28 \\\\
&=2^8-37 \\\\
&=256-37=219
\end{aligned}
$$ | mcq | jee-main-2015-offline |
ITWzRryGcqebgm3ACxWNT | maths | sets-and-relations | number-of-sets-and-relations | Let P = {$$\theta $$ : sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 \,\cos \theta $$} <br/><br/>and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then | [{"identifier": "A", "content": "P $$ \\subset $$ Q and Q $$-$$ P $$ \\ne $$ $$\\phi $$"}, {"identifier": "B", "content": "Q $$ \\not\\subset $$ P "}, {"identifier": "C", "content": "P $$ \\not\\subset $$ Q"}, {"identifier": "D", "content": "P = Q"}] | ["D"] | null | Given,
<br><br>sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 $$cos$$\theta $$
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $$\left( {\sqrt 2 + 1} \right)$$cos$$\theta $$
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $${{2 - 1} \over {\sqrt 2 - 1}}$$cos$$\theta $$
<br><br>$$ \Rightarrow $$ $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ = cos$$\theta $$ . . . (1)
<br><br>This is for set P.
<br><br>sin$$\theta $$ + cos$$\theta $$ = $${\sqrt 2 }$$sin$$\theta $$
<br><br>$$ \Rightarrow $$ cos$$\theta $$ = $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ . . . .(2)
<br><br>This is from set Q.
<br><br>So, P = Q | mcq | jee-main-2016-online-10th-april-morning-slot |
DFFepIdax5IrxuOX | maths | sets-and-relations | number-of-sets-and-relations | Two sets A and B are as under :
<br/><br/>A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};
<br/><br/>B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)<sup>2</sup> + 9(b - 5)<sup>2</sup> $$ \le $$ 36 };
<br/><br/>Then | [{"identifier": "A", "content": "neither A $$ \\subset $$ B nor B $$ \\subset $$ A"}, {"identifier": "B", "content": "B $$ \\subset $$ A"}, {"identifier": "C", "content": "A $$ \\subset $$ B"}, {"identifier": "D", "content": "A $$ \\cap $$ B = $$\\phi $$ ( an empty set )"}] | ["C"] | null | Given,
<br><br>$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$
<br><br>Let $$a - 6 = x$$ and $$b - 5 = y$$
<br><br>$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$
<br><br>$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$
<br><br>This is a equation of ellipse.
<br><br>This ellipse will look like this,
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264141/exam_images/ldgk0sluqe0qfofg0aga.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Sets and Relations Question 85 English Explanation 1">
<br><br>According to set A,
<br><br>$$\left| {a - 5} \right| < 1$$
<br><br>as $$a - 6 = x$$ then $$a - 5 = x + 1$$
<br><br>$$\therefore\,\,\,$$ $$\left| {x + 1} \right| < 1$$
<br><br>$$ \Rightarrow \,\,\, - 1 < x + 1 < 1$$
<br><br>$$ \Rightarrow \,\,\, - 2 < x < 0$$
<br><br>$$\left| {b - 5} \right| < 1$$
<br><br>as $$b - 5 = y$$
<br><br>$$\therefore\,\,\,$$ $$\left| y \right| < 1$$
<br><br>$$ \Rightarrow \,\,\,\, - 1 < y < 1$$
<br><br>This will look like this,
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264060/exam_images/ievtu0uoxd4s4ljgzkmw.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Sets and Relations Question 85 English Explanation 2">
<br><br>By combining both those graphs it will look like this,
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264557/exam_images/vr7gusba4nyfa18dc5vf.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Sets and Relations Question 85 English Explanation 3">
<br><br>To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$ and if this inequality satisfies, then it is inside B.
<br><br>By looking at the graph you can surely say (0, 1) or (0, $$-$$1) inside the graph. But to check point ($$-2$$, 1) or ($$-$$2, $$-$$1) is inside of the ellipse or not, put on the $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$$
<br><br>Putting ($$-$$2, 1) on the inequality,
<br><br>LHS $$ = {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$$
<br><br>$$ = {4 \over 9} + {1 \over 4}$$
<br><br>$$ = {{25} \over {36}} < 1$$
<br><br>$$\therefore\,\,\,$$ Inequality holds.
<br><br>So, $$\left( { - 2,1} \right)$$ is inside the ellipse.
<br><br>Similarly by checking we can see $$\left( { - 2, - 1} \right)$$ is also inside the ellipse.
<br><br>Hence, we can say entire set A is inside of the set B.
<br><br>$$\therefore\,\,\,\,$$ A $$ \subset $$ B | mcq | jee-main-2018-offline |
WCJJppkqk0JTLAItniIrv | maths | sets-and-relations | number-of-sets-and-relations | Let S = {1, 2, 3, … , 100}. The number of non-empty subsets A of S such that the product of elements in A is even is : | [{"identifier": "A", "content": "2<sup>50</sup> \u2013 1 "}, {"identifier": "B", "content": "2<sup>50</sup> (2<sup>50</sup> $$-$$ 1)"}, {"identifier": "C", "content": "2<sup>100</sup> $$-$$ 1"}, {"identifier": "D", "content": "2<sup>50</sup> + 1"}] | ["B"] | null | S = {1,2,3, . . . .100}
<br><br>= Total non empty subsets-subsets with product of element is odd
<br><br>= 2<sup>100</sup> $$-$$ 1 $$-$$ 1[(2<sup>50</sup> $$-$$ 1)]
<br><br>= 2<sup>100</sup> $$-$$ 2<sup>50</sup>
<br><br>= 2<sup>50</sup> (2<sup>50</sup> $$-$$ 1) | mcq | jee-main-2019-online-12th-january-morning-slot |
qjItDZ6yIHBa8CspKAyI4 | maths | sets-and-relations | number-of-sets-and-relations | Let Z be the set of integers.
