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1lgvqpf80 | maths | sequences-and-series | arithmetico-geometric-progression | <p>Suppose $$a_{1}, a_{2}, 2, a_{3}, a_{4}$$ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $$\frac{49}{2}$$, then $$a_{4}$$ is equal to __________.</p> | [] | null | 16 | Since, common ratio of A.G.P. is 2 therefore A.G.P. can be taken as
<br/><br/>$$
\begin{aligned}
& \frac{(a-2 d)}{4}, \frac{(c-d)}{2}, a, 2(a+d), 4(a+2 d) \\\\
& \text { or } a_1, a_2, 2, a_3, a_4 \text { (Given) } \\\\
& \Rightarrow a=2
\end{aligned}
$$
<br/><br/>also sum of thes A.G.P. is $\frac{49}{2}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{2-2 d}{4}+\frac{2-d}{2}+2+2(2+d)+4(2+2 d)=\frac{49}{2} \\\\
& \Rightarrow \frac{1}{4}[2-2 d+4-2 d+8+16+8 d+32+32 d]=\frac{49}{2} \\\\
& \Rightarrow 36 d+62=98
\end{aligned}
$$
<br/><br/>$$
\Rightarrow 36 d=36 \Rightarrow d=1
$$
<br/><br/>Hence, $a_4=4(a+2 d)=4(2+2 \times 1)=16$ | integer | jee-main-2023-online-10th-april-evening-shift |
1lgypzf6s | maths | sequences-and-series | arithmetico-geometric-progression | <p>Let $$0 < z < y < x$$ be three real numbers such that $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in an arithmetic progression and $$x, \sqrt{2} y, z$$ are in a geometric progression. If $$x y+y z+z x=\frac{3}{\sqrt{2}} x y z$$ , then $$3(x+y+z)^{2}$$ is equal to ____________.</p> | [] | null | 150 | $\because \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.
<br/><br/>$$
\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y}
$$ ........... (i)
<br/><br/>and $x, \sqrt{2} y, z$ are in G.P.
<br/><br/>$$
\Rightarrow 2 y^2=x z
$$ .......... (ii)
<br/><br/>from (i), $\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$
<br/><br/>$$
\Rightarrow 4 y=x+z
$$
<br/><br/>$$
\begin{aligned}
& \text { Also, } x y+y z+z x=\frac{3}{\sqrt{2}} x y z \\\\
& y(4 y)+x z=\frac{3}{\sqrt{2}}\left(2 y^2\right) y \\\\
& \Rightarrow 4 y^2+2 y^2=3 \sqrt{2} y^3 \\\\
& \Rightarrow 6 y^2=3 \sqrt{2} y^3 \Rightarrow y=\sqrt{2} \\\\
& \therefore 3(x+y+z)^2=3(5 y)^2=3(5 \sqrt{2})^2 \\\\
& =150
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-evening-shift |
lsblc3ma | maths | sequences-and-series | arithmetico-geometric-progression | If $8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2 p)+\frac{1}{4^3}(3+3 p)+\cdots \cdots \infty$, then the value of $p$ is ____________. | [] | null | 9 | <p>$$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}$$</p>
<p>$$\text { (sum of infinite terms of A.G.P }=\frac{a}{1-r}+\frac{d r}{(1-r)^2} \text { ) }$$</p>
<p>$$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
lv9s20lt | maths | sequences-and-series | arithmetico-geometric-progression | <p>If $$1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\ldots$$ upto $$\infty=2+\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right)$$, where a and b are integers with $$\operatorname{gcd}(a, b)=1$$, then $$\mathrm{11 a+18 b}$$ is equal to __________.</p> | [] | null | 76 | <p>$$\begin{aligned}
& S=1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\ldots \infty \\\\
& =1+\frac{(1-\sqrt{2} / \sqrt{3})}{2}+\frac{(1-\sqrt{2} / \sqrt{3})^2}{6}+\frac{(1-\sqrt{2} / \sqrt{3})^3}{12}+\ldots \infty
\end{aligned}$$</p>
<p>$$\text { let } 1-\frac{\sqrt{2}}{\sqrt{3}}=a$$</p>
<p>$$\begin{aligned}
& S=1+\frac{a}{2}+\frac{a^2}{6}+\frac{a^3}{12}+\ldots \\\\
& =1+\left(1-\frac{1}{2}\right) a+\left(\frac{1}{2}-\frac{1}{3}\right) a^2+\left(\frac{1}{3}-\frac{1}{4}\right) a^3+\ldots \\\\
& =1+\left(a+\frac{a^2}{2}+\frac{a^3}{3} \ldots \infty\right)+\frac{1}{a}\left(\frac{-a^2}{2}-\frac{a^3}{3}-\frac{a^4}{4} \ldots \infty\right) \\\\
& =-\ln (1-a)+\frac{1}{a}\left(-a-\frac{a^2}{2}-\frac{a^3}{3} \ldots \infty\right)+2 \\\\
& =-\ln (1-a)+\frac{1}{a} \ln (1-a)+2 \\\\
& =2+\left(\frac{1}{a}-1\right) \ln (1-a) \\\\
& =2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \ln \left(1-1+\sqrt{\frac{2}{3}}\right) \\\\
& =2+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2} \ln \sqrt{\frac{2}{3}} \\\\
& =2+\left(\frac{\sqrt{6}+2}{1} \cdot \frac{1}{2} \ln \frac{2}{3}\right) \\\\
& \therefore 2+\left(\sqrt{\frac{3}{2}}+1\right) \ln \frac{2}{3} \\\\
& \therefore 11 a+18 b=76
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-evening-shift |
lvc57uf2 | maths | sequences-and-series | arithmetico-geometric-progression | <p>Let the first term of a series be $$T_1=6$$ and its $$r^{\text {th }}$$ term $$T_r=3 T_{r-1}+6^r, r=2,3$$,
............ $$n$$. If the sum of the first $$n$$ terms of this series is $$\frac{1}{5}\left(n^2-12 n+39\right)\left(4 \cdot 6^n-5 \cdot 3^n+1\right)$$, then $$n$$ is equal to ___________.</p> | [] | null | 6 | <p>$$\begin{aligned}
& T_r=3 T_{r-1}+6^r \\
& \Rightarrow \text { solving homogenous part } \\
& T_r=3 T_{r-1} \\
& \Rightarrow x=3 \text { is the root }
\end{aligned}$$</p>
<p>$$\therefore T_r=a .3^r$$</p>
<p>Solving for particular part</p>
<p>$$\begin{aligned}
& T_r=b .6^r \\
& b .6^r=3 b 6^{r-1}+6^r \\
& \Rightarrow 6 b=3 b+6 \\
& \Rightarrow 3 b=6 \\
& \Rightarrow b=2 \\
& T_r=a^n+a^p \\
& T_r=a 3^{b r}+2.6^r \quad \text{.... (i)} \\
& T_r=3 T_{r-1}+6^r
\end{aligned}$$</p>
<p>Putting $$r=2$$</p>
<p>$$T_2=18+36=54 \quad \text{.... (ii)}$$</p>
<p>Using equation (i) and (ii)</p>
<p>$$\begin{aligned}
& 54=9 a+72 \Rightarrow-18=9 a \Rightarrow a=-2 \\
& \therefore T_r=2 \cdot 6^r-2 \cdot 3^r=2\left(6^r-3^r\right) \\
& \sum_{r=1}^n T_r=2 \sum 6^r-2 \sum 3^r \\
& =2 \cdot 6 \frac{\left(6^n-1\right)}{5}-2 \cdot 3 \frac{\left(3^n-1\right)}{2} \\
& =\frac{3}{5}\left(4 \cdot 6^n-5 \cdot 3^n+1\right) \\
& \therefore n^2-12 n+39=3 \\
& n^2-12 n+36=0 \\
& (n-6)^2=0 \\
& \therefore n=6
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-morning-shift |
jk3IWUpURQ7vO8z9 | maths | sequences-and-series | geometric-progression-(g.p) | Fifth term of a GP is 2, then the product of its 9 terms is | [{"identifier": "A", "content": "256"}, {"identifier": "B", "content": "512 "}, {"identifier": "C", "content": "1024"}, {"identifier": "D", "content": "none of these"}] | ["B"] | null | $$a{r^4} = 2$$
<br><br>$$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$$
<br><br>$$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$$ | mcq | aieee-2002 |
y3zqSDYlgLFRtylu | maths | sequences-and-series | geometric-progression-(g.p) | l, m, n are the $${p^{th}}$$, $${q^{th}}$$ and $${r^{th}}$$ term of a G.P all positive, $$then\,\left| {\matrix{
{\log \,l} & p & 1 \cr
{\log \,m} & q & 1 \cr
{\log \,n} & r & 1 \cr
} } \right|\,equals$$ | [{"identifier": "A", "content": "- 1"}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "1 "}, {"identifier": "D", "content": "0 "}] | ["D"] | null | $$l = A{R^{p - 1}}$$
<br><br>$$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$$
<br><br>$$m = A{R^{q - 1}}$$
<br><br>$$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$$
<br><br>$$n = A{R^{r - 1}}$$
<br><br>$$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$$
<br><br>Now, $$\left| {\matrix{
{\log l} & p & 1 \cr
{\log m} & q & 1 \cr
{\log n} & r & 1 \cr
} } \right|$$
<br><br>$$ = \left| {\matrix{
{\log A + \left( {p - 1} \right)\log R} & p & 1 \cr
{\log A + \left( {q - 1} \right)\log R} & q & 1 \cr
{\log A + \left( {r - 1} \right)\log R} & r & 1 \cr
} } \right|$$
<br><br>Operating $${C_1} - \left( {\log R} \right){C_2} + \left( {\log R - \log A} \right){C_3}$$
<br><br>$$ = \left| {\matrix{
0 & p & 1 \cr
0 & q & 1 \cr
0 & r & 1 \cr
} } \right| = 0$$ | mcq | aieee-2002 |
WsEUUUVc0kVsIZYL | maths | sequences-and-series | geometric-progression-(g.p) | Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is | [{"identifier": "A", "content": "5 "}, {"identifier": "B", "content": "3/5 "}, {"identifier": "C", "content": "8/5 "}, {"identifier": "D", "content": "1/5 "}] | ["B"] | null | Let $$a=$$ first team of $$G.P.$$ and $$r=$$ common ratio of $$G.P.;$$
<br><br>Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$
<br><br>Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$
<br><br>$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$
<br><br>Also $${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$$ to $$\infty = 100$$
<br><br>$$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$$
<br><br>$$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$$
<br><br>From $$(i),$$ $${a^2} = 400{\left( {1 - r} \right)^2};$$
<br><br>From $$(ii),$$ we get $$100\left( {1 - r} \right)\left( {1 + r} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$$
<br><br>$$ \Rightarrow 1 + r = 4 - 4r$$
<br><br>$$ \Rightarrow 5r = 3$$
<br><br>$$ \Rightarrow r = 3/5.$$ | mcq | aieee-2002 |
LpYUFoQKA1MVXCBb | maths | sequences-and-series | geometric-progression-(g.p) | In a geometric progression consisting of positive terms, each term equals the sum of the next two terns. Then the common ratio of its progression is equals | [{"identifier": "A", "content": "$${\\sqrt 5 }$$ "}, {"identifier": "B", "content": "$$\\,{1 \\over 2}\\left( {\\sqrt 5 - 1} \\right)$$ "}, {"identifier": "C", "content": "$${1 \\over 2}\\left( {1 - \\sqrt 5 } \\right)$$ "}, {"identifier": "D", "content": "$${1 \\over 2}\\sqrt 5 $$."}] | ["B"] | null | Let the series $$a,ar,$$ $$a{r^2},........$$ are in geometric progression.