<br/>If A = {x $$ \in $$ Z : 2<sup>(x + 2) (x<sup>2</sup> $$-$$ 5x + 6)</sup> = 1} and
<br/>B = {x $$ \in $$ Z : $$-$$ 3 < 2x $$-$$ 1 < 9},
<br/>then the number of subsets of the set A $$ \times $$ B, is | [{"identifier": "A", "content": "2<sup>12</sup>"}, {"identifier": "B", "content": "2<sup>18</sup>"}, {"identifier": "C", "content": "2<sup>10</sup>"}, {"identifier": "D", "content": "2<sup>15</sup>"}] | ["D"] | null | A ={x $$ \in $$ z : 2<sup>(x+2)(x<sup>2</sup> $$-$$ 5x + 6) </sup> = 1}
<br><br>2<sup>(x+2)(x<sup>2</sup> $$-$$ 5x + 6)</sup> = 2<sup>0</sup> $$ \Rightarrow $$ x = $$-$$ 2, 2, 3
<br><br>A = {$$-$$2, 2, 3}
<br><br>B = {x $$\varepsilon $$ Z : $$-$$ < 2x $$-$$ 1 < 9}
<br><br>B = {0, 1, 2, 3, 4}
<br><br>A $$ \times $$ B has is 15 elements so number of subsets of A $$ \times $$ B is 2<sup>15</sup>. | mcq | jee-main-2019-online-12th-january-evening-slot |
mW5EIn2EI2rmTjurp27k9k2k5foidxf | maths | sets-and-relations | number-of-sets-and-relations | Let X = {n $$ \in $$ N : 1 $$ \le $$ n $$ \le $$ 50}. If
<br/>A = {n $$ \in $$ X: n is a multiple of 2} and
<br/>B = {n $$ \in $$ X: n is a multiple of 7}, then the number of elements in the smallest subset of X
containing both A and B is ________. | [] | null | 29 | X = {1, 2, 3, 4, …, 50}
<br><br>A = {2, 4, 6, 8, …, 50} = 25 elements
<br><br>B = {7, 14, 21, 28, 35, 42, 49} = 7 elements
<br><br>Here n(A$$ \cup $$B) = n(A) + n(B) – n(A$$ \cap $$B)
<br><br>= 25 + 7 – 3 = 29 | integer | jee-main-2020-online-7th-january-evening-slot |
uh9zBvgdVXVFTfGA1mjgy2xukewrzn2p | maths | sets-and-relations | number-of-sets-and-relations | If R = {(x, y) : x, y
$$ \in $$ Z, x<sup>2</sup> + 3y<sup>2</sup>
$$ \le $$ 8} is a relation
on the set of integers Z, then the domain of R<sup>–1</sup> is : | [{"identifier": "A", "content": "{0, 1} "}, {"identifier": "B", "content": "{\u20132, \u20131, 1, 2}"}, {"identifier": "C", "content": "{\u20131, 0, 1}"}, {"identifier": "D", "content": "{\u20132, \u20131, 0, 1, 2}"}] | ["C"] | null | Given R = {(x, y) : x, y
$$ \in $$ Z, x<sup>2</sup> + 3y<sup>2</sup>
$$ \le $$ 8}
<br><br>So R = {(0,1), (0,–1), (1,0), (–1,0), (1,1), (1,-1)
<br>(-1,1), (-1,-1), (2,0), (-2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}
<br><br>$$ \Rightarrow $$ R : { -2, -1, 0, 1, 2} $$ \to $$ {-1, 0, 1}
<br><br>$$ \therefore $$ R<sup>-1</sup> : {-1, 0, 1} $$ \to $$ { -2, -1, 0, 1, 2}
<br><br>$$ \therefore $$ Domain of R<sup>–1</sup> = {-1, 0, 1} | mcq | jee-main-2020-online-2nd-september-morning-slot |
B2HRGK7dEczQF7C94Njgy2xukf0qgrp2 | maths | sets-and-relations | number-of-sets-and-relations | Consider the two sets :
<br/>A = {m $$ \in $$ R : both the roots of<br/> x<sup>2</sup>
– (m + 1)x + m + 4 = 0 are real}
<br/>and B = [–3, 5).
<br/>Which of the following is not true? | [{"identifier": "A", "content": "A $$ \\cap $$ B = {\u20133}"}, {"identifier": "B", "content": "B \u2013 A = (\u20133, 5)"}, {"identifier": "C", "content": "A $$ \\cup $$ B = R"}, {"identifier": "D", "content": "A - B = ($$ - $$$$ \\propto $$, $$ - $$3) $$ \\cup $$ (5, $$ \\propto $$)"}] | ["D"] | null | As roots are real so, $$D \ge 0$$<br><br>$${(m + 1)^2} - 4(m + 4) \ge 0$$<br><br>$$ \Rightarrow {m^2} - 2m - 15 \ge 0$$<br><br>$$ \Rightarrow $$ $$(m - 5)(m + 3) \ge 0$$<br><br>$$m\, \in \,$$($$ - $$$$ \propto $$, $$ - $$3] $$ \cup $$ [5, $$ \propto $$)<br><br>$$A= ( - $$$$ \propto $$, $$ - $$3] $$ \cup $$ [5, $$ \propto $$)<br><br>Given B = [$$ - $$3, 5)<br><p>Now, let's examine the options.</p>
<p><b>Option A :</b> A ∩ B = {–3}
The intersection of sets A and B would be the set of elements common to both sets. In this case, the only common element is -3. So, option A is true.</p>
<p><b>Option B :</b> B – A = (–3, 5)
The subtraction (or difference) of sets A from B is the set of elements that are in B but not in A. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting A from B would leave an open interval (-3, 5), not including -3 and 5. So, option B is also true.</p>
<p><b>Option C :</b> A ∪ B = R
The union of sets A and B is the set of elements that are in A, or B, or both. Here, A U B would cover all real numbers. So, option C is true.</p>
<p><b>Option D :</b> A - B = (-∞, -3) ∪ (5, ∞)
The subtraction (or difference) of set B from A is the set of elements that are in A but not in B. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting B from A would leave (-∞, -3) U [5, ∞), not including -3 and 5. But according to the convention for writing intervals, it should be (-∞, -3) U (5, ∞). So, option D is not true.</p> | mcq | jee-main-2020-online-3rd-september-morning-slot |
1p4U7IKhG25B9jE6nSjgy2xukfakf6tf | maths | sets-and-relations | number-of-sets-and-relations | Let $$\mathop \cup \limits_{i = 1}^{50} {X_i} = \mathop \cup \limits_{i = 1}^n {Y_i} = T$$ where each X<sub>i</sub> contains 10 elements and each Y<sub>i</sub> contains 5 elements. If each element of the set T is an element of exactly 20 of sets X<sub>i</sub>’s and exactly 6 of sets Y<sub>i</sub>’s, then n is equal to :
| [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "45"}] | ["A"] | null | $$\mathop \cup \limits_{i = 1}^{50} {X_i} = $$ X<sub>1</sub>, X<sub>2</sub>,....., X<sub>50</sub> = 50 sets. Given each sets having 10 elements.
<br><br>So total elements = 50 $$ \times $$ 10
<br><br>$$\mathop \cup \limits_{i = 1}^n {Y_i} =$$ $$ Y<sub>1</sub>, Y<sub>2</sub>,....., Y<sub>n</sub> = n sets. Given each sets having 5 elements.
<br><br>So total elements = 5 $$ \times $$ n
<br><br>Now each element of set T contains exactly 20 of sets X<sub>i</sub>.
<br><br>So number of effective elements in set T = $${{50 \times 10} \over {20}}$$
<br><br>Also each element of set T contains exactly 6 of sets Y<sub>i</sub>.
<br><br>So number of effective elements in set T = $${{50 \times 10} \over {20}}$$
<br><br>$$ \therefore $$ $${{50 \times 10} \over {20}}$$ = $${{5 \times n} \over {20}}$$
<br><br>$$ \Rightarrow $$ n = 30 | mcq | jee-main-2020-online-4th-september-evening-slot |
3iz5nbJtSww5lL4Sg9jgy2xukfxgtuou | maths | sets-and-relations | number-of-sets-and-relations | Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more
than the total number of subsets of B, then the value of m.n is ______. | [] | null | 28 | Number of subsets of A = 2<sup>m</sup>
<br><br>Number of subsets of B = 2<sup>n</sup>
<br><br>Given = 2<sup>m</sup> – 2<sup>n</sup>
= 112
<br><br>$$ \therefore $$ m = 7, n = 4 (2<sup>7</sup> – 2<sup>4</sup>
= 112)
<br><br>$$ \therefore $$ m $$ \times $$ n = 7 $$ \times $$ 4 = 28 | integer | jee-main-2020-online-6th-september-morning-slot |
iQP8FXc5rG0uqHHQlB1klrjjed5 | maths | sets-and-relations | number-of-sets-and-relations | Let A = {n $$ \in $$ N: n is a 3-digit number}<br/><br>
B = {9k + 2: k $$ \in $$ N}
<br/><br>and C = {9k + $$l$$: k $$ \in $$ N} for some $$l ( 0 < l < 9)$$<br/><br>
If the sum of all the elements of the set A $$ \cap $$ (B $$ \cup $$ C) is 274 $$ \times $$ 400, then $$l$$ is equal to ________.</br></br></br> | [] | null | 5 | In this problem, we're dealing with 3-digit numbers in set $A$, and subsets $B$ and $C$ which represent numbers of specific forms.
<br/><br/>1. First, we consider the numbers of the form $9k + 2$ (Set $B$) within the 3-digit range, which starts at 101 and ends at 992.
<br/><br/>2. We calculate the sum of these numbers, denoted as $s_1$. To calculate $s_1$, you use the formula for the sum of an arithmetic series :
<br/><br/> $$(n/2) \times (\text{{first term}} + \text{{last term}})$$
<br/><br/> Here, $n$ is the total count of such numbers. These are 3-digit numbers of the form $9k + 2$, and we can find the total count by subtracting the smallest such number (101) from the largest (992), dividing the result by 9 (because we're considering numbers with a difference of 9), and then adding 1.
<br/><br/> The sum $s_1$ is calculated as follows :
<br/><br/> $$(100/2) \times (101 + 992) = 54650$$
<br/><br/>3. According to the problem, the sum of all elements of the set $A \cap (B \cup C)$ is $274 \times 400 = 109600$.
<br/><br/> Since the set $A \cap (B \cup C)$ is the union of two disjoint sets (the set of all three-digit numbers of form $9k + 2$ and the set of all three-digit numbers of form $9k + l$), we can write this sum as :
<br/><br/> $$s_1 $$(for numbers of the form 9k + 2) + $$s_2$$ (for numbers of the form 9k + l) = 109600
<br/><br/>4. Solving this equation for $s_2$ (the sum of numbers of the form $9k + l$), we get :
<br/><br/> $$s_2 = 109600 - s_1 = 109600 - 54650 = 54950$$
<br/><br/>5. The sum $s_2$ can be expressed as $(n/2) \times (\text{{first term}} + \text{{last term}})$, where $n$ is the count of numbers of the form $9k + l$. The first term here is the smallest 3-digit number of this form, which is $99 + l$, and the last term is the largest such number, which is $990 + l$.
<br/><br/> We equate this to $s_2$ to solve for $l$ :
<br/><br/> $$54950 = (100/2)[(99 + l) + (990 + l)]$$
<br/><br/>6. Simplifying this equation, we get :
<br/><br/> $$2l + 1089 = 1099$$
<br/><br/> Solving for $l$, we find :
<br/><br/> $$l = 5$$
<br/><br/>So, the correct answer is 5. | integer | jee-main-2021-online-24th-february-morning-slot |
ElJhwaKkT1JdIO9sfl1kmhw02h6 | maths | sets-and-relations | number-of-sets-and-relations | The number of elements in the set {x $$\in$$ R : (|x| $$-$$ 3) |x + 4| = 6} is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["B"] | null | <b>Case 1 :</b><br><br>x $$ \le $$ $$-$$4<br><br>($$-$$x $$-$$ 3)($$-$$x $$-$$ 4) = 6<br><br>$$ \Rightarrow $$ (x + 3)(x + 4) = 6<br><br>$$ \Rightarrow $$ x<sup>2</sup> + 7x + 6 = 0<br><br>$$ \Rightarrow $$ x = $$-$$1 or $$-$$6<br><br>but x $$ \le $$ $$-$$4<br><br>x = $$-$$6<br><br><b>Case 2 :</b><br><br>x $$\in$$ ($$-$$4, 0)<br><br>($$-$$x $$-$$ 3)(x + 4) = 6<br><br>$$ \Rightarrow $$ $$-$$x<sup>2</sup> $$-$$ 7x $$-$$ 12 $$-$$ 6 = 0<br><br>$$ \Rightarrow $$ x<sup>2</sup> + 7x + 18 = 0<br><br>D < 0 No solution<br><br><b>Case 3 :</b><br><br>x $$ \ge $$ 0<br><br>(x $$-$$ 3)(x + 4) = 6<br><br>$$ \Rightarrow $$ x<sup>2</sup> + x $$-$$ 12 $$-$$ 6 = 0<br><br>$$ \Rightarrow $$ x<sup>2</sup> + x $$-$$ 18 = 0<br><br>x = $${{ - 1 \pm \sqrt {1 + 72} } \over 2}$$<br><br>$$ \therefore $$ x = $${{\sqrt {73} - 1} \over 2}$$ only | mcq | jee-main-2021-online-16th-march-morning-shift |
Cy64jNuK5iXgalKmT51kmiwrnp6 | maths | sets-and-relations | number-of-sets-and-relations | Let A = {2, 3, 4, 5, ....., 30} and '$$ \simeq $$' be an equivalence relation on A $$\times$$ A, defined by (a, b) $$ \simeq $$ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "7"}] | ["D"] | null | ad = bc<br><br>(a, b) R (4, 3) $$ \Rightarrow $$ 3a = 4b<br><br>a = $${4 \over 3}$$b<br><br>b must be multiple of 3<br><br>b = {3, 6, 9 ..... 30}<br><br>(a, b) = {(4, 3), (8, 16), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)}<br><br>$$ \Rightarrow $$ 7 ordered pair | mcq | jee-main-2021-online-16th-march-evening-shift |
1krygqgjj | maths | sets-and-relations | number-of-sets-and-relations | Let A = {n $$\in$$ N | n<sup>2</sup> $$\le$$ n + 10,000}, B = {3k + 1 | k$$\in$$ N} an dC = {2k | k$$\in$$N}, then the sum of all the elements of the set A $$\cap$$(B $$-$$ C) is equal to _____________. | [] | null | 832 | B $$-$$ C $$ \equiv $$ {7, 13, 19, ......, 97, .......}<br><br>Now, n<sup>2</sup> $$-$$ n $$\le$$ 100 $$\times$$ 100<br><br>$$\Rightarrow$$ n(n $$-$$ 1) $$\le$$ 100 $$\times$$ 100<br><br>$$\Rightarrow$$ A = {1, 2, ......., 100}.<br><br>So, A$$\cap$$(B $$-$$ C) = {7, 13, 19, ......., 97}<br><br>Hence, sum = $${{16} \over 2}(7 + 97) = 832$$ | integer | jee-main-2021-online-27th-july-evening-shift |
1ktep0ng3 | maths | sets-and-relations | number-of-sets-and-relations | If A = {x $$\in$$ R : |x $$-$$ 2| > 1}, <br/>B = {x $$\in$$ R : $$\sqrt {{x^2} - 3} $$ > 1}, <br/>C = {x $$\in$$ R : |x $$-$$ 4| $$\ge$$ 2} and Z is the set of all integers, then the number of subsets of the <br/>set (A $$\cap$$ B $$\cap$$ C)<sup>c</sup> $$\cap$$ Z is ________________. | [] | null | 256 | A = ($$-$$$$\infty$$, 1) $$\cup$$ (3, $$\infty$$)<br><br>B = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ (2, $$\infty$$)<br><br>C = ($$-$$$$\infty$$, 2] $$\cup$$ [6, $$\infty$$)<br><br>So, A $$\cap$$ B $$\cap$$ C = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ [6, $$\infty$$)<br><br>z $$\cap$$ (A $$\cap$$ B $$\cap$$ C)' = {$$-$$2, $$-$$1, 0, $$-$$1, 2, 3, 4, 5}<br><br>Hence, no. of its subsets = 2<sup>8</sup> = 256. | integer | jee-main-2021-online-27th-august-morning-shift |
1l567giit | maths | sets-and-relations | number-of-sets-and-relations | <p>Let R<sub>1</sub> and R<sub>2</sub> be relations on the set {1, 2, ......., 50} such that</p>
<p>R<sub>1</sub> = {(p, p<sup>n</sup>) : p is a prime and n $$\ge$$ 0 is an integer} and</p>
<p>R<sub>2</sub> = {(p, p<sup>n</sup>) : p is a prime and n = 0 or 1}.</p>
<p>Then, the number of elements in R<sub>1</sub> $$-$$ R<sub>2</sub> is _______________.</p> | [] | null | 8 | Given, ${R}_1=\left\{\left(p, p^n\right): p\right.$ is a Prime and $n \geq 0$ is an integer $\}$
<br/><br/>and, set $A=\{1,2,3 \ldots \ldots .50\}$
<br/><br/>$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47
<br/><br/>$\therefore$ We can calculate no. of elements in $\mathrm{R}_1$
<br/><br/>$$
\mathrm{R}_1=\left(2,2^0\right),\left(2,2^1\right),\left(2,2^2\right)\left(2,2^3\right) \ldots \ldots\left(2,2^5\right)=6$$ number of ordered pairs
<br/><br/>$\left(3,3^0\right),\left(3,3^1\right),\left(3,3^2\right) \ldots \ldots . .\left(3,3^3\right)=4$ number of order paris
<br/><br/>$\left(5,5^0\right),\left(5,5^1\right),\left(5,5^2\right) \ldots \ldots \ldots . .=3$ number of order paris
<br/><br/>$\left(7,7^0\right) \ldots \ldots .\left(7,7^2\right) \ldots \ldots \ldots=3$ number of order paris
<br/><br/>$\left(11,11^0\right)$ and $\left(11,11^1\right)=2$ number of order paris
<br/><br/>$\left(13,13^0\right)$ and $\left(13,13^1\right)=2$ number of order paris
<br/><br/>$$ \therefore $$ For the 11 prime numbers ($11,13,17,19,23,29,31,37,41,43$ and 47), $n$ can only be 0, 1 (two pairs each).