<br><br>given, $$a = ar + a{r^2}$$
<br><br>$$ \Rightarrow 1 = r + {r^2}$$
<br><br>$$ \Rightarrow {r^2} + r - 1 = 0$$
<br><br>$$ \Rightarrow r = {{ - 1 \mp \sqrt {1 - 4 \times - 1} } \over 2}$$
<br><br>$$ \Rightarrow r = {{ - 1 \pm \sqrt 5 } \over 2}$$
<br><br>$$ \Rightarrow r = {{\sqrt 5 - 1} \over 2}$$
<br><br>[ As terms of $$G.P.$$ are positive
<br><br>$$\therefore$$ $$r$$ should be positive] | mcq | aieee-2007 |
dsXDAGYKTZx0KfLz | maths | sequences-and-series | geometric-progression-(g.p) | The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is | [{"identifier": "A", "content": "- 4"}, {"identifier": "B", "content": "- 12"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "4"}] | ["B"] | null | As per question,
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}$$
<br><br>$$ \Rightarrow {r^2} = 4, \Rightarrow r = - 2$$
<br><br>(As terms are $$=+ve$$ and $$-ve$$ alternately)
<br><br>$$ \Rightarrow a = - 12$$ | mcq | aieee-2008 |
cbq6hH68znDtn5vz | maths | sequences-and-series | geometric-progression-(g.p) | Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is : | [{"identifier": "A", "content": "$$2 - \\sqrt 3 $$ "}, {"identifier": "B", "content": "$$2 + \\sqrt 3 $$ "}, {"identifier": "C", "content": "$$\\sqrt 2 + \\sqrt 3 $$ "}, {"identifier": "D", "content": "$$3 + \\sqrt 2 $$ "}] | ["B"] | null | Let $$a,ar,a{r^2}$$ are in $$G.P.$$
<br><br>According to the question
<br><br>$$a,2ar,a{r^2}$$ are in $$A.P.$$
<br><br>$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$
<br><br>$$ \Rightarrow 4r = 1 + {r^2}$$
<br><br>$$ \Rightarrow {r^2} - 4r + 1 = 0$$
<br><br>$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$
<br><br>Since $$r > 1$$
<br><br>$$\therefore$$ $$\pi = 2 - \sqrt 3 $$ is rejected
<br><br>Hence, $$r = 2 + \sqrt 3 $$ | mcq | jee-main-2014-offline |
t8Hle6z9XMRBLBrD | maths | sequences-and-series | geometric-progression-(g.p) | If the $${2^{nd}},{5^{th}}\,and\,{9^{th}}$$ terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is : | [{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "$${7 \\over 4}$$ "}, {"identifier": "C", "content": "$${8 \\over 5}$$ "}, {"identifier": "D", "content": "$${4 \\over 3}$$"}] | ["D"] | null | <p>The terms of an Arithmetic Progression (A.P.) are given by $a$, $a + d$, $a + 2d$, ..., where $a$ is the first term and $d$ is the common difference.</p>
<p>Given that the 2nd, 5th and 9th terms of an A.P. are in Geometric Progression (G.P.), we can denote them as follows :</p>
<p>2nd term = $a + d$</p>
<p>5th term = $a + 4d$</p>
<p>9th term = $a + 8d$</p>
<p>For three numbers to be in G.P., the square of the middle term must be equal to the product of the other two terms. So,</p>
<p>$(a + 4d)^2 = (a + d)(a + 8d)$</p>
<p>Expanding and simplifying :</p>
<p>$a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2$</p>
<p>$8ad + 16d^2 = a^2 + 9ad + 8d^2$</p>
<p>$8ad - 9ad = 8d^2 - 16d^2$</p>
<p>$-ad = -8d^2$</p>
<p>$a = 8d$</p>
<p>The common ratio of the G.P. is the ratio of the 5th term to the 2nd term, or $(a + 4d) / (a + d)$. Substituting $a = 8d$ gives :</p>
<p>$(8d + 4d) / (8d + d) = 12d / 9d = 4 / 3$</p>
<p>So, the common ratio of the G.P. is $4 / 3$. The answer is option D.</p>
| mcq | jee-main-2016-offline |
r0ajdDNJvO6Ela0EA6Qzo | maths | sequences-and-series | geometric-progression-(g.p) | If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval : | [{"identifier": "A", "content": "($$-$$ $$\\infty $$, $$-$$10]"}, {"identifier": "B", "content": "($$-$$10, 0)"}, {"identifier": "C", "content": "(0, 10)"}, {"identifier": "D", "content": "[10, $$\\infty $$)"}] | ["C"] | null | Sum of infinite G.P,<br><br>
S = $$b \over {1-r}$$ where $$\left| r \right| < 1$$<br><br>
$$ \Rightarrow $$ 5 = $$b \over {1-r}$$<br><br>
$$ \Rightarrow $$ 1 - r = $$b \over 5$$<br><br>
$$ \Rightarrow $$ b = 5(1 - r)<br><br>
as $$\left| r \right| < 1$$<br><br>
$$ \therefore $$ -1 < r < 1<br><br>
$$ \Rightarrow $$ 1 > -r > -1<br><br>
$$ \Rightarrow $$ 2 > 1-r > 0<br><br>
$$ \Rightarrow $$ 10 > 5(1-r) > 0<br><br>
$$ \Rightarrow $$ 10 > b > 0<br><br>
$$ \therefore $$ interval of b = (0, 10) | mcq | jee-main-2018-online-15th-april-morning-slot |
4jkMKIeMvDgkctyNVmMvG | maths | sequences-and-series | geometric-progression-(g.p) | If a, b, c are in A.P. and a<sup>2</sup>, b<sup>2</sup>, c<sup>2</sup> are in G.P. such that
<br/>a < b < c and a + b + c = $${3 \over 4},$$ then the value of a is : | [{"identifier": "A", "content": "$${1 \\over 4} - {1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over 4} - {1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over 4} - {1 \\over {2\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 4} - {1 \\over {\\sqrt 2 }}$$"}] | ["C"] | null | $$ \because $$$$\,\,\,$$a, b, c are in A.P. then
<br><br>a + c = 2b
<br><br>also it is given that,
<br><br>a + b + c = $${{3 \over 4}}$$ . . . .(1)
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2b + b = $${{3 \over 4}}$$ $$ \Rightarrow $$$$\,\,\,$$ b = $${{1 \over 4}}$$ . . . . .(2)
<br><br>Again it is given that, a<sup>2</sup>, b<sup>2</sup>, c<sup>2</sup> are in G.P. then
<br><br>(b<sup>2</sup>)<sup>2</sup> = a<sup>2</sup>c<sup>2</sup> $$ \Rightarrow $$ ac = $$ \pm $$ $${{1 \over {16}}}$$ . . . . (3)
<br><br>From (1), (2) and (3), we get;
<br><br>$$a \pm {1 \over {16a}}$$ = $${1 \over 2}$$ $$ \Rightarrow $$ 16a<sup>2</sup> $$-$$ 8a $$ \pm $$ 1 = 0
<br><br><b>Case I </b>: 16a<sup>2</sup> $$-$$ 8a + 1 = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$a = $${1 \over 4}$$ (not possible as a < b)
<br><br><b>Case II:</b> 16a<sup>2</sup> $$-$$ 8a $$-$$ 1 = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ a = $${{8 \pm \sqrt {128} } \over {32}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ a = $${1 \over 4} \pm {1 \over {2\sqrt 2 }}$$
<br><br>$$ \therefore $$$$\,\,\,$$ a = $${1 \over 4} - {1 \over {2\sqrt 2 }}$$ ($$ \because $$ a < b) | mcq | jee-main-2018-online-15th-april-evening-slot |
m3xNcW58RdFGXh2LILIzd | maths | sequences-and-series | geometric-progression-(g.p) | If three distinct numbers a, b, c are in G.P. and the
equations ax<sup>2</sup>
+ 2bx + c = 0 and
dx<sup>2</sup>
+ 2ex + ƒ = 0 have a common root, then
which one of the following statements is
correct? | [{"identifier": "A", "content": "$$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in G.P."}, {"identifier": "B", "content": "d, e, \u0192 are in A.P"}, {"identifier": "C", "content": "d, e, \u0192 are in G.P"}, {"identifier": "D", "content": "$$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in A.P."}] | ["D"] | null | Given, a, b, c are in G.P.
<br><br>$$ \therefore $$ b<sup>2</sup> = ac
<br><br>In this equation ax<sup>2</sup>
+ 2bx + c = 0,
<br><br>Discrimant, D = 4b<sup>2</sup> - 4ac
<br><br>= 4ac - 4ac
<br><br>= 0
<br><br>Discrimant = 0 meand roots of the equation are equal.
<br><br>Let both the roots of the equation = $$\alpha $$
<br><br>$$ \therefore $$ 2$$\alpha $$ = $$ - {{2b} \over a}$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $$ - {b \over a}$$
<br><br>As both the
equations ax<sup>2</sup>
+ 2bx + c = 0 and
dx<sup>2</sup>
+ 2ex + ƒ = 0 have a common root,
<br><br>so $$ - {b \over a}$$ is also root of the equation dx<sup>2</sup>
+ 2ex + ƒ = 0.
<br><br>$$ \therefore $$ $$ - {b \over a}$$ satisfy the equation dx<sup>2</sup>
+ 2ex + ƒ = 0.
<br><br>$$ \therefore $$ $$d{\left( { - {b \over a}} \right)^2} + 2e\left( { - {b \over a}} \right) + f = 0$$
<br><br>$$ \Rightarrow $$ $$d{b^2} - 2aeb + f{a^2} = 0$$
<br><br>$$ \Rightarrow $$ $$dac - 2aeb + f{a^2} = 0$$
<br><br>$$ \Rightarrow $$ $$dc - 2eb + fa = 0$$
<br><br>$$ \Rightarrow $$ $${{dc} \over {ac}} - {{2eb} \over {ac}} + {{fa} \over {ac}} = 0$$
<br><br>$$ \Rightarrow $$ $${{dc} \over {ac}} - {{2eb} \over {{b^2}}} + {{fa} \over {ac}} = 0$$
<br><br>$$ \Rightarrow $$ $${d \over a} - {{2e} \over b} + {f \over c} = 0$$
<br><br>$$ \Rightarrow $$ $${{2e} \over b} = {d \over a} + {f \over c}$$
<br><br>$$ \therefore $$
$$d \over a$$, $$e \over b$$, $$f \over c$$ are in A.P. | mcq | jee-main-2019-online-8th-april-evening-slot |
jOGbNFYRMg8wfAaE013rsa0w2w9jx2b2rv6 | maths | sequences-and-series | geometric-progression-(g.p) | Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$ \ne $$ 0 and 0 < r $$ \le $$ $${1 \over 2}$$
. If 3$$a$$, 7b and 15c are the first three
terms of an A.P., then the 4<sup>th</sup> term of this A.P. is : | [{"identifier": "A", "content": "$$a$$"}, {"identifier": "B", "content": "$${7 \\over 3}a$$"}, {"identifier": "C", "content": "5$$a$$"}, {"identifier": "D", "content": "$${2 \\over 3}a$$"}] | ["A"] | null | a = a, b = ar and c = ar<sup>2</sup><br><br>
3a, 7b, 15c $$ \to $$ A.P.<br><br>
14b = 3a + 15c<br><br>
14(ar) = 3a + 15(ar<sup>2</sup>)<br><br>
15r<sup>2</sup> – 14r + 3 = 0<br><br>
$$ \Rightarrow r = {1 \over 3},{3 \over 5}(rejected)$$<br><br>
Common difference = 7b – 3a<br><br>
= 7ar – 3a<br><br>
$$ \Rightarrow $$ $${{7a} \over 3} - 3a = - {2 \over 3}a$$<br><br>
4<sup>th</sup> term is $$ \Rightarrow $$ $$15c - {2 \over 3}a = {{15} \over 9}a - {2 \over 3}a = a$$ | mcq | jee-main-2019-online-10th-april-evening-slot |
Fnn2C32pKxBkXMQh8M8iF | maths | sequences-and-series | geometric-progression-(g.p) | The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is : | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "28"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "24"}] | ["B"] | null | Let terms are $${a \over r},a,ar \to G.P$$
<br><br>$$ \therefore $$ $${a^3}$$ = 512 $$ \Rightarrow $$ a = 8
<br><br>$${8 \over r} + 4,12,8r \to A.P.$$
<br><br>24 = $${8 \over r} + 4 + 8r$$
<br><br>r = 2, r = $${1 \over 2}$$
<br><br>r = 2(4, 8, 16)
<br><br>r = $${1 \over 2}$$ (16, 8, 4)
<br><br>Sum = 28 | mcq | jee-main-2019-online-12th-january-morning-slot |
JASdaZ2MclBZMbjgwCTi1 | maths | sequences-and-series | geometric-progression-(g.p) | Let a<sub>1</sub>, a<sub>2</sub>, . . . . . ., a<sub>10</sub> be a G.P. If $${{{a_3}} \over {{a_1}}} = 25,$$ then $${{{a_9}} \over {{a_5}}}$$ equals | [{"identifier": "A", "content": "5<sup>3</sup>"}, {"identifier": "B", "content": "2(5<sup>2</sup>)"}, {"identifier": "C", "content": "4(5<sup>2</sup>)"}, {"identifier": "D", "content": "5<sup>4</sup>"}] | ["D"] | null | a<sub>1</sub>, a<sub>2</sub>, . . . . ., a<sub>10</sub> are in G.P.,
<br><br>Let the common ratio be r
<br><br>$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$
<br><br>$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
bCZmTgEn73Gkw9JiQJNZG | maths | sequences-and-series | geometric-progression-(g.p) | The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $${{27} \over {19}}$$.Then the common ratio of this series is : | [{"identifier": "A", "content": "$${4 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over 9}$$"}] | ["C"] | null | $${a \over {1 - r}} = 3\,\,\,\,\,\,\,.....(1)$$
<br><br>$${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$$
<br><br>$$ \Rightarrow 6{r^2} - 13r + 6 = 0$$
<br><br>$$ \Rightarrow r = {2 \over 3}\,\,$$
<br><br>as $$\left| r \right| < 1$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
ccIa2Uu7f8juJqgTd5Eiy | maths | sequences-and-series | geometric-progression-(g.p) | Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... a<sub>10</sub> be in G.P. with a<sub>i</sub> > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k $$ \in $$ N (the set of natural numbers) for which
<br/><br/>$$\left| {\matrix{
{{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} & {{{\log }_e}\,{a_3}^r{a_4}^k} \cr
{{{\log }_e}\,{a_4}^r{a_5}^k} & {{{\log }_e}\,{a_5}^r{a_6}^k} & {{{\log }_e}\,{a_6}^r{a_7}^k} \cr
{{{\log }_e}\,{a_7}^r{a_8}^k} & {{{\log }_e}\,{a_8}^r{a_9}^k} & {{{\log }_e}\,{a_9}^r{a_{10}}^k} \cr
} } \right|$$ $$=$$ 0.
<br/><br/>Then the number of elements in S, is - | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "infinitely many"}] | ["D"] | null | Apply
<br><br>C<sub>3</sub> $$ \to $$ C<sub>3</sub><sub></sub> $$-$$ C<sub>2</sub>
<br><br>C<sub>2</sub><sub></sub> $$ \to $$ C<sub>2</sub> $$-$$ C<sub>1</sub>
<br><br>We get D = 0 | mcq | jee-main-2019-online-10th-january-evening-slot |
sIssaJfB1HmzkLOvnfDcl | maths | sequences-and-series | geometric-progression-(g.p) | If a, b, c be three distinct real numbers in G.P. and a + b + c = xb , then x <b>cannot</b> be | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "-3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "-2"}] | ["A"] | null | a, b, c are in G.P.