<br/><br/>$$
\therefore n\left(\mathrm{R}_1\right)=6+4+3+3+(2 \times 11)=38
$$
<br/><br/>$\mathrm{R}_2=\left(p, p^n\right) $, where n = 0 or 1
<br/><br/>$$
\left(2,2^0\right),\left(2,2^1\right)\left(3,3^0\right)\left(3,3^1\right) \ldots . .\left(47,47^0\right)\left(47,47^1\right)
$$
<br/><br/>Two ordered pairs of each element $n\left({R}_2\right)=2 \times 15=30$ elements
<br/><br/>Hence $ R_1-R_2=38-30=8$ | integer | jee-main-2022-online-28th-june-morning-shift |
1l58apr54 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let A = {n $$\in$$ N : H.C.F. (n, 45) = 1} and</p>
<p>Let B = {2k : k $$\in$$ {1, 2, ......., 100}}. Then the sum of all the elements of A $$\cap$$ B is ____________.</p> | [] | null | 5264 | <p>Sum of all elements of A $$\cap$$ B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]</p>
<p>$$ = 2\left[ {{{100 \times 101} \over 2} - 3\left( {{{33 \times 34} \over 2}} \right) - 5\left( {{{20 \times 21} \over 2}} \right) + 15\left( {{{6 \times 7} \over 2}} \right)} \right]$$</p>
<p>$$ = 10100 - 3366 - 2100 + 630$$</p>
<p>$$ = 5264$$</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l58aqy4b | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\min \,\{ i,j\} } } $$ and $$B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \,\{ i,j\} } } $$. Then A + B is equal to _____________.</p> | [] | null | 1100 | <p>$$\sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\{ i,\,j\} } } $$</p>
<p>= {1, 1} {1, 2} {1, 3} ..... {1, 10}</p>
<p>{2, 1} {2, 2} {2, 3} ..... {2, 10}</p>
<p>{3, 1} {3, 2} {3, 3} ..... {3, 10}</p>
<p>$$ \vdots $$</p>
<p>{10, 1} {10, 2} {10, 3} ..... {10, 10}</p>
<p>Now, $$A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {min\{ i,\,j\} } } $$</p>
<p>= minimum between i and j in all sets and summation of all those values.</p>
<p>and $$B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \{ i,\,j\} } } $$</p>
<p>= maximum between i and j in all sets and summation of all those values.</p>
<p>For 1 :</p>
<p>1 is minimum in sets =</p>
<p>{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {1, 9}, {1, 10}, {2, 1}, {3, 1}, {4, 1}, {5, 1}, {6, 1}, {7, 1}, {8, 1}, {9, 1}, {10, 1}</p>
<p>$$\therefore$$ 1 is minimum in 19 sets</p>
<p>1 is maximum in {1, 1} sets.</p>
<p>$$\therefore$$ 1 is maximum and minimum in total 20 sets.</p>
<p>$$\therefore$$ Sum of 1 in all those sets = 1 $$\times$$ 20 = 20</p>
<p>For 2 :</p>
<p>2 is minimum in sets =</p>
<p>{2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {2, 9}, {2, 10}, {3, 2}, {4, 2}, {5, 2}, {6, 2}, {7, 2}, {8, 2}, {9, 2}, {10, 2}</p>
<p>$$\therefore$$ 2 is minimum in 17 sets</p>
<p>2 is maximum in sets = {1, 2}, {2, 1}, {2, 2}</p>
<p>$$\therefore$$ 2 is maximum and minimum in 20 sets.</p>
<p>$$\therefore$$ Sum of 2 in all those sets = 2 $$\times$$ 20 = 40</p>
<p>Similarly 3 is maximum and minimum in 20 sets.</p>
<p>$$\therefore$$ Sum of 3 in all those sets = 20 $$\times$$ 3 = 60</p>
<p>$$ \vdots $$</p>
<p>Similarly, 10 is maximum and minimum in 20 sets.</p>
<p>$$\therefore$$ Sum of 10 in all those sets = 20 $$\times$$ 10 = 200</p>
<p>$$\therefore$$ A + B = 20 + 20 $$\times$$ 2 + 20 $$\times$$ 3 + ....... + 20 $$\times$$ 10</p>
<p>= 20(1 + 2 + 3 + ...... + 10)</p>
<p>= 20 $$\times$$ $${{10 \times 11} \over 2}$$</p>
<p>= 1100</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l5bb3057 | maths | sets-and-relations | number-of-sets-and-relations | <p>The sum of all the elements of the set $$\{ \alpha \in \{ 1,2,.....,100\} :HCF(\alpha ,24) = 1\} $$ is __________.</p> | [] | null | 1633 | <p>The numbers upto 24 which gives g.c.d. with 24 equals to 1 are 1, 5, 7, 11, 13, 17, 19 and 23.</p>
<p>Sum of these numbers = 96</p>
<p>There are four such blocks and a number 97 is there upto 100.</p>
<p>$$\therefore$$ Complete sum</p>
<p>= 96 + (24 $$\times$$ 8 + 96) + (48 $$\times$$ 8 + 96) + (72 $$\times$$ 8 + 96) + 97</p>
<p>= 1633</p> | integer | jee-main-2022-online-24th-june-evening-shift |
1l6f35cin | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3,4,5,6,7\}$$. Define $$B=\{T \subseteq A$$ : either $$1 \notin T$$ or $$2 \in T\}$$ and $$C=\{T \subseteq A: T$$ the sum of all the elements of $$T$$ is a prime number $$\}$$. Then the number of elements in the set $$B \cup C$$ is ________________.</p> | [] | null | 107 | <p>$$\because$$ $$(B \cup C)' = B'\, \cap C'$$</p>
<p>B' is a set containing sub sets of A containing element 1 and not containing 2.</p>
<p>And C' is a set containing subsets of A whose sum of elements is not prime.</p>
<p>So, we need to calculate number of subsets of {3, 4, 5, 6, 7} whose sum of elements plus 1 is composite.</p>
<p>Number of such 5 elements subset = 1</p>
<p>Number of such 4 elements subset = 3 (except selecting 3 or 7)</p>
<p>Number of such 3 elements subset = 6 (except selecting {3, 4, 5}, {3, 6, 7}, {4, 5, 7} or {5, 6, 7})</p>
<p>Number of such 2 elements subset = 7 (except selecting {3, 7}, {4, 6}, {5, 7})</p>
<p>Number of such 1 elements subset = 3 (except selecting {4} or {6})</p>
<p>Number of such 0 elements subset = 1</p>
<p>$$n(B'\, \cap C') = 21 \Rightarrow n(B \cup C) = {2^7} - 21 = 107$$</p> | integer | jee-main-2022-online-25th-july-evening-shift |
1l6hzials | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3,4,5,6,7\}$$ and $$B=\{3,6,7,9\}$$. Then the number of elements in the set $$\{C \subseteq A: C \cap B \neq \phi\}$$ is ___________.</p> | [] | null | 112 | <p>As C $$\cap$$ B $$\ne$$ $$\phi$$, c must be not be formed by {1, 2, 4, 5}</p>
<p>$$\therefore$$ Number of subsets of A = 2<sup>7</sup> = 128</p>
<p>and number of subsets formed by {1, 2, 4, 5} = 16</p>
<p>$$\therefore$$ Required no. of subsets = 2<sup>7</sup> $$-$$ 2<sup>4</sup> = 128 $$-$$ 16 = 112</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6p0j53w | maths | sets-and-relations | number-of-sets-and-relations | <p>Let R be a relation from the set $$\{1,2,3, \ldots, 60\}$$ to itself such that $$R=\{(a, b): b=p q$$, where $$p, q \geqslant 3$$ are prime numbers}. Then, the number of elements in R is :</p> | [{"identifier": "A", "content": "600"}, {"identifier": "B", "content": "660"}, {"identifier": "C", "content": "540"}, {"identifier": "D", "content": "720"}] | ["B"] | null | <p>We have a set S = {1, 2, 3, ..., 60}, and a relation R defined on the set S. An element (a, b) belongs to the relation R if and only if b can be expressed as the product of two prime numbers p and q, where both p and q are greater than or equal to 3.</p>
<p>In terms of number theory, prime numbers are integers greater than 1 that have no divisors other than 1 and themselves. We are interested in prime numbers that are greater than or equal to 3, because p and q must both be greater than or equal to 3.</p>
<p>The primes greater than or equal to 3 and less than or equal to 60 are {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59}.</p>
<p>We need to find all possible values of b = p $$ \times $$ q such that b belongs to the set S. We start by choosing the smallest prime number (which is 3) and keep multiplying it with all the prime numbers until the product exceeds 60 :</p>
<ul>
<li>1. If we start with p = 3, we can choose q to be 3, 5, 7, 11, 13, 17, or 19. This gives us 7 valid products that are less than or equal to 60.</li>
<li>2. If we start with p = 5, we can choose q to be 5, 7, or 11. This gives us 3 valid products that are less than or equal to 60.</li>
<li>3. If we start with p = 7, we can choose q to be 7. This gives us 1 valid product that is less than or equal to 60.</li>
</ul>
<p>So, we have a total of 7 + 3 + 1 = 11 possible values for b = p $$ \times $$ q that satisfy the conditions.</p>
<p>Since a can be any number in the set S, there are 60 possible values for a for each of the 11 values of b. Therefore, the total number of elements in the relation R is 60 $$ \times $$ 11 = 660.</p>
| mcq | jee-main-2022-online-29th-july-morning-shift |
1l6p3rlat | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$S=\{4,6,9\}$$ and $$T=\{9,10,11, \ldots, 1000\}$$. If $$A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in \mathbf{N}, a_{1}, a_{2}, a_{3}, \ldots, a_{k}\right.$$ $$\epsilon S\}$$, then the sum of all the elements in the set $$T-A$$ is equal to __________.</p> | [] | null | 11 | <p>Here $$S = \{ 4,6,9\} $$</p>
<p>And $$T = \{ 9,10,11,\,\,......,\,\,1000\} $$.</p>
<p>We have to find all numbers in the form of $$4x + 6y + 9z$$, where $$x,y,z \in \{ 0,1,2,\,......\} $$.</p>
<p>If a and b are coprime number then the least number from which all the number more than or equal to it can be express as $$ax + by$$ where $$x,y \in \{ 0,1,2,\,......\} $$ is $$(a - 1)\,.\,(b - 1)$$.</p>
<p>Then for $$6y + 9z = 3(2y + 3z)$$</p>
<p>All the number from $$(2 - 1)\,.\,(3 - 1) = 2$$ and above can be express as $$2x + 3z$$ (say t).</p>
<p>Now $$4x + 6y + 9z = 4x + 3(t + 2)$$</p>
<p>$$ = 4x + 3t + 6$$</p>
<p>again by same rule $$4x + 3t$$, all the number from $$(4 - 1)\,(3 - 1) = 6$$ and above can be express from $$4x + 3t$$.<p>Then $$4x + 6y + 9z$$ express all the numbers from 12 and above.</p>
<p>again 9 and 10 can be express in form $$4x + 6y + 9z$$.</p>
<p>Then set $$A = \{ 9,10,12,13,\,....,\,1000\} .$$</p>
<p>Then $$T - A = \{ 11\} $$</p>
<p>Only one element 11 is there.</p>
<p>Sum of elements of $$T - A = 11$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
1ldv2tpr1 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____________.</p> | [] | null | 43 | Elements of the type $3 \mathrm{k}=3$<br/><br/>
Elements of the type $3 \mathrm{k}+1=1,7,9$<br/><br/>
Elements of the type $3 \mathrm{k}+2=2,5,11$<br/><br/>
Subsets containing one element $S_1=1$<br/><br/>
Subsets containing two elements<br/><br/>
$$
S_2={ }^3 C_1 \times{ }^3 C_1=9
$$<br/><br/>
Subsets containing three elements<br/><br/>
$$
\mathrm{S}_3={ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1+1+1=11
$$<br/><br/>
Subsets containing four elements<br/><br/>
$$<br/><br/>
\mathrm{S}_4={ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2=11
$$<br/><br/>
Subsets containing five elements<br/><br/>
$$
\mathrm{S}_5={ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2 \times 1=9
$$<br/><br/>
Subsets containing six elements $\mathrm{S}_6=1$<br/><br/>
Subsets containing seven elements $\mathrm{S}_7=1$<br/><br/>
$$
\Rightarrow \text { sum }=43
$$ | integer | jee-main-2023-online-25th-january-morning-shift |
lgnzb2hr | maths | sets-and-relations | number-of-sets-and-relations | The number of elements in the set <br/><br/>$\left\{n \in \mathbb{N}: 10 \leq n \leq 100\right.$ and $3^{n}-3$ is a multiple of 7$\}$ is ___________. | [] | null | 15 | To determine the number of elements in the given set, we need to find how many natural numbers $n$ between $10$ and $100$ (inclusive) satisfy the condition that $3^n - 3$ is a multiple of $7$.
<br/><br/>Recall that for any integers $a$ and $b$, $a$ is a multiple of $b$ if there exists an integer $k$ such that $a = bk$. So in our case, we need to find how many $n$ satisfy the equation $3^n - 3 = 7k$ for some integer $k$.
<br/><br/>Notice that $3^n - 3 = 3(3^{n-1} - 1)$. We want this expression to be a multiple of 7. Let's explore a few powers of 3 modulo 7:
<br/><br/>$3^1 \equiv 3 \pmod{7}$
<br/><br/>$3^2 \equiv 9 \equiv 2 \pmod{7}$
<br/><br/>$3^3 \equiv 27 \equiv 6 \pmod{7}$
<br/><br/>$3^4 \equiv 81 \equiv 4 \pmod{7}$
<br/><br/>$3^5 \equiv 243 \equiv 5 \pmod{7}$
<br/><br/>$3^6 \equiv 729 \equiv 1 \pmod{7}$
<br/><br/>We observe that $3^n \pmod{7}$ follows a cycle of length 6. So, $3^{n-1} \pmod{7}$ also follows the same cycle, but shifted:
<br/><br/>$3^0 \equiv 1 \pmod{7}$
<br/><br/>$3^1 \equiv 3 \pmod{7}$
<br/><br/>$3^2 \equiv 2 \pmod{7}$
<br/><br/>$3^3 \equiv 6 \pmod{7}$
<br/><br/>$3^4 \equiv 4 \pmod{7}$
<br/><br/>$3^5 \equiv 5 \pmod{7}$
<br/><br/>We want $3(3^{n-1} - 1) \equiv 0 \pmod{7}$, which means that $3^{n-1} - 1 \equiv 0 \pmod{7}$. From the cycle above, we see that this is true when $n-1$ is a multiple of 6, or equivalently, when $n$ is one more than a multiple of 6.