<br><br>So, b = ar
<br><br>and c = ar<sup>2</sup>
<br><br>given a + b + c = xb
<br><br>$$ \Rightarrow $$ a + br + ar<sup>2</sup> = x(ar)
<br><br>$$ \Rightarrow $$ 1 + r + r<sup>2</sup> = xr
<br><br>$$ \Rightarrow $$ x = 1 + r + $${1 \over r}$$
<br><br>let sum of r + $${1 \over r}$$ = M
<br><br>$$ \therefore $$ r<sup>2</sup> + 1 = Mr
<br><br>$$ \Rightarrow $$ r<sup>2</sup> $$-$$ Mr + 1 = 0
<br><br>this quadratic equation will have
<br><br>real solution when discriminant is $$ \ge $$ 0
<br><br>$$ \therefore $$ b<sup>2</sup> $$-$$ 4ac $$ \ge $$ 0
<br><br>M<sup>2</sup> $$-$$ 4.1.1 $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ M<sup>2</sup> $$ \ge $$ 4
<br><br>M $$ \ge $$ 2 or M $$ \le $$ $$-$$ 2
<br><br>$$ \therefore $$ M $$ \in $$ ($$-$$ $$ \propto $$, $$-$$ 2] $$ \cup $$ [2, $$ \propto $$)
<br><br>As x = 1 + r + $${1 \over r}$$
<br><br>= 1 + M
<br><br>$$ \therefore $$ x $$ \in $$ ($$-$$ $$ \propto $$, $$-$$ 1] $$ \cup $$ [3, $$ \propto $$)
<br><br>$$ \therefore $$ x can't be 0, 1, 2. | mcq | jee-main-2019-online-9th-january-morning-slot |
4x6CObgJQXDZB4of137k9k2k5fj4i71 | maths | sequences-and-series | geometric-progression-(g.p) | Let $${a_1}$$
, $${a_2}$$
, $${a_3}$$
,....... be a G.P. such that <br/>$${a_1}$$
< 0, $${a_1}$$
+ $${a_2}$$
= 4 and $${a_3}$$
+ $${a_4}$$
= 16. <br/>If $$\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $$, then $$\lambda $$ is
equal to: | [{"identifier": "A", "content": "171"}, {"identifier": "B", "content": "-171"}, {"identifier": "C", "content": "-513"}, {"identifier": "D", "content": "$${{511} \\over 3}$$"}] | ["B"] | null | $${a_1}$$
+ $${a_2}$$
= 4
<br><br>$$ \Rightarrow $$ $${a_1}$$
+ $${a_1}$$r
= 4 ...(1)
<br><br>$${a_3}$$
+ $${a_4}$$
= 16
<br><br>$$ \Rightarrow $$ $${a_1}$$r<sup>2</sup>
+ $${a_1}$$r<sup>3</sup>
= 16 ...(2)
<br><br>Doing (1) $$ \div $$ (2), we get
<br><br>r = $$ \pm $$ 2
<br><br>If r = 2, then a<sub>1</sub> = $${4 \over 3}$$
<br><br>If r = -2, then a<sub>1</sub> = -4
<br><br>Given $${a_1}$$
< 0
<br><br>$$ \therefore $$ a<sub>1</sub> = -4
<br><br>$$ \therefore $$ $$\sum\limits_{i = 1}^9 {{a_i}}$$ = $${{a\left( {{r^9} - 1} \right)} \over {r - 1}}$$ = 4$$\lambda $$
<br><br>$$ \Rightarrow $$ $${{ - 4\left( {{{\left( { - 2} \right)}^9} - 1} \right)} \over { - 2 - 1}}$$ = 4$$\lambda $$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = -171 | mcq | jee-main-2020-online-7th-january-evening-slot |
94MqkDoVjiepQoAoLy7k9k2k5khy3qy | maths | sequences-and-series | geometric-progression-(g.p) | Let a<sub>n</sub> be the n<sup>th</sup> term of a G.P. of positive terms.<br/><br/>
$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$ and $$\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $$,
<br/><br/>
then $$\sum\limits_{n = 1}^{200} {{a_n}} $$ is equal to : | [{"identifier": "A", "content": "150"}, {"identifier": "B", "content": "175"}, {"identifier": "C", "content": "225"}, {"identifier": "D", "content": "300"}] | ["A"] | null | $$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$
<br><br>$$ \Rightarrow $$ a<sub>3</sub> + a<sub>5</sub> + a<sub>7</sub> + .... + a<sub>201</sub> = 200
<br><br>$$ \Rightarrow $$ $$a{r^2}{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}$$ = 200 ....(1)
<br><br>$$\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $$
<br><br>$$ \Rightarrow $$ a<sub>2</sub> + a<sub>4</sub> + a<sub>6</sub> + ... + a<sub>200</sub> = 100
<br><br>$$ \Rightarrow $$ $$ar{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}$$ = 100 ....(2)
<br><br>dividing (1) by (2)
<br><br>we get, r = 2
<br><br>adding both (1) and (2), we get
<br><br>a<sub>2</sub> + a<sub>3</sub> + a<sub>4</sub> + a<sub>5</sub> + ..... + a<sub>201</sub> = 300
<br><br>$$ \Rightarrow $$ r(a<sub>1</sub> + a<sub>2</sub> + ..... + a<sub>200</sub>) = 300
<br><br>$$ \Rightarrow $$ a<sub>1</sub> + a<sub>2</sub> + ..... + a<sub>200</sub> = $${{300} \over r}$$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{n = 1}^{200} {{a_n}} $$ = $${{300} \over 2}$$ = 150 | mcq | jee-main-2020-online-9th-january-evening-slot |
rg1izz7VKPhZqiYMlojgy2xukews3ll8 | maths | sequences-and-series | geometric-progression-(g.p) | The sum of the first three terms of a G.P. is S and
their product is 27. Then all such S lie in : | [{"identifier": "A", "content": "[-3, $$\\infty $$)"}, {"identifier": "B", "content": "(-$$ \\propto $$, 9]"}, {"identifier": "C", "content": "(-$$ \\propto $$, -9] $$ \\cup $$ [-3, $$\\infty $$)"}, {"identifier": "D", "content": "(-$$ \\propto $$, -3] $$ \\cup $$ [9, $$\\infty $$)"}] | ["D"] | null | Let three terms of G.P. are $${a \over r}$$, a, ar
<br><br>$$ \therefore $$ $$a\left( {{1 \over r} + 1 + r} \right)$$ = S ...(1)
<br><br>and a<sup>3</sup>
= 27
<br><br>$$ \Rightarrow $$ a = 3
<br><br>$$ \therefore $$ $$3\left( {{1 \over r} + 1 + r} \right)$$ = S
<br><br>$$ \Rightarrow $$ $${{1 \over r} + r = {S \over 3} - 1}$$
<br><br>$$ \Rightarrow $$ As $${{1 \over r} + r \ge 2}$$ or $${{1 \over r} + r \le - 2}$$
<br><br>$$ \therefore $$ $${{S \over 3} - 1 \ge 2}$$ or $${{S \over 3} - 1 \le - 2}$$
<br><br>$$ \Rightarrow $$ $${{S \over 3} \ge 3}$$ or $${{S \over 3} \le - 1}$$
<br><br>$$ \Rightarrow $$ S $$ \ge $$ 9 or S$$ \le $$ -3
<br><br>$$ \therefore $$ S $$ \in $$ (-$$ \propto $$, -3] $$ \cup $$ [9, $$\infty $$) | mcq | jee-main-2020-online-2nd-september-morning-slot |
jvyQgYboQpU3UgCLRYjgy2xukf0znhpx | maths | sequences-and-series | geometric-progression-(g.p) | The value of $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$ is equal to ______. | [] | null | 4 | Given, $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$
<br><br>As sum of GP upto infinity = $${a \over {1 - r}}$$
<br><br>$$ \therefore $$ $${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $$ = $${{{1 \over 3}} \over {1 - {1 \over 3}}}$$ = $${1 \over 2}$$
<br><br>$$ \therefore $$ $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$
<br><br>= $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$$
<br><br>= $${\left( {{{16} \over {100}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$$
<br><br>= $${\left( {{4 \over {10}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$$
<br><br>= $${\left[ {{{\left( {{{10} \over 4}} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$$
<br><br>= $${\left[ {{{\left( {2.5} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$$
<br><br>= $${{{\left( {2.5} \right)}^{ - 2{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}}$$
<br><br>= $${{{\left( {{1 \over 2}} \right)}^{ - 2}}}$$ = 4 | integer | jee-main-2020-online-3rd-september-morning-slot |
uGscaepOHl8oit0YBgjgy2xukfqat3me | maths | sequences-and-series | geometric-progression-(g.p) | If the sum of the second, third and fourth terms
of a positive term G.P. is 3 and the sum of its
sixth, seventh and eighth terms is 243, then the
sum of the first 50 terms of this G.P. is : | [{"identifier": "A", "content": "$${2 \\over {13}}\\left( {{3^{50}} - 1} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over {13}}\\left( {{3^{50}} - 1} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over {26}}\\left( {{3^{49}} - 1} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over {26}}\\left( {{3^{50}} - 1} \\right)$$"}] | ["D"] | null | Let first term = a > 0
<br><br>Common ratio = r > 0
<br><br>ar + ar<sup>2</sup>
+ ar<sup>3</sup>
= 3 ....(i)
<br><br>ar<sup>5</sup>
+ ar<sup>6</sup>
+ ar<sup>7</sup>
= 243 ....(ii)
<br><br>$$ \Rightarrow $$ r<sup>4</sup>(ar + ar<sup>2</sup> + ar<sup>3</sup>) = 243
<br><br>$$ \Rightarrow $$ r<sup>4</sup>(3) = 243
<br><br>$$ \Rightarrow $$ r = 3 as r > 0
<br><br>from (i)
<br><br>3a + 9a + 27a = 3
<br><br>$$ \Rightarrow $$ a = $${1 \over {13}}$$
<br><br>$$ \therefore $$ S<sub>50</sub> = $${{a\left( {{r^{50}} - 1} \right)} \over {\left( {r - 1} \right)}}$$
<br><br>= $${1 \over {26}}\left( {{3^{50}} - 1} \right)$$ | mcq | jee-main-2020-online-5th-september-evening-slot |
eCdigtUaSTPW33VdAAjgy2xukfuv26o5 | maths | sequences-and-series | geometric-progression-(g.p) | Let a , b, c , d and p be any non zero distinct real numbers such that
<br/>(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>)p<sup>2</sup> – 2(ab + bc + cd)p + (b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) = 0. Then : | [{"identifier": "A", "content": "a, c, p are in G.P."}, {"identifier": "B", "content": "a, b, c, d are in G.P."}, {"identifier": "C", "content": "a, b, c, d are in A.P. "}, {"identifier": "D", "content": "a, c, p are in A.P."}] | ["B"] | null | (a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>)p<sup>2</sup> – 2(ab + bc + cd)p + (b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) = 0
<br><br>$$ \Rightarrow $$ (a<sup>2</sup>p<sup>2</sup>
+ 2abp + b<sup>2</sup>
) + (b<sup>2</sup>p<sup>2</sup>
+ 2bcp + c<sup>2</sup>
) + (c<sup>2</sup>
p<sup>2</sup>
+ 2cdp + d<sup>2</sup>) = 0
<br><br>$$ \Rightarrow $$ (ab + b)<sup>2</sup>
+ (bp + c)<sup>2</sup>
+ (cp + d)<sup>2</sup>
= 0
<br><br><b>Note :</b> If sum of two or more positive quantity is zero then they are all zero.