<br/><br/>Now let's find the multiples of 6 between 10 and 100:
<br/><br/>$12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96$
<br/><br/>Adding 1 to each of these values, we get the set of natural numbers $n$ that satisfy the given condition:
<br/><br/>$13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97$
<br/><br/>There are 15 elements in this set. Therefore, the number of elements in the given set is $\boxed{15}$. | integer | jee-main-2023-online-15th-april-morning-shift |
lgnzfh40 | maths | sets-and-relations | number-of-sets-and-relations | Let $A=\{1,2,3,4\}$ and $\mathrm{R}$ be a relation on the set $A \times A$ defined by <br/><br/>$R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$. Then the number of elements in $\mathrm{R}$ is ____________. | [] | null | 6 | $$
2a + 3b = 4c + 5d
$$
<br/><br/>
Given A = {1, 2, 3, 4}, the maximum value of $2a + 3b$ is 20, when (a, b) = (4, 4), and the minimum value of $4c + 5d$ is 9, when (c, d) = (1, 1). Therefore, the possible values for $2a + 3b = 4c + 5d$ are 9, 13, 14, 17, 18, and 19.
<br/><br/>
Now, let's find the combinations of (a, b), (c, d) that satisfy the given equation:
<br/><br/>
1. $2a + 3b = 9 \Rightarrow (a, b) = (3, 1) \Rightarrow (c, d) = (1, 1)$<br/><br/>
2. $2a + 3b = 13 \Rightarrow (a, b) = (2, 3) \Rightarrow (c, d) = (2, 1)$<br/><br/>
3. $2a + 3b = 14 \Rightarrow (a, b) = (4, 2) \Rightarrow (c, d) = (1, 2)$<br/><br/>
4. $2a + 3b = 14 \Rightarrow (a, b) = (1, 4) \Rightarrow (c, d) = (1, 2)$<br/><br/>
5. $2a + 3b = 17 \Rightarrow (a, b) = (4, 3) \Rightarrow (c, d) = (3, 1)$<br/><br/>
6. $2a + 3b = 18 \Rightarrow (a, b) = (3, 4) \Rightarrow (c, d) = (2, 2)$<br/><br/>
There are a total of 6 elements in the relation R for the given equation with the specified values of a, b, c, and d. | integer | jee-main-2023-online-15th-april-morning-shift |
1lgsumb17 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$\mathrm{A}=\{1,3,4,6,9\}$$ and $$\mathrm{B}=\{2,4,5,8,10\}$$. Let $$\mathrm{R}$$ be a relation defined on $$\mathrm{A} \times \mathrm{B}$$ such that $$\mathrm{R}=\left\{\left(\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)\right): a_{1} \leq b_{2}\right.$$ and $$\left.b_{1} \leq a_{2}\right\}$$. Then the number of elements in the set R is :</p> | [{"identifier": "A", "content": "180"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "52"}, {"identifier": "D", "content": "160"}] | ["D"] | null | Given that the sets are $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$, for the relation $\mathrm{R}$ on the set $A \times B$, we need to find the combinations of pairs that satisfy the conditions $a_1 \leq b_2$ and $b_1 \leq a_2$.
<br/><br/>We find the number of combinations by considering the possible values for $b_2$ for each $a_1$ and the possible values for $a_2$ for each $b_1$ :
<br/><br/>For each $a_1$ in $A = \{1, 3, 4, 6, 9\}$, the number of valid $b_2$ values in $B = \{2, 4, 5, 8, 10\}$ are :
<br/><br/>- For $a_1 = 1$, there are 5 choices for $b_2$.
<br/><br/>- For $a_1 = 3$, there are 4 choices for $b_2$.
<br/><br/>- For $a_1 = 4$, there are 4 choices for $b_2$.
<br/><br/>- For $a_1 = 6$, there are 2 choices for $b_2$.
<br/><br/>- For $a_1 = 9$, there is 1 choice for $b_2$.
<br/><br/>This results in a total of $5+4+4+2+1 = 16$ possible pairs $(a_1, b_2)$.
<br/><br/>Similarly, for each $b_1$ in $B$, the number of valid $a_2$ values in $A$ are :
<br/><br/>- For $b_1 = 2$, there are 4 choices for $a_2$.
<br/><br/>- For $b_1 = 4$, there are 3 choices for $a_2$.
<br/><br/>- For $b_1 = 5$, there are 2 choices for $a_2$.
<br/><br/>- For $b_1 = 8$, there is 1 choice for $a_2$.
<br/><br/>- For $b_1 = 10$, there are no choices for $a_2$.
<br/><br/>This results in a total of $4+3+2+1+0 = 10$ possible pairs $(b_1, a_2)$.
<br/><br/>Therefore, the total number of elements in the relation $\mathrm{R}$, which satisfies the given conditions, is $16 \times 10 = 160$.