<br><br>$$ \therefore $$ ap + b = 0 and bp + c = 0 and cp + d = 0
<br><br>p = $$ - {b \over a}$$ = $$ - {c \over b}$$ = $$ - {d \over c}$$
<br><br>or $${b \over a}$$ = $${c \over b}$$ = $${d \over c}$$
<br><br>$$ \therefore $$ a, b, c, d are in G.P.
| mcq | jee-main-2020-online-6th-september-morning-slot |
C38geiuYcGLB2P3ZAY1klrmzqlk | maths | sequences-and-series | geometric-progression-(g.p) | The sum of first four terms of a geometric progression (G. P.) is $${{65} \over {12}}$$ and the sum of their respective reciprocals is $${{65} \over {18}}$$. If the product of first three terms of the G.P. is 1, and the third term is $$\alpha$$, then 2$$\alpha$$ is _________. | [] | null | 3 | Let the terms are $$a,ar,a{r^2},a{r^3}$$<br><br>$$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$$ ..........(1)<br><br>$${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$$<br><br>$${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$$ ...............(2)<br><br>Doing $${{(1)} \over {(2)}},$$
<br><br>$${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$$<br><br>Also given, $${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$$<br><br>$${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$$<br><br>$$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$$<br><br>$$2\alpha = 3$$ | integer | jee-main-2021-online-24th-february-evening-slot |
qJBoOi84bbZpOZlOUr1kls5n6t7 | maths | sequences-and-series | geometric-progression-(g.p) | Let A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, ....... be squares such that for each n $$ \ge $$ 1, the length of the side of A<sub>n</sub> equals the length of diagonal of A<sub>n+1</sub>. If the length of A<sub>1</sub> is 12 cm, then the smallest value of n for which area of A<sub>n</sub> is less than one, is __________. | [] | null | 9 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266517/exam_images/rfwj0y3zp4ddisjyhq5w.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Mathematics - Sequences and Series Question 146 English Explanation">
<br><br>$$ \therefore $$ Side lengths are in G.P.<br><br>$${T_n} = {{12} \over {{{\left( {\sqrt 2 } \right)}^{n - 1}}}}$$<br><br>$$ \therefore $$ Area $$ = {{144} \over {{2^{n - 1}} }}$$ < 1<br><br>$$ \Rightarrow {2^{n - 1}} > 144$$<br><br>Smallest n = 9 | integer | jee-main-2021-online-25th-february-morning-slot |
9AKsFRmmitCwyDhceD1kluhfa4b | maths | sequences-and-series | geometric-progression-(g.p) | In an increasing geometric series, the sum of the second and the sixth term is $${{25} \over 2}$$ and the product of the third and fifth term is 25. Then, the sum of 4<sup>th</sup>, 6<sup>th</sup> and 8<sup>th</sup> terms is equal to : | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "26"}, {"identifier": "D", "content": "35"}] | ["D"] | null | a, ar, ar<sup>2</sup>, .....<br><br>$${T_2} + {T_6} = {{25} \over 2} \Rightarrow ar(1 + {r^4}) = {{25} \over 2}$$<br><br>$${a^2}{r^2}{(1 + {r^4})^2} = {{625} \over 4}$$ .... (1)<br><br>$${T_3}.{T_5} = 25 \Rightarrow (a{r^2})(a{r^4}) = 25$$<br><br>$${a^2}{r^6} = 25$$ .....(2)<br><br>On dividing (1) by (2)<br><br>$${{{{(1 + {r^4})}^2}} \over {{r^4}}} = {{25} \over 4}$$<br><br>$$4{r^8} - 14{r^4} + 4 = 0$$<br><br>$$(4{r^4} - 1)({r^4} - 4) = 0$$<br><br>$${r^4} = {1 \over 4},4 \Rightarrow {r^4} = 4$$ (an increasing geometric series)<br><br>$${a^2}{r^6} = 25 \Rightarrow {(a{r^3})^2} = 25$$<br><br>$${T_4} + {T_6} + {T_8} = a{r^3} + a{r^5} + a{r^7}$$<br><br>$$ = a{r^3}(1 + {r^2} + {r^4})$$<br><br>$$ = 5(1 + 2 + 4) = 35$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
y5OVsMxprZsXF3XuT31kmhzkrrq | maths | sequences-and-series | geometric-progression-(g.p) | Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ___________. | [] | null | 3 | A.P. from the set will be 11, 16, 21, 26 .....
<br><br>G.P. from the set will be 4, 8, 16, 32, 64, 128, 256,
512, 1024, 2048, 4096, 8192 .....
<br><br>So common terms are 16, 256, 4096. | integer | jee-main-2021-online-16th-march-morning-shift |
SVaD9jBaTrJAspxagr1kmizc846 | maths | sequences-and-series | geometric-progression-(g.p) | Let $${1 \over {16}}$$, a and b be in G.P. and $${1 \over a}$$, $${1 \over b}$$, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ___________. | [] | null | 14 | $${a^2} = {b \over {16}}$$ and $${2 \over b} = {1 \over a} + 6$$<br><br>Solving, we get $$a = {1 \over {12}}$$ or $$a = - {1 \over 4}$$ [rejected]<br><br>if $$a = {1 \over {12}} \Rightarrow b = {1 \over 9}$$<br><br>$$ \therefore $$ $$72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$$ | integer | jee-main-2021-online-16th-march-evening-shift |
1ktbfqwpt | maths | sequences-and-series | geometric-progression-(g.p) | If the sum of an infinite GP a, ar, ar<sup>2</sup>, ar<sup>3</sup>, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar<sup>2</sup>, ar<sup>4</sup>, ar<sup>6</sup>, ....... is : | [{"identifier": "A", "content": "$${5 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${25 \\over 2}$$"}, {"identifier": "D", "content": "$${9 \\over 2}$$"}] | ["B"] | null | Sum of infinite terms :<br><br>$${a \over {1 - r}} = 15$$ ..... (i)<br><br>Series formed by square of terms :<br><br>a<sup>2</sup>, a<sup>2</sup>r<sup>2</sup>, a<sup>2</sup>r<sup>4</sup>, a<sup>2</sup>r<sup>6</sup> .......<br><br>Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$<br><br>$$ \Rightarrow {a \over {1 - r}}.{a \over {1 + r}} = 150 \Rightarrow 15.{a \over {1 + r}} = 150$$<br><br>$$ \Rightarrow {a \over {1 + r}} = 10$$ ...... (ii)<br><br>by (i) and (ii), a = 12; r = $${1 \over 5}$$<br><br>Now, series : ar<sup>2</sup>, ar<sup>4</sup>, ar<sup>6</sup><br><br>Sum = $${{a{r^2}} \over {1 - {r^2}}} = {{12.\left( {{1 \over {25}}} \right)} \over {1 - {1 \over {25}}}} = {1 \over 2}$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktd3ojty | maths | sequences-and-series | geometric-progression-(g.p) | Let a<sub>1</sub>, a<sub>2</sub>, ......., a<sub>10</sub> be an AP with common difference $$-$$ 3 and b<sub>1</sub>, b<sub>2</sub>, ........., b<sub>10</sub> be a GP with common ratio 2. Let c<sub>k</sub> = a<sub>k</sub> + b<sub>k</sub>, k = 1, 2, ......, 10. If c<sub>2</sub> = 12 and c<sub>3</sub> = 13, then $$\sum\limits_{k = 1}^{10} {{c_k}} $$ is equal to _________. | [] | null | 2021 | $$a_{1}, a_{2}, a_{3}, \ldots, a_{10}$$ are in AP common difference $$=-3$$<br/><br/> $$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$$ are in GP common ratio $$=2$$<br/><br/> Since, $$c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$$<br/><br/>
$$\therefore c_{2} =a_{2}+b_{2}=12$$<br/><br/>
$$ c_{3} =a_{3}+b_{3}=13$$<br/><br/>
Now, $$\mathrm{C}_{3}-\mathrm{C}_{2}=1$$<br/><br/>
$$
\begin{array}{ll}
\Rightarrow & \left(a_{3}-a_{2}\right)+\left(b_{3}-b_{2}\right) \neq 1 \Rightarrow-3+\left(2 b_{2}-b_{2}\right) \neq 1 \\
\Rightarrow & b_{2}=4 \\
\therefore & a_{2}=8
\end{array}
$$<br/><br/>
So, AP is $$11,8,5, \ldots$$.
<br/><br/>
Now, $$\sum_{k=1}^{10} C_{k}=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$$<br/><br/>
$$
\begin{aligned}
&=\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\
&=5(22-27)+2(1023)=2046-25 \\
&=2021
\end{aligned}
$$ | integer | jee-main-2021-online-26th-august-evening-shift |
1ktipg9jo | maths | sequences-and-series | geometric-progression-(g.p) | Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r<sup>2</sup>, then r<sup>2</sup> $$-$$ d is equal to : | [{"identifier": "A", "content": "7 $$-$$ 7$$\\sqrt 3 $$"}, {"identifier": "B", "content": "7 + $$\\sqrt 3 $$"}, {"identifier": "C", "content": "7 $$-$$ $$\\sqrt 3 $$"}, {"identifier": "D", "content": "7 + 3$$\\sqrt 3 $$"}] | ["B"] | null | Let numbers be $${a \over r}$$, a, ar $$\to$$ G.P.<br><br>$${a \over r}$$, 2a, ar $$\to$$ A.P. $$\Rightarrow$$ 4a = $${a \over r}$$ + ar $$\Rightarrow$$ r + $${1 \over r}$$ = 4<br><br>r = 2 $$\pm$$ $$\sqrt 3 $$<br><br>4<sup>th</sup> form of G.P. = 3r<sup>2</sup> $$\Rightarrow$$ ar<sup>2</sup> = 3r<sup>2</sup> $$\Rightarrow$$ a = 3<br><br>r = 2 + $$\sqrt 3 $$, a = 3, d = 2a $$-$$ $${a \over r}$$ = 3$$\sqrt 3 $$<br><br>r<sup>2</sup> $$-$$ d = (2 + $$\sqrt 3 $$)<sup>2</sup> $$-$$ 3$$\sqrt 3 $$<br><br>= 7 + 4$$\sqrt 3 $$ $$-$$ 3$$\sqrt 3 $$<br><br>= 7 + $$\sqrt 3 $$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1l55j37xx | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let for n = 1, 2, ......, 50, S<sub>n</sub> be the sum of the infinite geometric progression whose first term is n<sup>2</sup> and whose common ratio is $${1 \over {{{(n + 1)}^2}}}$$. Then the value of <br/><br/>$${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$ is equal to ___________.</p> | [] | null | 41651 | <p>$${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$$</p>
<p>Now $${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$</p>
<p>$$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\left\{ {({n^2} - n) + 2\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} \right\}} $$</p>
<p>$$ = {1 \over {26}} + {{50 \times 51 \times 101} \over 6} - {{50 \times 51} \over 2} + 2\left( {{1 \over 2} - {1 \over {52}}} \right)$$</p>
<p>$$ = 1 + 25 \times 17(101 - 3)$$</p>
<p>$$ = 41651$$</p> | integer | jee-main-2022-online-28th-june-evening-shift |
1l566am8y | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, ....... be an increasing geometric progression of positive real numbers. If A<sub>1</sub>A<sub>3</sub>A<sub>5</sub>A<sub>7</sub> = $${1 \over {1296}}$$ and A<sub>2</sub> + A<sub>4</sub> = $${7 \over {36}}$$, then the value of A<sub>6</sub> + A<sub>8</sub> + A<sub>10</sub> is equal to</p> | [{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "37"}, {"identifier": "C", "content": "43"}, {"identifier": "D", "content": "47"}] | ["C"] | null | <p>$${{{A_4}} \over {{r^3}}}.\,{{{A_4}} \over r}.\,{A_4}r\,.\,{A_4}{r^3} = {1 \over {1296}}$$</p>
<p>$${A_4} = {1 \over 6}$$</p>
<p>$${A_2} = {7 \over {36}} - {1 \over 6} = {1 \over {36}}$$</p>
<p>So $${A_6} + {A_8} + {A_{10}} = 1 + 6 + 36 = 43$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l58h3pwb | maths | sequences-and-series | geometric-progression-(g.p) | <p>If a<sub>1</sub> (> 0), a<sub>2</sub>, a<sub>3</sub>, a<sub>4</sub>, a<sub>5</sub> are in a G.P., a<sub>2</sub> + a<sub>4</sub> = 2a<sub>3</sub> + 1 and 3a<sub>2</sub> + a<sub>3</sub> = 2a<sub>4</sub>, then a<sub>2</sub> + a<sub>4</sub> + 2a<sub>5</sub> is equal to ___________.</p> | [] | null | 40 | <p>Let G.P. be a<sub>1</sub> = a, a<sub>2</sub> = ar, a<sub>3</sub> = ar<sup>2</sup>, .........</p>
<p>$$\because$$ 3a<sub>2</sub> + a<sub>3</sub> = 2a<sub>4</sub></p>
<p>$$\Rightarrow$$ 3ar + ar<sup>2</sup> = 2ar<sup>3</sup></p>
<p>$$\Rightarrow$$ 2ar<sup>2</sup> $$-$$ r $$-$$ 3 = 0</p>
<p>$$\therefore$$ r = $$-$$1 or $${3 \over 2}$$</p>
<p>$$\because$$ a<sub>1</sub> = a > 0 then r $$\ne$$ $$-$$1</p>
<p>Now, a<sub>2</sub> + a<sub>4</sub> = 2a<sub>3</sub> + 1</p>
<p>ar + ar<sup>3</sup> = 2ar<sup>2</sup> + 1</p>
<p>$$a\left( {{3 \over 2} + {{27} \over 8} - {9 \over 2}} \right) = 1$$</p>
<p>$$\therefore$$ a = $${8 \over 3}$$</p>
<p>$$\therefore$$ a<sub>2</sub> + a<sub>4</sub> + 2a<sub>5</sub> = a(r + r<sup>3</sup> + 2r<sup>4</sup>)</p>
<p>$$ = {8 \over 3}\left( {{3 \over 2} + {{27} \over 8} + {{81} \over 8}} \right) = 40$$</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1l6kjt9ea | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be $$\frac{98}{25}$$. Then the sum of the first 21 terms of an AP, whose first term is $$10\mathrm{a r}, \mathrm{n}^{\text {th }}$$ term is $$\mathrm{a}_{\mathrm{n}}$$ and the common difference is $$10 \mathrm{ar}^{2}$$, is equal to :</p> | [{"identifier": "A", "content": "$$21 \\,\\mathrm{a}_{11}$$"}, {"identifier": "B", "content": "$$22 \\,\\mathrm{a}_{11}$$"}, {"identifier": "C", "content": "$$15 \\,\\mathrm{a}_{16}$$"}, {"identifier": "D", "content": "$$14 \\,\\mathrm{a}_{16}$$"}] | ["A"] | null | <p>Let first term of G.P. be a and common ratio is r</p>
<p>Then, $${a \over {1 - r}} = 5$$ ...... (i)</p>
<p>$$a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}$$</p>
<p>$$\therefore$$ $${r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3 \over 5}}}$$</p>
<p>$$\therefore$$ Then, $${S_{21}} = {{21} \over 2}\left[ {2 \times 10ar + 20 \times 10a{r^2}} \right]$$</p>
<p>$$ = 21\left[ {10ar + 10\,.\,10a{r^2}} \right]$$</p>
<p>$$ = 21\,{a_{11}}$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1ldprmen7 | maths | sequences-and-series | geometric-progression-(g.p) | <p>If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}] | ["A"] | null | $\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}, \mathrm{ar}^{3}(\mathrm{a}, \mathrm{r}>0)$
<br/><br/>$a^{4} r^{6}=1296$
<br/><br/>$a^{2} r^{3}=36$
<br/><br/>$a=\frac{6}{r^{3 / 2}}$
<br/><br/>$a+a r+a r^{2}+a r^{3}=126$
<br/><br/>$\frac{1}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^{2}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^{3}}{\mathrm{r}^{3 / 2}}=\frac{126}{6}=21$
<br/><br/>$\left(r^{-3 / 2}+r^{3 / 2}\right)+\left(r^{1 / 2}+r^{-1 / 2}\right)=21$
<br/><br/>$\mathrm{r}^{1 / 2}+\mathrm{r}^{-1 / 2}=\mathrm{A}$
<br/><br/>$\mathrm{r}^{-3 / 2}+\mathrm{r}^{3 / 2}+3 \mathrm{~A}=\mathrm{A}^{3}$
<br/><br/>$\mathrm{A}^{3}-3 \mathrm{~A}+\mathrm{A}=21$
<br/><br/>$\mathrm{A}^{3}-2 \mathrm{~A}=21$
<br/><br/>$A=3$
<br/><br/>$\sqrt{\mathrm{r}}+\frac{1}{\sqrt{\mathrm{r}}}=3$
<br/><br/>$\mathrm{r}+1=3 \sqrt{\mathrm{r}}$
<br/><br/>$r^{2}+2 r+1=9 r$
<br/><br/>$r^{2}-7 r+1=0$
<br/><br/>$\Rightarrow r_{1}+r_{2}=7$ | mcq | jee-main-2023-online-31st-january-morning-shift |
ldqw0prh | maths | sequences-and-series | geometric-progression-(g.p) | Let $a, b, c>1, a^3, b^3$ and $c^3$ be in A.P., and $\log _a b, \log _c a$ and $\log _b c$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\frac{a+4 b+c}{3}$ and the common difference is $\frac{a-8 b+c}{10}$ is $-444$, then $a b c$ is equal to : | [{"identifier": "A", "content": "343"}, {"identifier": "B", "content": "216"}, {"identifier": "C", "content": "$\\frac{343}{8}$"}, {"identifier": "D", "content": "$\\frac{125}{8}$"}] | ["B"] | null | <p>$$2{b^3} = {a^3} + {c^3}$$</p>
<p>$${\left( {{{\log a} \over {\log c}}} \right)^2} = \left( {{{\log b} \over {\log a}}} \right)\left( {{{\log c} \over {\log b}}} \right)$$</p>
<p>$$ \Rightarrow {(\log a)^3} = {(\log c)^3}$$</p>
<p>$$ \Rightarrow \log a = \log c$$</p>
<p>$$ \Rightarrow a = c$$</p>
<p>$$ \Rightarrow a = b = c$$</p>
<p>$${T_1} = 2a,d = - {{3a} \over 5}$$</p>
<p>$${S_{20}} = - 444$$</p>
<p>$$ \Rightarrow {{20} \over 2}\left( {2(2a) + (19)\left( { - {{3a} \over 5}} \right)} \right) = - 444$$</p>
<p>$$ \Rightarrow 10{{(20a - 57a)} \over 5} = - 444$$</p>
<p>$$ \Rightarrow 37a = 222$$</p>
<p>$$ \Rightarrow a = 6$$</p>
<p>$$ \Rightarrow abc = {(6)^3} = 216$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldsgdvt5 | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$\{ {a_k}\} $$ and $$\{ {b_k}\} ,k \in N$$, be two G.P.s with common ratios $${r_1}$$ and $${r_2}$$ respectively such that $${a_1} = {b_1} = 4$$ and $${r_1} < {r_2}$$. Let $${c_k} = {a_k} + {b_k},k \in N$$. If $${c_2} = 5$$ and $${c_3} = {{13} \over 4}$$ then $$\sum\limits_{k = 1}^\infty {{c_k} - (12{a_6} + 8{b_4})} $$ is equal to __________.</p> | [] | null | 9 | <p>$$\{ {a_k}\} $$ be a G.P. with $${a_1} = 4,r = {r_1}$$</p>
<p>And</p>
<p>$$\{ {b_k}\} $$ be G.P. with $${b_1} = 4,r = {r_2}$$ $$({r_1} < {r_2})$$</p>
<p>Now</p>
<p>$${C_k} = {a_k} + {b_k}$$</p>
<p>$${c_1} = 4 + 4 = 8$$ and $${c_2} = 5$$<p>
<p>$${a_2} + {b_2} = 5$$</p>
<p>$$\therefore$$ $${r_1} + {r_2} = {5 \over 4}$$</p>
<p>and $${c_3} = {{13} \over 4} \Rightarrow r_4^2 + r_2^2 = {{13} \over {16}}$$</p>
<p>$$\therefore$$ $${{25} \over {16}} - 2{r_1}{r_2} = {{13} \over {16}} \Rightarrow 2{r_1}{r_2} = {3 \over 4}$$</p>
<p>$$\therefore$$ $${r_2} - {r_1} = \sqrt {{{25} \over {16}} - {3 \over 2}} = {1 \over 4}$$</p>
<p>$$\therefore$$ $${r_2} = {3 \over 4},{r_1} = {1 \over 2}$$</p>
<p>$$\therefore$$ $${a_6} = 4 \times {1 \over {{2^5}}} = {1 \over 8},{b_4} = 4 \times {{27} \over {64}} = {{27} \over {16}}$$</p>
<p>and $$\sum\limits_{K = 1}^\infty {{C_K} = 4\left[ {{1 \over {1 - {1 \over 2}}} + {1 \over {1 - {3 \over 4}}}} \right] = 24} $$</p>
<p>$$\therefore$$ $$\sum\limits_{K = 1}^\infty {{C_K} - (12{a_6} + 8{b_4}) = 09} $$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldswm61u | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$a_1,a_2,a_3,...$$ be a $$GP$$ of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $$a_1a_9+a_2a_4a_9+a_5+a_7$$ is equal to __________.</p> | [] | null | 60 | Let $r$ be the common ratio of the G.P
<br/><br/>
$\therefore a_{1} r^{3} \times a_{1} r^{5}=9$
<br/><br/>
$a_{1}^{2} r^{8}=9 \Rightarrow a_{1} r^{4}=3$
<br/><br/>
And
<br/><br/>
$$
\begin{aligned}
& a_{1}\left(r^{4}+r^{6}\right)=24 \\\\
\Rightarrow & 3\left(1+r^{2}\right)=24 \\\\
\therefore & r^{2}=7 \text { and } a_{1}=\frac{3}{49}
\end{aligned}
$$
<br/><br/>
Now
<br/><br/>
$$
\begin{aligned}
& a_{1} a_{9}+a_{2} a_{4} a_{9}+a_{5}+a_{7} \\\\
& =a_{1}^{2} r^{8}+a_{1}^{3} r^{12}+24 \\\\
& =24+\frac{9}{7^{4}} \times 7^{4}+\frac{27}{7^{6}} \cdot 7^{6}=60
\end{aligned}
$$ | integer | jee-main-2023-online-29th-january-morning-shift |
1ldu65bcm | maths | sequences-and-series | geometric-progression-(g.p) | <p>For the two positive numbers $$a,b,$$ if $$a,b$$ and $$\frac{1}{18}$$ are in a geometric progression, while $$\frac{1}{a},10$$ and $$\frac{1}{b}$$ are in an arithmetic progression, then $$16a+12b$$ is equal to _________.</p> | [] | null | 3 | $$
\begin{aligned}
& \mathrm{a}, \mathrm{b}, \frac{1}{18} \rightarrow \mathrm{GP} \\\\
& \frac{\mathrm{a}}{18}=\mathrm{b}^2\quad...(i) \\\\
& \frac{1}{\mathrm{a}}, 10, \frac{1}{\mathrm{~b}} \rightarrow \mathrm{AP} \\\\
& \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=20 \\\\
& \Rightarrow \mathrm{a}+\mathrm{b}=20 \mathrm{ab}, \text { from eq. (i) } ; \text { we get } \\\\
& \Rightarrow 18 \mathrm{~b}^2+\mathrm{b}=360 \mathrm{~b}^3 \\\\
& \Rightarrow 360 \mathrm{~b}^2-18 \mathrm{~b}-1=0 \quad\{\because \mathrm{b} \neq 0\} \\\\
& \Rightarrow \mathrm{b}=\frac{18 \pm \sqrt{324+1440}}{720} \\\\
& \Rightarrow b =\frac{18+\sqrt{1764}}{720} \quad\{\because \mathrm{b}>0\} \\\\
& \Rightarrow b = \frac{1}{12} \\\\
& \Rightarrow \mathrm{a}=18 \times \frac{1}{144}=\frac{1}{8} \\\\
& \text{Now},16\mathrm{a}+12 \mathrm{~b}=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-evening-shift |
1ldyc0ddl | maths | sequences-and-series | geometric-progression-(g.p) | <p>The 4$$^\mathrm{th}$$ term of GP is 500 and its common ratio is $$\frac{1}{m},m\in\mathbb{N}$$. Let $$\mathrm{S_n}$$ denote the sum of the first n terms of this GP. If $$\mathrm{S_6 > S_5 + 1}$$ and $$\mathrm{S_7 < S_6 + \frac{1}{2}}$$, then the number of possible values of m is ___________</p> | [] | null | 12 | $T_{4}=500$
<br/><br/>
$$
a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}}
$$
<br/><br/>
Now,
<br/><br/>
$$
\begin{aligned}
& S_{6} > S_{5}+1 \\\\
& \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\
& a r^{5} > 1 \\\\
& \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}} \\\\
& \Rightarrow \quad m^{2} < 500 \\\\
& \because m > 0 \Rightarrow m \in(0,10 \sqrt{5}) \\\\
& \quad S_{7} < S_{6}+\frac{1}{2} \\\\
& \quad \frac{a\left(1-r^{6}\right)}{1-r}<\frac{a\left(1-r^{6}\right)}{1-r}+\frac{1}{2} \\\\
& \quad a r^{6}<\frac{1}{2}
\end{aligned}
$$
<br/><br/>
$\because r=\frac{1}{m}$ and $a=\frac{500}{r^{5}}$
<br/><br/>
$$
\frac{1}{m^{3}}<\frac{1}{1000}
$$
<br/><br/>
$\Rightarrow m \in(10, \infty)$
<br/><br/>
Possible values of $m$ is $\{11,12,....22 \}$
<br/><br/>
$\because m \in N$
<br/><br/>
Total 12 values | integer | jee-main-2023-online-24th-january-morning-shift |
1lgow1w1l | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let a$$_1$$, a$$_2$$, a$$_3$$, .... be a G.P. of increasing positive numbers. Let the sum of its 6<sup>th</sup> and 8<sup>th</sup> terms be 2 and the product of its 3<sup>rd</sup> and 5<sup>th</sup> terms be $$\frac{1}{9}$$. Then $$6(a_2+a_4)(a_4+a_6)$$ is equal to</p> | [{"identifier": "A", "content": "2$$\\sqrt2$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3$$\\sqrt3$$"}, {"identifier": "D", "content": "3"}] | ["D"] | null |
<p>Given the conditions :</p>
<ol>
<li>$a_6 + a_8 = 2 \Rightarrow a r^5 + a r^7 = 2$</li>
<li>$a_3 \cdot a_5 = \frac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4 = \frac{1}{9} \Rightarrow a r^3 = \frac{1}{3}$</li>
</ol>
<p>From this, we can form the equation $\frac{r^2}{3} + \frac{r^4}{3} = 2$, which simplifies to $r^4 + r^2 = 6$.</p>
<p>This can be factored to give $\left(r^2 - 2\right)\left(r^2 + 3\right) = 0$, yielding $r^2 = 2$ (since $r^2$ cannot be $-3$ for real $r$).</p>
<p>So, we have $r = \sqrt{2}$.</p>
<p>Substituting $r = \sqrt{2}$ into the equation $a r = \frac{1}{6}$, we get $a = \frac{1}{6\sqrt{2}}$.</p>
<p>Now, we find the value of $6(a_2+a_4)(a_4+a_6)$:</p>
<p>$6(a_2+a_4)(a_4+a_6) = 6\left(a r + a r^3\right)\left(a r^3 + a r^5\right)$</p>
<p>$= 6\left(\frac{1}{6\sqrt{2}} + \frac{1}{3\sqrt{2}}\right)\left(\frac{1}{3\sqrt{2}} + \frac{2}{3\sqrt{2}}\right)$</p>
<p>$= 6 \cdot \frac{1}{2} \cdot 1 = 3$.</p>
| mcq | jee-main-2023-online-13th-april-evening-shift |
1lgxw30j8 | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let the first term $$\alpha$$ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to</p> | [{"identifier": "A", "content": "241"}, {"identifier": "B", "content": "231"}, {"identifier": "C", "content": "220"}, {"identifier": "D", "content": "210"}] | ["B"] | null | Given that the first term $a$ and common ratio $r$ of a geometric progression be positive integer. So, their 1st three terms are $a, a r, a r^2$
<br/><br/>According to the question, $a^2+a^2 r^2+a^2 r^4=33033$
<br/><br/>$$
\begin{aligned}
\Rightarrow a^2\left(1+r^2+r^4\right) & =3 \times 7 \times 11 \times 11 \times 13 \\
& =3 \times 7 \times 13 \times 11^2
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \quad a^2=11^2 \\\\
& \Rightarrow \quad a=11 \\\\