<br/><br/>So, the correct answer is 160. | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgvqdfo3 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$\mathrm{A}=\{2,3,4\}$$ and $$\mathrm{B}=\{8,9,12\}$$. Then the number of elements in the relation
$$\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right.$$ divides $$\mathrm{b}_{2}$$ and $$\mathrm{a}_{2}$$ divides $$\left.\mathrm{b}_{1}\right\}$$ is :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "12"}] | ["C"] | null |
<p>Given sets :
<br/>$ A = {2,3,4} $
<br/>$ B = {8,9,12} $</p>
<p>We want to find the number of elements of the form $( (a_1, b_1), (a_2, b_2) )$ such that :</p>
<ol>
<li>$ a_1 $ divides $ b_2 $</li>
<li>$ a_2 $ divides $ b_1 $</li>
</ol>
<p>For the first condition :
<br/>$ a_1 $ divides $ b_2 $
<br/>Given $ a_1 \in A $ and $ b_2 \in B $, we can list the pairs:
<br/>$ (a_1, b_2) \in {(2,8),(2,12),(3,9),(3,12),(4,8),(4,12)} $
<br/>This gives 6 pairs.</p>
<p>For the second condition, the pairs are the same, because it's just the reversed relation. So :
<br/>$ a_2 $ divides $ b_1 $
<br/>Again has 6 valid pairs.</p>
<p>Now, for every pair from the first condition, we can have any pair from the second condition. This leads to :
<br/>$ 6 \times 6 = 36 $
relations.</p>
| mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxwachw | maths | sets-and-relations | number-of-sets-and-relations | <p>The number of elements in the set $$\{ n \in Z:|{n^2} - 10n + 19| < 6\} $$ is _________.</p> | [] | null | 6 | Given, $\left|n^2-10 n+19\right|<6$
<br/><br/>$\Rightarrow-6 < n^2-10 n+19 < 6$
<br/><br/>Take, $-6 < n^2-10 n+19$ and $n^2-10 n+19 < 6$
<br/><br/>$$
\begin{array}{ll}
\Rightarrow n^2-10 n+25 > 0 & \text { and }\quad n^2-10 n+13 < 0 \\\\
\Rightarrow(n-5)^2 > 0 & \text { and } n=\frac{10 \pm \sqrt{100-52}}{2}<0
\end{array}
$$
<br/><br/>$\Rightarrow n \in \mathbb{Z}-\{5\}$
<br/><br/>$$
\begin{array}{lr}
& \therefore n \in[5-2 \sqrt{3}, 5+2 \sqrt{3}] \\\\
& \therefore n \in[13,8.3] \\\\
& \therefore n=2,3,4,5,6,7,8
\end{array}
$$
<br/><br/>Thus, number of element in the set is ' 6 ' | integer | jee-main-2023-online-10th-april-morning-shift |
1lh23u6jh | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$\mathrm{A}=\{1,2,3,4, \ldots ., 10\}$$ and $$\mathrm{B}=\{0,1,2,3,4\}$$. The number of elements in the relation $$R=\left\{(a, b) \in A \times A: 2(a-b)^{2}+3(a-b) \in B\right\}$$ is ___________.</p> | [] | null | 18 | <p>Given sets :
<br/><br/>A={1,2,3,4, ............,10}
<br/><br/> B={0,1,2,3,4}
<br/><br/>We are looking for pairs $(a,b) \in A \times A$ such that :
<br/><br/>$ 2(a-b)^2 + 3(a-b) \in B $</p>
<p>Let's break down the relation :</p>
<p><strong>Case 1 :</strong> $ a-b = 0 $
<br/><br/>$ 2(a-b)^2 + 3(a-b) = 0 $
<br/><br/>Pairs : $(1,1), (2,2), (3,3), \ldots, (10,10)$ which gives 10 pairs.</p>
<p><strong>Case 2 :</strong> $ a-b = 1 $
<br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(1) + 3(1) = 5 $
<br/><br/>But 5 is not in B, so no pairs for this case.</p>
<p><strong>Case 3 :</strong> $ a-b = -1 $
<br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(1) - 3(1) = -1 $
<br/><br/>This value is not in B, so no pairs for this case.</p>
<p><strong>Case 4 :</strong> $ a-b = 2 $
<br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(4) + 3(2) = 8+6 = 14 $
<br/><br/>Again, 14 is not in B, so no pairs for this case.</p>
<p><strong>Case 5 :</strong> $ a-b = -2 $
<br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(4) - 3(2) = 8 - 6 = 2 $
<br/><br/>Pairs : $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10)$ which gives 8 pairs.</p>
<p>For any other $ a-b $ value, the quadratic will grow larger than the maximum value in B, so we don't need to consider them.</p>
<p>In total, we have $ 10 + 8 = 18 $ pairs in the relation $ R $.</p>
<p>Therefore, the number of elements in the relation $ R $ is 18.</p>
| integer | jee-main-2023-online-6th-april-morning-shift |
lsaqc1qh | maths | sets-and-relations | number-of-sets-and-relations | Let $A=\{1,2,3, \ldots, 20\}$. Let $R_1$ and $R_2$ two relation on $A$ such that
<br/><br/>$R_1=\{(a, b): b$ is divisible by $a\}$
<br/><br/>$R_2=\{(a, b): a$ is an integral multiple of $b\}$.
<br/><br/>Then, number of elements in $R_1-R_2$ is equal to _____________. | [] | null | 46 | <p>To determine the number of elements in $R_1 - R_2$, let's first articulate the meaning of both relations on set $A = \{1, 2, 3, \ldots, 20\}$:</p><p>$R_1$ includes pairs $(a, b)$ where $b$ is divisible by $a$. This includes pairs like $(1,1), (1,2), \ldots, (1,20)$ for $1$; similar series for $2$ up to $(2,20)$ (excluding odd numbers); for $3$ up to $(3,18)$; and so on, reflecting the divisibility condition.</p>
<br/>$\begin{aligned} & R_1:\left\{\begin{array}{l}(1,1),(1,2) \ldots,(1,20), \\ (2,2),(2,4) \ldots,(2,20), \\ (3,3),(3,6) \ldots,(3,18), \\ (4,4),(4,8) \ldots,(4,20), \\ (5,5)(5,10) \ldots,(5,20), \\ (6,6),(6,12),(6,18),(7,7),(7,14), \\ (8,8),(8,16),(9,9),(9,18)(10,10), \\ (10,20),(11,11),(12,12) \ldots,(20,20)\end{array}\right\} \\\\ & n\left(R_1\right)=66\end{aligned}$
<p>$R_2$ consists of pairs $(a, b)$ where $a$ is an integral multiple of $b$. Essentially, this relationship is the reverse of $R_1$. However, for the essence of $R_1 - R_2$, the key overlap comes with pairs where $a = b$, since those are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in $A$ from 1 to 20, resulting in the common elements between $R_1$ and $R_2$ (the intersection $R_1 \cap R_2$) being 20 pairs.</p>
<br/>$\begin{aligned} & \mathrm{R}_1 \cap \mathrm{R}_2=\{(1,1),(2,2), \ldots(20,20)\} \\\\ & \mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right)=20 \end{aligned}$
<p>The difference $R_1 - R_2$ seeks elements present in $R_1$ but not in $R_2$. Given that $R_1$ and $R_2$ share 20 elements that are identical, to find $R_1 - R_2$, we subtract these 20 common elements from the total in $R_1$, resulting in $66 - 20 = 46$ pairs.</p>
<br/>$\begin{aligned} & \mathrm{n}\left(\mathrm{R}_1-\mathrm{R}_2\right)=\mathrm{n}\left(\mathrm{R}_1\right)-\mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right) \\\\ & =\mathrm{n}\left(\mathrm{R}_1\right)-20 \\\\ & =66-20 \\\\ & \mathrm{R}_1-\mathrm{R}_2=46 \text { Pair }\end{aligned}$ | integer | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lscn4dk2 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A$$ and $$B$$ be two finite sets with $$m$$ and $$n$$ elements respectively. The total number of subsets of the set $$A$$ is 56 more than the total number of subsets of $$B$$. Then the distance of the point $$P(m, n)$$ from the point $$Q(-2,-3)$$ is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["B"] | null | <p>$$\begin{aligned}
& 2^{\mathrm{m}}-2^{\mathrm{n}}=56 \\
& 2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 \\
& 2^{\mathrm{n}}=2^3 \text { and } 2^{\mathrm{m}-\mathrm{n}}-1=7 \\
& \Rightarrow \mathrm{n}=3 \text { and } 2^{\mathrm{m}-\mathrm{n}}=8 \\
& \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}-\mathrm{n}=3 \\
& \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}=6 \\
& \mathrm{P}(6,3) \text { and } \mathrm{Q}(-2,-3) \\
& \mathrm{PQ}=\sqrt{8^2+6^2}=\sqrt{100}=10
\end{aligned}$$</p>
<p>Hence option (1) is correct</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd55mch | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3, \ldots \ldots \ldots \ldots, 100\}$$. Let $$R$$ be a relation on $$\mathrm{A}$$ defined by $$(x, y) \in R$$ if and only if $$2 x=3 y$$. Let $$R_1$$ be a symmetric relation on $$A$$ such that $$R \subset R_1$$ and the number of elements in $$R_1$$ is $$\mathrm{n}$$. Then, the minimum value of $$\mathrm{n}$$ is _________.</p> | [] | null | 66 | <p>$$\begin{aligned}
& \mathrm{R}=\{(3,2),(6,4),(9,6),(12,8), \ldots \ldots \ldots .(99,66)\} \\
& \mathrm{n}(\mathrm{R})=33 \\
& \therefore 66
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse5zdng | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3,4\}$$ and $$R=\{(1,2),(2,3),(1,4)\}$$ be a relation on $$\mathrm{A}$$. Let $$\mathrm{S}$$ be the equivalence relation on $$\mathrm{A}$$ such that $$R \subset S$$ and the number of elements in $$\mathrm{S}$$ is $$\mathrm{n}$$. Then, the minimum value of $$n$$ is __________.</p> | [] | null | 16 | $$
\begin{aligned}
& A=\{1,2,3,4\} \\\\
& R=\{(1,2),(2,3),(1,4)\}
\end{aligned}
$$
<br/><br/>$S$ is equivalence
for $R < S$ and reflexive
<br/><br/>$$
\{(1,1),(2,2),(3,3),(4,4)\}
$$
<br/><br/>for symmetric
<br/><br/>$$
\{(2,1),(4,1),(3,2)\}
$$
<br/><br/>for transitive
<br/><br/>$$
\{(1,3),(3,1),(4,2),(2,4)\}
$$
<br/><br/>Now set $S=\{(1,1),(2,2),(3,3),(4,4),(1,2)$, $(2, 3),(1,4),(4,3),(3,4),(2,1),(4,1),(3,2),(1,3),(3$, 1), $(4,2),(2,4)\}$
<br/><br/>$$
n(S)=16
$$ | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsfkhne6 | maths | sets-and-relations | number-of-sets-and-relations | <p>If R is the smallest equivalence relation on the set $$\{1,2,3,4\}$$ such that $$\{(1,2),(1,3)\} \subset \mathrm{R}$$, then the number of elements in $$\mathrm{R}$$ is __________.</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "8"}] | ["B"] | null | <p>Given set $$\{1,2,3,4\}$$</p>
Minimum order pairs are</p>
<p>$$(1,1),(2,2),(3,3),(4,4),(3,1),(2,1),(2,3),(3,2),(1,3),(1,2)$$</p>
<p>Thus no. of elements $$=10$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg59rbo | maths | sets-and-relations | number-of-sets-and-relations | <p>The number of symmetric relations defined on the set $$\{1,2,3,4\}$$ which are not reflexive is _________.</p> | [] | null | 960 | <p>To find the number of symmetric relations on the set $$\{1,2,3,4\}$$ that are not reflexive, we first calculate the total number of symmetric relations and then subtract the count of those that are both symmetric and reflexive.</p><p>A symmetric relation involves pairs where if a pair (x, y) is in the relation, then (y, x) is also in the relation. For a set with $$n$$ elements, there are $$\frac{n(n+1)}{2}$$ slots in the relation matrix that can independently be occupied or not, corresponding to a total of $$2^{\frac{n(n+1)}{2}}$$ possible symmetric relations.</p><p>A relation is reflexive if every element is related to itself, requiring all diagonal slots of the relation matrix (n of them) to be filled. The remaining $$\frac{n(n-1)}{2}$$ slots can be filled in any manner, leading to $$2^{\frac{n(n-1)}{2}}$$ reflexive (and possibly symmetric) relations.</p><p>For the set $$\{1,2,3,4\}$$ ($$n=4$$):</p><ul><li>Total symmetric relations: $$2^{\frac{4(4+1)}{2}} = 2^{10} = 1024$$</li><li>Symmetric and reflexive relations: $$2^{\frac{4(4-1)}{2}} = 2^{6} = 64$$</li></ul><p>Therefore, the number of symmetric relations that are not reflexive: $$1024 - 64 = 960$$.</p> | integer | jee-main-2024-online-30th-january-evening-shift |
luy9cl8m | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{2,3,6,7\}$$ and $$B=\{4,5,6,8\}$$. Let $$R$$ be a relation defined on $$A \times B$$ by $$(a_1, b_1) R(a_2, b_2)$$ if and only if $$a_1+a_2=b_1+b_2$$. Then the number of elements in $$R$$ is __________.</p> | [] | null | 25 | <p>To find the number of elements in the relation $$R$$ defined on $$A \times B$$, we need to determine all pairs $$((a_1, b_1), (a_2, b_2))$$ such that $$a_1 + a_2 = b_1 + b_2$$, where $$a_1, a_2 \in A$$ and $$b_1, b_2 \in B$$.</p>
<p>First, consider all possible sums of pairs from set $$A$$ and set $$B$$.</p>
<p>Possible sums from set $$A = \{2, 3, 6, 7\}$$:</p>
<ul>
<li>$$2 + 2 = 4$$</li>
<li>$$2 + 3 = 5$$</li>
<li>$$2 + 6 = 8$$</li>
<li>$$2 + 7 = 9$$</li>
<li>$$3 + 2 = 5$$</li>
<li>$$3 + 3 = 6$$</li>
<li>$$3 + 6 = 9$$</li>
<li>$$3 + 7 = 10$$</li>
<li>$$6 + 2 = 8$$</li>
<li>$$6 + 3 = 9$$</li>
<li>$$6 + 6 = 12$$</li>
<li>$$6 + 7 = 13$$</li>
<li>$$7 + 2 = 9$$</li>
<li>$$7 + 3 = 10$$</li>
<li>$$7 + 6 = 13$$</li>
<li>$$7 + 7 = 14$$</li>
</ul>
<p>Possible sums from set $$B = \{4, 5, 6, 8\}$$:</p>
<ul>
<li>$$4 + 4 = 8$$</li>
<li>$$4 + 5 = 9$$</li>
<li>$$4 + 6 = 10$$</li>
<li>$$4 + 8 = 12$$</li>
<li>$$5 + 4 = 9$$</li>
<li>$$5 + 5 = 10$$</li>
<li>$$5 + 6 = 11$$</li>
<li>$$5 + 8 = 13$$</li>
<li>$$6 + 4 = 10$$</li>
<li>$$6 + 5 = 11$$</li>
<li>$$6 + 6 = 12$$</li>
<li>$$6 + 8 = 14$$</li>
<li>$$8 + 4 = 12$$</li>
<li>$$8 + 5 = 13$$</li>
<li>$$8 + 6 = 14$$</li>
<li>$$8 + 8 = 16$$</li>
</ul>
<p>Now, identify the common sums from both sets:</p>
<p>Common sums: $$8, 9, 10, 12, 13, 14$$</p>
<p>For each common sum, count the pairs from set $$A$$ and set $$B$$ that produce these sums:</p>
<ul>
<li>Sum = 8: From $$A$$: {(2,6), (6,2)} - 2 pairs; From $$B$$: {(4,4)} - 1 pair; Hence, 2 * 1 = 2 pairs</li>
<li>Sum = 9: From $$A$$: {(2,7), (3,6), (6,3), (7,2)} - 4 pairs; From $$B$$: {(4,5), (5,4)} - 2 pairs; Hence, 4 * 2 = 8 pairs</li>
<li>Sum = 10: From $$A$$: {(3,7), (7,3)} - 2 pairs; From $$B$$: {(4,6), (5,5), (6,4)} - 3 pairs; Hence, 2 * 3 = 6 pairs</li>
<li>Sum = 12: From $$A$$: {(6,6)} - 1 pair; From $$B$$: {(4,8), (6,6), (8,4)} - 3 pairs; Hence, 1 * 3 = 3 pairs</li>
<li>Sum = 13: From $$A$$: {(6,7), (7,6)} - 2 pairs; From $$B$$: {(5,8), (8,5)} - 2 pairs; Hence, 2 * 2 = 4 pairs</li>
<li>Sum = 14: From $$A$$: {(7,7)} - 1 pair; From $$B$$: {(6,8), (8,6)} - 2 pairs; Hence, 1 * 2 = 2 pairs</li>
</ul>
<p>Adding all these, we get the number of elements in the relation $$R$$:</p>
<p>2 + 8 + 6 + 3 + 4 + 2 = 25</p>
<p>Thus, the number of elements in $$R$$ is 25.</p> | integer | jee-main-2024-online-9th-april-morning-shift |
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