& \text { and } 1+r^2+r^4=273 \\\\
& \Rightarrow r^2+r^4=272 \\\\
& \Rightarrow r^4+r^2-272=0 \\\\
& \Rightarrow\left(r^2+17\right)\left(r^2-16\right)=0 \\\\
& \Rightarrow r^2=-17(not ~possible),\\\\
& \Rightarrow r^2-16=0 \\\\
& \text { } \Rightarrow r= \pm 4 \\\\
& \Rightarrow r=4 ~~( \because r > 0)
\end{aligned}
$$
<br/><br/>So, sum of these first three terms is $a+a r+a r^2$ $=11+44+176=231$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lh2z2v5w | maths | sequences-and-series | geometric-progression-(g.p) | <p>If
<br/><br/>$$(20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}=k(20)^{19}$$,
<br/><br/>then $$k$$ is equal to ___________.</p> | [] | null | 400 | $\begin{aligned} &(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17} \\ & \quad+\ldots \ldots+20(21)^{19}=k(20)^{19} \\\\ & \Rightarrow(20)^{19}\left[1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right]=k(20)^{19} \\\\ & \Rightarrow k=1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19} ..........(i)\end{aligned}$
<br/><br/>Now,
<br/><br/>$\begin{aligned} k\left(\frac{21}{20}\right)=\left(\frac{21}{20}\right) & +2\left(\frac{21}{20}\right)^2+3\left(\frac{21}{20}\right)^3 +\ldots+20\left(\frac{21}{20}\right)^{20}\end{aligned}$ ........(ii)
<br/><br/>On subtracting Equation (ii) from Equation (i), we get
<br/><br/>$$
\begin{aligned}
& k\left(\frac{-1}{20}\right)=1+\frac{21}{20}+\left(\frac{21}{20}\right)^2+\ldots \ldots+\left(\frac{21}{20}\right)^{19}-20\left(\frac{21}{20}\right)^{20} \\\\
& \Rightarrow k\left(\frac{-1}{20}\right)=\frac{1\left(\left(\frac{21}{20}\right)^{20}-1\right)}{\frac{21}{20}-1}-20\left(\frac{21}{20}\right)^{20} \\\\
& \Rightarrow k\left(\frac{-1}{20}\right)=20\left(\left(\frac{21}{20}\right)^{20}-1\right)-20\left(\frac{21}{20}\right)^{20} \\\\
& =20\left(\frac{21}{20}\right)^{20}-20-20\left(\frac{21}{20}\right)^{20} \\\\
& \Rightarrow k\left(\frac{-1}{20}\right)=-20 \\\\
& \Rightarrow k=400
\end{aligned}
$$ | integer | jee-main-2023-online-6th-april-evening-shift |
lsan9avi | maths | sequences-and-series | geometric-progression-(g.p) | If three successive terms of a G.P. with common ratio $\mathrm{r}(\mathrm{r}>1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to _____________. | [] | null | 1 | <p>To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $ar^2$, where $a$ is the first term and $r > 1$. These three terms represent the lengths of the sides of a triangle.</p>
<p>According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:</p>
<p>$$
1) \ a + ar > ar^2 \\\\
$$
<br/><br/>$$2) \ a + ar^2 > ar $$
<br/><br/>$$
3) \ ar + ar^2 > a
$$</p>
<p>Given that $r > 1$, inequalities 2 and 3 will always hold because:</p>
<p>$$
ar < ar^2 \ \text{and} \ a < ar,
$$</p>
<p>indicating that both $a + ar^2$ and $ar + ar^2$ will be greater than $ar$ and $a$ respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:</p>
<p>$$
a + ar > ar^2
$$</p>
<p>Simplifying this, we get:</p>
<p>$$
a(1 + r) > a r^2
$$</p>
<p>Since $a$ is positive (as it represents the length of a side of a triangle), we can divide both sides of the inequality by $a$ without changing the direction of the inequality:</p>
<p>$$
1 + r > r^2
$$</p>
<p>We can then subtract $r$ from both sides:</p>
<p>$$
1 > r^2 - r
$$</p>
<p>Simplifying the right side by factoring $r$:</p>
<p>$$
1 > r(r - 1)
$$</p>
<p>Given that $r > 1$, the quantity $(r - 1)$ is positive; hence, $r(r - 1)$ is also positive. This means the actual value for $r$ to satisfy the inequality is within the interval $(1, \sqrt{2})$ because $r(r - 1)$ increases with increasing $r$, and it would be 1 when $r = \sqrt{2}$. It should be greater than 1, and less than $\sqrt{2}$ such that $r^2 - r$ stays below 1.</p>
<p>Now let’s consider the expressions $[r]$ and $[-r]$. The symbol $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function).</p>
<p>Since $1 < r < \sqrt{2}$, $[r] = 1$, because 1 is the greatest integer less than $r$ within that interval.</p>
<p>For $[-r]$, we need the greatest integer less than or equal to $-r$. Since $-r$ is negative and less than $-1$ (because $r > 1$), $[-r] = -2$, as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\sqrt{2}$ and $-1$).</p>
<p>Now we can substitute these values into the expression:</p>
<p>$$
3[r] + [-r] = 3 \cdot 1 + (-2) = 3 - 2 = 1
$$</p>
<p>Therefore $3[r] + [-r]$ is equal to 1.</p>
| integer | jee-main-2024-online-1st-february-evening-shift |
jaoe38c1lsd4o6jt | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$2^{\text {nd }}, 8^{\text {th }}$$ and $$44^{\text {th }}$$ terms of a non-constant A. P. be respectively the $$1^{\text {st }}, 2^{\text {nd }}$$ and $$3^{\text {rd }}$$ terms of a G. P. If the first term of the A. P. is 1, then the sum of its first 20 terms is equal to -</p> | [{"identifier": "A", "content": "990"}, {"identifier": "B", "content": "980"}, {"identifier": "C", "content": "960"}, {"identifier": "D", "content": "970"}] | ["D"] | null | <p>$$\begin{aligned}
& 1+d, \quad 1+7 d, 1+43 d \text { are in GP } \\
& (1+7 d)^2=(1+d)(1+43 d) \\
& 1+49 d^2+14 d=1+44 d+43 d^2 \\
& 6 d^2-30 d=0 \\
& d=5 \\
& S_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] \\
& \quad=10[2+95] \\
& \quad=970
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lseyfkoa | maths | sequences-and-series | geometric-progression-(g.p) | <p>If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["B"] | null | <p>$$\begin{aligned}
& a+a r+a r^2+a r^3+\ldots .+a r^{63} \\
& =7\left(a+a r^2+a r^4 \ldots .+a r^{62}\right) \\
& \Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2} \\
& r=6
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfktybb | maths | sequences-and-series | geometric-progression-(g.p) | <p>If each term of a geometric progression $$a_1, a_2, a_3, \ldots$$ with $$a_1=\frac{1}{8}$$ and $$a_2 \neq a_1$$, is the arithmetic mean of the next two terms and $$S_n=a_1+a_2+\ldots . .+a_n$$, then $$S_{20}-S_{18}$$ is equal to</p> | [{"identifier": "A", "content": "$$-2^{15}$$\n"}, {"identifier": "B", "content": "$$2^{15}$$\n"}, {"identifier": "C", "content": "$$-2^{18}$$\n"}, {"identifier": "D", "content": "$$2^{18}$$"}] | ["A"] | null | <p>Let $$r^{\prime}$$th term of the GP be $$a^{n-1}$$. Given,</p>
<p>$$\begin{aligned}
& 2 a_r=a_{r+1}+a_{r+2} \\
& 2 a r^{n-1}=a r^n+a r^{n+1} \\
& \frac{2}{r}=1+r \\
& r^2+r-2=0
\end{aligned}$$</p>
<p>Hence, we get, $$r=-2$$ (as $$r \neq 1$$)</p>
<p>So, $$\mathrm{S}_{20}-\mathrm{S}_{18}=$$ (Sum upto 20 terms) $$-$$ (Sum upto 18 terms) $$=\mathrm{T}_{19}+\mathrm{T}_{20}$$</p>
<p>$$\mathrm{~T}_{19}+\mathrm{T}_{20}=\mathrm{ar}^{18}(1+\mathrm{r})$$</p>
<p>Putting the values $$\mathrm{a}=\frac{1}{8}$$ and $$\mathrm{r}=-2$$;</p>
<p>we get $$T_{19}+T_{20}=-2^{15}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg3p684 | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$a$$ and $$b$$ be be two distinct positive real numbers. Let $$11^{\text {th }}$$ term of a GP, whose first term is $$a$$ and third term is $$b$$, is equal to $$p^{\text {th }}$$ term of another GP, whose first term is $$a$$ and fifth term is $$b$$. Then $$p$$ is equal to</p> | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "25"}] | ["C"] | null | <p>The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.</p><p>For the first GP, with first term $$a$$ and third term $$b$$:</p><ul><li>The third term is given by $$t_3 = ar^2 = b$$, leading to $$r^2 = \frac{b}{a}$$.</li><li>The eleventh term is $$t_{11} = ar^{10} = a\left(\frac{b}{a}\right)^5$$.</li></ul><p>For the second GP, with first term $$a$$ and fifth term $$b$$:</p><ul><li>The fifth term is $$T_5 = ar^4 = b$$, yielding $$r^4 = \frac{b}{a}$$ and $$r = \left(\frac{b}{a}\right)^{1/4}$$.</li><li>The $$p^{\text{th}}$$ term is thus $$T_p = ar^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$.</li></ul><p>Equating the eleventh term of the first GP to the $$p^{\text{th}}$$ term of the second GP gives: $$a\left(\frac{b}{a}\right)^5 = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$.</p><p>Simplifying, we find that $$5 = \frac{p-1}{4}$$, leading to $$p = 21$$, which is the solution.</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
luxwd2ht | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$a, a r, a r^2$$, ............ be an infinite G.P. If $$\sum_\limits{n=0}^{\infty} a r^n=57$$ and $$\sum_\limits{n=0}^{\infty} a^3 r^{3 n}=9747$$, then $$a+18 r$$ is equal to</p> | [{"identifier": "A", "content": "27"}, {"identifier": "B", "content": "38"}, {"identifier": "C", "content": "31"}, {"identifier": "D", "content": "46"}] | ["C"] | null | <p>$$\begin{array}{ll}
\sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\
\sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)}
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\
& \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\
& \Rightarrow \quad 18 r^2-39 r+18=0 \\
& \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\
& \therefore \quad a=19 \\
& \quad a+18 r \\
& \quad=19+12=31
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
lv0vxcyd | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let the first three terms 2, p and q, with $$q \neq 2$$, of a G.P. be respectively the $$7^{\text {th }}, 8^{\text {th }}$$ and $$13^{\text {th }}$$ terms of an A.P. If the $$5^{\text {th }}$$ term of the G.P. is the $$n^{\text {th }}$$ term of the A.P., then $n$ is equal to:</p> | [{"identifier": "A", "content": "151"}, {"identifier": "B", "content": "177"}, {"identifier": "C", "content": "163"}, {"identifier": "D", "content": "169"}] | ["C"] | null | <p>$$\begin{aligned}
& \text { Let } p=2 r, q=2 r^2 \\
& T_7=2, T_8=2 r, T_{13}=2 r^2 \\
& d=2 r-2=2(r-1) \\
& 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\
& \Rightarrow r^2-6 r+5=0 \\
& \Rightarrow(r-1)(r-5)=0 \\
& \therefore r=1,5 \\
& r=1 \text { (rejected) as } q \neq 2 \\
& \therefore r=5
\end{aligned}$$</p>
<p>$$5^{\text {th }}$$ term of G.P $$=2 . r^4=2.5^4$$</p>
<p>Let $$1^{\text {st }}$$ term of A.P $$b a=a, d=8$$</p>
<p>$$2=a+(6)(8) \Rightarrow a=-46$$</p>
<p>$$\mathrm{n}^{\text {th }}$$ term of A.P $$=-46+(n-1) 8=8 n-54$$</p>
<p>$$\begin{aligned}
& 2.5^4=8 n-54 \\
& \Rightarrow 1250+54=8 n \\
& \Rightarrow n=\frac{1304}{8}=163
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv3veby6 | maths | sequences-and-series | geometric-progression-(g.p) | <p>In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $$\frac{70}{3}$$ and the product of the third and fifth terms is 49. Then the sum of the $$4^{\text {th }}, 6^{\text {th }}$$ and $$8^{\text {th }}$$ terms is equal to:</p> | [{"identifier": "A", "content": "78"}, {"identifier": "B", "content": "96"}, {"identifier": "C", "content": "91"}, {"identifier": "D", "content": "84"}] | ["C"] | null | <p>Let's denote the first term of the geometric progression by $$a$$ and the common ratio by $$r$$. The terms of the geometric progression can be written as follows:</p>
<p>First term: $$a$$</p>
<p>Second term: $$ar$$</p>
<p>Third term: $$(ar^2)$$</p>
<p>Fourth term: $$(ar^3)$$</p>
<p>Fifth term: $$(ar^4)$$</p>
<p>Sixth term: $$(ar^5)$$</p>
<p>Eighth term: $$(ar^7)$$</p>
<p>We are given two key pieces of information:</p>
<p>1. The sum of the second and sixth terms is $$\frac{70}{3}$$:</p>
<p>$$ar + ar^5 = \frac{70}{3}$$</p>
<p>2. The product of the third and fifth terms is 49:</p>
<p>$$(ar^2) \cdot (ar^4) = 49$$</p>
<p>$$a^2 r^6 = 49$$</p>
<p>$$a^2 = \frac{49}{r^6}$$</p>
<p>$$a = \frac{7}{r^3}$$</p>
<p>Substituting $$a = \frac{7}{r^3}$$ into the first equation:</p>
<p>$$\frac{7}{r^3} \cdot r + \frac{7}{r^3} \cdot r^5 = \frac{70}{3}$$</p>
<p>$$\frac{7r}{r^3} + \frac{7r^5}{r^3} = \frac{70}{3}$$</p>
<p>$$\frac{7}{r^2} + \frac{7r^2}{1} = \frac{70}{3}$$</p>
<p>Let $$x = r^2$$. Then:</p>
<p>$$\frac{7}{x} + 7x = \frac{70}{3}$$</p>
<p>Multiply through by 3x to clear the denominator:</p>
<p>$$21 + 21x^2 = 70x$$</p>
<p>Rearrange into a standard quadratic equation:</p>
<p>$$21x^2 - 70x + 21 = 0$$</p>
<p>Divide by 7 to simplify:</p>
<p>$$3x^2 - 10x + 3 = 0$$</p>
<p>Solve this quadratic equation using the quadratic formula:</p>
<p>$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$</p>
<p>Where $$a = 3$$, $$b = -10$$, and $$c = 3$$. Thus:</p>
<p>$$x = \frac{10 \pm \sqrt{100 - 36}}{6}$$</p>
<p>$$x = \frac{10 \pm \sqrt{64}}{6}$$</p>
<p>$$x = \frac{10 \pm 8}{6}$$</p>
<p>$$x = 3$$ or $$x = \frac{1}{3}$$</p>
<p>Since $$x = r^2$$ and $$r$$ is positive, we get $$r = \sqrt{3}$$ or $$r = \frac{1}{\sqrt{3}}$$. We need to choose the value that results in positive, increasing terms:</p>
<p>If $$r = \sqrt{3}$$:</p>
<p>$$a = \frac{7}{r^3} = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7}{3} \cdot \frac{1}{\sqrt{3}} = \frac{7\sqrt{3}}{9}$$</p>
<p>Now we can determine the sum of the 4th, 6th, and 8th terms:</p>
<p>The 4th term is: $$ar^3 = \frac{7\sqrt{3}}{9} \cdot 3\sqrt{3} = 7$$</p>
<p>The 6th term is: $$ar^5 = \frac{7\sqrt{3}}{9} \cdot 9\sqrt{3} = 21$$</p>
<p>The 8th term is: $$ar^7 = \frac{7\sqrt{3}}{9} \cdot 27(\sqrt{3}) = 49$$</p>
<p>Adding these together:</p>
<p>$$(4th + 6th + 8th terms) = 7 + 21 + 63 = 91$$</p>
<p>Therefore, the sum of the 4th, 6th, and 8th terms is 91.</p>
<p>Correct answer:</p>
<p><strong>Option C: 91</strong></p>
<p></p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lvb294nj | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$A B C$$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $$A B C$$ and the same process is repeated infinitely many times. If $$\mathrm{P}$$ is the sum of perimeters and $$Q$$ is be the sum of areas of all the triangles formed in this process, then :</p> | [{"identifier": "A", "content": "$$\\mathrm{P}^2=72 \\sqrt{3} \\mathrm{Q}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{P}^2=36 \\sqrt{3} \\mathrm{Q}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{P}=36 \\sqrt{3} \\mathrm{Q}^2$$\n"}, {"identifier": "D", "content": "$$\\mathrm{P}^2=6 \\sqrt{3} \\mathrm{Q}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwac04ho/c0187de2-d2dd-45ec-8dc9-5238b7084f24/51988fc0-141b-11ef-aad1-15919f5484e3/file-1lwac04hp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwac04ho/c0187de2-d2dd-45ec-8dc9-5238b7084f24/51988fc0-141b-11ef-aad1-15919f5484e3/file-1lwac04hp.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 4 English Explanation"></p>
<p>$$\triangle A B C$$ is an equilateral triangle having side $$=a$$ unit</p>
<p>Now, perimeter $$=$$ sum of all sides $$=3 a$$</p>
<p>Area $$=\frac{\sqrt{3}}{4} a^2$$</p>
<p>Now, $$\ln \triangle D E F, D E=\frac{a}{2}=E F=D F$$</p>
<p>$$\begin{aligned}
& \text { Perimeter }=3 \times \frac{a}{2}=\frac{3 a}{2} \\
& \text { Area }=\frac{\sqrt{3}}{4} \times\left(\frac{a}{2}\right)^2=\frac{\sqrt{3} a^2}{16}
\end{aligned}$$</p>
<p>Now, $$P=3 a+\frac{3 a}{2}+\frac{3 a}{4}+\cdots$$</p>
<p>$$Q=\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{16} a^2+\frac{\sqrt{3}}{64} a^2+\cdots$$</p>
<p>$$P=\frac{3 a}{1-\frac{1}{2}}=3 a \times 2 \Rightarrow P=6 a \quad \text{.... (i)}$$</p>
<p>$$Q=\frac{\frac{\sqrt{3}}{4} a^2}{1-\frac{1}{4}}=\frac{4}{3} \times \frac{\sqrt{3}}{4} a^2 \quad Q=\frac{\sqrt{3}}{3} a^2 \quad \text{.... (ii)}$$</p>
<p>From equation (i) & (ii)</p>
<p>$$\begin{aligned}
& P=6 a \\
& Q=\frac{\sqrt{3}}{3} a^2 \\
& Q=\frac{\sqrt{3}}{3} \times \frac{P^2}{36} \\
& P^2=36 \sqrt{3} Q
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
UKf6HoT18gPVSASW | maths | sequences-and-series | harmonic-progression-(h.p) | If $$x = \sum\limits_{n = 0}^\infty {{a^n},\,\,y = \sum\limits_{n = 0}^\infty {{b^n},\,\,z = \sum\limits_{n = 0}^\infty {{c^n},} } } \,\,$$ where a, b, c are in A.P and $$\,\left| a \right| < 1,\,\left| b \right| < 1,\,\left| c \right| < 1$$ then x, y, z are in | [{"identifier": "A", "content": "G.P."}, {"identifier": "B", "content": "A.P."}, {"identifier": "C", "content": "Arithmetic-Geometric Progression"}, {"identifier": "D", "content": "H.P."}] | ["D"] | null | $$x = \sum\limits_{n = 0}^\infty {{a^n}} = {1 \over {1 - a}}\,\,\,\,\,\,\,\,\,\,a = 1 - {1 \over x}$$
<br><br>$$y = \sum\limits_{n = 0}^\infty {{b^n}} = {1 \over {1 - b}}\,\,\,\,\,\,\,\,\,\,b = 1 - {1 \over y}$$
<br><br>$$z = \sum\limits_{n = 0}^\infty {{c^n}} = {1 \over {1 - c}}\,\,\,\,\,\,\,\,\,\,c = 1 - {1 \over z}$$
<br><br>$$a,b,c$$ are in $$A.P.$$ OR $$2b = a + c$$
<br><br>$$2\left( {1 - {1 \over y}} \right) = 1 - {1 \over x} + 1 - {1 \over y}$$
<br><br>$${2 \over y} = {1 \over x} + {1 \over z} \Rightarrow x,y,z$$ are in $$H.P.$$ | mcq | aieee-2005 |
VdoDcdBuFZIwxtlr | maths | sequences-and-series | harmonic-progression-(h.p) | If $${{a_1},{a_2},....{a_n}}$$ are in H.P., then the expression $${{a_1}\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$$ is equal to | [{"identifier": "A", "content": "$$n({a_1}\\, - {a_n})$$ "}, {"identifier": "B", "content": "$$(n - 1)({a_1}\\, - {a_n})$$ "}, {"identifier": "C", "content": "$$n{a_1}{a_n}$$ "}, {"identifier": "D", "content": "$$(n - 1)\\,\\,{a_1}{a_n}$$ "}] | ["D"] | null | $${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$$ (say)
<br><br>Then $${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}$$
<br><br>$$\therefore$$ $${a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}$$
<br><br>$$ = {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}$$
<br><br>$$ = {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}$$
<br><br>Also, $${1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d$$
<br><br>$$ \Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d$$
<br><br>$$ \Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}$$
<br><br>Which is the required result. | mcq | aieee-2006 |
1l57nygjf | maths | sequences-and-series | harmonic-progression-(h.p) | <p>$$x = \sum\limits_{n = 0}^\infty {{a^n},y = \sum\limits_{n = 0}^\infty {{b^n},z = \sum\limits_{n = 0}^\infty {{c^n}} } } $$, where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc $$\ne$$ 0, then :</p> | [{"identifier": "A", "content": "x, y, z are in A.P."}, {"identifier": "B", "content": "x, y, z are in G.P."}, {"identifier": "C", "content": "$${1 \\over x}$$, $${1 \\over y}$$, $${1 \\over z}$$ are in A.P."}, {"identifier": "D", "content": "$${1 \\over x}$$ + $${1 \\over y}$$ + $${1 \\over z}$$ = 1 $$-$$ (a + b + c)"}] | ["C"] | null | <p>$$x = \sum\limits_{n = 0}^\infty {{a^n} = {1 \over {1 - a}};\,y = \sum\limits_{n = 0}^\infty {{b^n} = {1 \over {1 - b}};\,z = \sum\limits_{n = 0}^\infty {{c^n} = {1 \over {1 - c}}} } } $$</p>
<p>Now,</p>
<p>a, b, c $$\to$$ AP</p>
<p>1 $$-$$ a, 1 $$-$$ b, 1 $$-$$ c $$\to$$ AP</p>
<p>$${1 \over {1 - a}},\,{1 \over {1 - b}},\,{1 \over {1 - c}} \to HP$$</p>
<p>x, y, z $$\to$$ HP</p>
<p>$$\therefore$$ $${1 \over x},{1 \over y},{1 \over z} \to AP$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
U1oLUueqALYqitPf | maths | sequences-and-series | summation-of-series | $${1^3} - \,\,{2^3} + {3^3} - {4^3} + ... + {9^3} = $$ | [{"identifier": "A", "content": "425"}, {"identifier": "B", "content": "- 425"}, {"identifier": "C", "content": "475"}, {"identifier": "D", "content": "- 475"}] | ["A"] | null | $${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$$
<br><br>$$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$$
<br><br>$$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$$
<br><br>$$ = {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$$
<br><br>$$ = 2025 - 1600 = 425$$ | mcq | aieee-2002 |
r5pz1x4onggAO3Gd | maths | sequences-and-series | summation-of-series | The value of $$\,{2^{1/4}}.\,\,{4^{1/8}}.\,{8^{1/16}}...\infty $$ is | [{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "3/2 "}, {"identifier": "D", "content": "4"}] | ["B"] | null | The product is $$p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$$
<br><br>$$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$$
<br><br>Now let
<br><br>$$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$$
<br><br>$${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)$$
<br><br>Subtracting $$(2)$$ from $$(1)$$
<br><br>$$ \Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty $$
<br><br>or $${1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1$$
<br><br>$$\therefore$$ $$P = {2^S} = 2$$ | mcq | aieee-2002 |
lUgZFx9XzBXy4yRV | maths | sequences-and-series | summation-of-series | The sum of the serier $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..............$$ up to $$\infty $$ is equal to | [{"identifier": "A", "content": "$$\\log {\\,_e}\\left( {{4 \\over e}} \\right)\\,\\,$$ "}, {"identifier": "B", "content": "$$2\\,\\log {\\,_e}2$$ "}, {"identifier": "C", "content": "$$\\log {\\,_e}2 - 1\\,$$ "}, {"identifier": "D", "content": "$$\\log {\\,_e}2$$ "}] | ["A"] | null | $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $$
<br><br>$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$
<br><br>$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$
<br><br>$$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$$
<br><br>$$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$$
<br><br>$$ = 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$$
<br><br>$$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$$ | mcq | aieee-2003 |
PL99UcmXmNOIsJPr | maths | sequences-and-series | summation-of-series | The sum of the first n terms of the series $${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}$$ when n is even. When n is odd the sum is | [{"identifier": "A", "content": "$${\\left[ {{{n(n + 1)} \\over 2}} \\right]^2}$$ "}, {"identifier": "B", "content": "$${{{n^2}(n + 1)} \\over 2}$$ "}, {"identifier": "C", "content": "$${{n{{(n + 1)}^2}} \\over 4}$$ "}, {"identifier": "D", "content": "$$\\,{{3n(n + 1)} \\over 2}$$ "}] | ["B"] | null | If $$n$$ is odd, the required sum is
<br><br>$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$
<br><br>$$ = {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$$
<br><br>[ As $$\left( {n - 1} \right)$$ is even
<br><br>$$\therefore$$ using given formula for the sum of $$\left( {n - 1} \right)$$ terms.]
<br><br>$$ = \left( {{{n - 1} \over 2} + 1} \right){n^2} = {{{n^2}\left( {n + 1} \right)} \over 2}$$ | mcq | aieee-2004 |
n9ha6c4bh8MQnUGo | maths | sequences-and-series | summation-of-series | The sum of series $${1 \over {2\,!}} + {1 \over {4\,!}} + {1 \over {6\,!}} + ........$$ is | [{"identifier": "A", "content": "$${{\\left( {{e^2} - 2} \\right)} \\over e}\\,$$ "}, {"identifier": "B", "content": "$${{{{\\left( {e - 1} \\right)}^2}} \\over {2e}}$$ "}, {"identifier": "C", "content": "$${{\\left( {{e^2} - 1} \\right)} \\over {2e}}\\,$$ "}, {"identifier": "D", "content": "$${{\\left( {{e^2} - 1} \\right)} \\over 2}$$ "}] | ["B"] | null | We know that
<br><br>$$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$$
<br><br>and
<br><br>$${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$$
<br><br>$$\therefore$$ $$e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]$$
<br><br>$$\therefore$$ $${1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + .......$$
<br><br>$$ = {{e + {e^{ - 1}}} \over 2} - 1$$
<br><br>$$ = {{{e^2} + 1 - 2e} \over {2e}}$$
<br><br>$$ = {{{{\left( {e - 1} \right)}^2}} \over {2e}}$$ | mcq | aieee-2004 |
M5gWDJeqiWRmB7hX | maths | sequences-and-series | summation-of-series | The sum of the series $$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$ ad inf. is | [{"identifier": "A", "content": "$${{e - 1} \\over {\\sqrt e }}\\,$$ "}, {"identifier": "B", "content": "$${{e + 1} \\over {\\sqrt e }}$$ "}, {"identifier": "C", "content": "$${{e - 1} \\over {2\\sqrt e }}$$ "}, {"identifier": "D", "content": "$${{e + 1} \\over {2\\sqrt e }}$$ "}] | ["D"] | null | $${{{e^x} + {e^{ - x}}} \over 2}$$
<br><br>$$ = 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........$$
<br><br>Putting $$x = {1 \over 2}$$ we get
<br><br>$$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$
<br><br>$$\infty = {{{e^{{1 \over 2}}} + {e^{{{ - 1} \over 2}}}} \over 2} = {{\sqrt e + {1 \over {\sqrt e }}} \over 2}$$
<br><br>$$ = {{e + 1} \over {2\sqrt e }}$$ | mcq | aieee-2005 |
4spZL4XKlmGNYvXy | maths | sequences-and-series | summation-of-series | The sum of series $${1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - .......$$ upto infinity is | [{"identifier": "A", "content": "$${e^{ - {1 \\over 2}}}$$ "}, {"identifier": "B", "content": "$${e^{ + {1 \\over 2}}}$$"}, {"identifier": "C", "content": "$${e^{ - 2}}$$ "}, {"identifier": "D", "content": "$${e^{ - 1}}$$"}] | ["D"] | null | We know that $${e^x} = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ........\infty $$
<br><br>Put $$x=-1$$
<br><br>$$\therefore$$ $${e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty $$
<br><br>$$\therefore$$ $${e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - {1 \over {5!}}........\infty $$ | mcq | aieee-2007 |
dEB3aGohwKvD17AB | maths | sequences-and-series | summation-of-series | The sum to infinite term of the series $$1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + .....$$ is | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["A"] | null | We have
<br><br>$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........\infty \,\,\,\,\,...\left( 1 \right)$$
<br><br>Multiplying both sides by $${1 \over 3}$$ we get
<br><br>$${1 \over 3}S = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + {{10} \over {{3^4}}} + .......\,\,\,\,\,...\left( 2 \right)$$
<br><br>Subtracting eqn. $$(2)$$ from eqn. $$(1)$$ we get
<br><br>$${2 \over 3}S = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $$
<br><br>$$ \Rightarrow {2 \over 3}S = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $$
<br><br>$$ \Rightarrow {2 \over 3}S = {{{4 \over 3}} \over {1 - {1 \over 3}}} = {4 \over 3} \times {3 \over 2}$$
<br><br>$$ \Rightarrow S - 3$$ | mcq | aieee-2009 |
HGW0sDmHzjbMJRmV | maths | sequences-and-series | summation-of-series | <p> <b> Statement-1: </b> The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +.....+ (361 + 380 + 400) is 8000.
</p><p> <b> Statement-2: </b> $$\sum\limits_{k = 1}^n {\left( {{k^3} - {{(k - 1)}^3}} \right)} = {n^3}$$, for any natural number n.</p>
| [{"identifier": "A", "content": "Statement-1 is false, Statement-2 is true."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}, {"identifier": "C", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "D", "content": "Statement-1 is true, Statement-2 is false."}] | ["B"] | null | $$n$$<sup>th </sup> term of the given series
<br><br>$$ = {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$$
<br><br>$$ = {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$$
<br><br>$$ = {n^3} - {\left( {n - 1} \right)^3}$$
<br><br>$$ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]} $$
<br><br>$$ \Rightarrow 8000 = {n^3}$$
<br><br>$$ \Rightarrow n = 20\,\,$$ which is a natural number.
<br><br>Now, put $$n = 1,2,3,.....20$$
<br><br>$${T_1} = {1^3} - {0^3}$$
<br><br>$${T_2} = {2^3} - {1^3}$$
<br><br>.
<br><br>.
<br><br>.
<br><br>$${T_{20}} = {20^3} - {19^3}$$
<br><br>Now, $${T_1} + {T_2} + ..... + {T_{20}} = {S_{20}}$$
<br><br>$$ \Rightarrow {S_{20}} = {20^3} - {0^3} = 8000$$
<br><br>Hence, both the given statements are true and statement $$2$$ supports statement $$1.$$ | mcq | aieee-2012 |
lh5wc3i1Vj06rYEy | maths | sequences-and-series | summation-of-series | The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,........,is | [{"identifier": "A", "content": "$${7 \\over {81}}\\left( {179 - {{10}^{ - 20}}} \\right)$$ "}, {"identifier": "B", "content": "$$\\,{7 \\over 9}\\left( {99 - {{10}^{ - 20}}} \\right)$$ "}, {"identifier": "C", "content": "$${7 \\over {81}}\\left( {179 + {{10}^{ - 20}}} \\right)$$ "}, {"identifier": "D", "content": "$${7 \\over 9}\\left( {99 + {{10}^{ - 20}}} \\right)$$ "}] | ["C"] | null | Given sequence can be written as
<br><br>$${7 \over {10}} + {{77} \over {100}} + {{777} \over {{{10}^3}}} + ..... + $$ up to $$20$$ terms
<br><br>$$ = 7\left[ {{1 \over {10}} + {{11} \over {100}} + {{111} \over {{{10}^3}}} + ...... + } \right.\,\,$$ up to $$20$$ terms ]
<br><br>Multiply and divide by $$9$$
<br><br>$$ = {7 \over 9}\left[ {{9 \over {10}} + {{99} \over {100}} + {{999} \over {1000}} + ......} \right.\,\,$$ $$+$$ up to $$20$$ terms ]
<br><br>$$ = {7 \over 9}\left[ {\left( {1 - {1 \over {10}}} \right)} \right. + \left( {1 - {1 \over {{{10}^2}}}} \right) + \left( {1 - {1 \over {{{10}^3}}}} \right) + ......$$ $$+$$ up to $$20$$ terms ]
<br><br>$$ = {7 \over 9}\left[ {20 - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^{20}}} \right)} \over {1 - {1 \over {10}}}}} \right]$$
<br><br>$$ = {7 \over 9}\left[ {{{179} \over 9} + {1 \over 9}{{\left( {{1 \over {10}}} \right)}^{20}}} \right]$$
<br><br>$$ = {7 \over {81}}\left[ {179 + {{\left( {10} \right)}^{ - 20}}} \right]$$ | mcq | jee-main-2013-offline |
4PDGIePDoIDBwloQ | maths | sequences-and-series | summation-of-series | If $${(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + ......... + 10{(11)^9} = k{(10)^9},$$, then k is equal to : | [{"identifier": "A", "content": "100 "}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "$${{121} \\over {10}}$$ "}, {"identifier": "D", "content": "$${{441} \\over {100}}$$ "}] | ["A"] | null | Let $${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$
<br><br>Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$$
<br><br>Multiplied by $${{11} \over {10}}$$ on both the sides
<br><br>$${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$$
<br><br>$$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$$
<br><br>$$ \Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$$
<br><br>$$ \Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$$
<br><br>$$ \Rightarrow x = {10^{11}} = k{.10^9}$$
<br><br>Given $$ \Rightarrow k = 100$$ | mcq | jee-main-2014-offline |
ZdSGnALvwlIDEdX3 | maths | sequences-and-series | summation-of-series | The sum of first 9 terms of the series.
<br/><br/>$${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......$$ | [{"identifier": "A", "content": "142"}, {"identifier": "B", "content": "192"}, {"identifier": "C", "content": "71"}, {"identifier": "D", "content": "96"}] | ["D"] | null | $${n^{th}}$$ term of series
<br><br>$$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$
<br><br>Sum of $$n$$ term $$ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$
<br><br>$$ = {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]$$
<br><br>$$ = {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]$$
<br><br>Sum of $$9$$ terms
<br><br>$$ = {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$$ | mcq | jee-main-2015-offline |
FaNkL9CszrtfuIWo | maths | sequences-and-series | summation-of-series | If the sum of the first ten terms of the series $${\left( {1{3 \over 5}} \right)^2} + {\left( {2{2 \over 5}} \right)^2} + {\left( {3{1 \over 5}} \right)^2} + {4^2} + {\left( {4{4 \over 5}} \right)^2} + .......is\,{{16} \over 5}m,$$ then m is equal to : | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "99"}, {"identifier": "C", "content": "102"}, {"identifier": "D", "content": "101"}] | ["D"] | null | $${\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}$$
<br><br>$$S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4^2} + ...... + {{11}^2}} \right)$$
<br><br>$$ = {{16} \over {25}}\left( {{{11\left( {11 + 1} \right)\left( {22 + 1} \right)} \over 6} - 1} \right)$$
<br><br>$$ = {{16} \over {25}} \times 505 = {{16} \over 5} \times 101$$
<br><br>$$ \Rightarrow {{16} \over 5}m = {{16} \over 5} \times 101$$
<br><br>$$ \Rightarrow m = 101.$$ | mcq | jee-main-2016-offline |
2JtY6v5Vo1VAChH2hqDWK | maths | sequences-and-series | summation-of-series | Let z = 1 + ai be a complex number, a > 0, such that z<sup>3</sup> is a real number.
<br/><br/>Then the sum 1 + z + z<sup>2</sup> + . . . . .+ z<sup>11</sup> is equal to : | [{"identifier": "A", "content": "$$ - 1250\\,\\sqrt 3 \\,i$$ "}, {"identifier": "B", "content": "$$ 1250\\,\\sqrt 3 \\,i$$ "}, {"identifier": "C", "content": "$$1365\\,\\sqrt 3 i$$"}, {"identifier": "D", "content": "$$-$$ $$1365\\,\\sqrt 3 i$$"}] | ["D"] | null | z = 1 + ai
<br><br>z<sup>2</sup> = 1 $$-$$ a<sup>2</sup> + 2ai
<br><br>z<sup>2</sup> . z = {(1 $$-$$ a<sup>2</sup>) + 2ai} {1 + ai}
<br><br>= (1 $$-$$ a<sup>2</sup>) + 2ai + (1 $$-$$ a<sup>2</sup>) ai $$-$$ 2a<sup>2</sup>
<br><br>$$ \because $$ z<sup>3</sup> is real $$ \Rightarrow $$ 2a + (1 $$-$$ a<sup>2</sup>) a = 0
<br><br>a (3 $$-$$ a<sup>2</sup>) = 0 $$ \Rightarrow $$ a = $$\sqrt 3 $$ (a > 0)
<br><br>1 + z + z<sup>2</sup> . . . . . . . z<sup>11</sup> = $${{{z^{12}} - 1} \over {z - 1}} = {{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {1 + \sqrt 3 i - 1}}$$
<br><br>= $${{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {\sqrt 3 i}}$$
<br><br>(1 + $${\sqrt 3 i}$$)<sup>12</sup> = 2<sup>12</sup> $${\left( {{1 \over 2} + {{\sqrt 3 } \over 2}i} \right)^{12}}$$
<br><br>= 2<sup>12</sup> (cos$${\pi \over 3}$$ + isin$${\pi \over 3}$$)<sup>12</sup> = 2<sup>12</sup> (cos4$$\pi $$ + isin4$$\pi $$) = 2<sup>12</sup>
<br><br>$$ \Rightarrow $$ $${{{2^{12}} - 1} \over {\sqrt 3 i}} = {{4095} \over {\sqrt 3 i}} = - {{4095} \over 3}\sqrt 3 i = - 1365\sqrt 3 i$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
xuwfSpE7OV4mQBpm | maths | sequences-and-series | summation-of-series | For any three positive real numbers a, b and c,
<br/><br/>9(25$${a^2}$$ + b<sup>2</sup>) + 25(c<sup>2</sup> - 3$$a$$c) = 15b(3$$a$$ + c).
<br/>Then | [{"identifier": "A", "content": "b, c and $$a$$ are in G.P."}, {"identifier": "B", "content": "b, c and $$a$$ are in A.P."}, {"identifier": "C", "content": "$$a$$, b and c are in A.P."}, {"identifier": "D", "content": "$$a$$, b and c are in G.P."}] | ["B"] | null | 9(25$${a^2}$$ + b<sup>2</sup>) + 25(c<sup>2</sup> - 3$$a$$c) = 15b(3$$a$$ + c)
<br><br> $$ \Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc$$
<br><br>$$ \Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac = 45ab + 15bc$$
<br><br>$$ \Rightarrow $$ $${1 \over 2}\left[ {{{\left( {15a - 3b} \right)}^2} + {{\left( {3b - 5c} \right)}^2} + {{\left( {5c - 15a} \right)}^2}} \right] = 0$$
<br><br>it is possible when 15a – 3b = 0, 3b – 5 c = 0 and
5c – 15a = 0
<br><br>$$ \Rightarrow $$ 15a = 3b = 5c
<br><br>$$ \Rightarrow $$ b = $${{5c} \over 3}$$, a = $${c \over 3}$$
<br><br>$$ \Rightarrow $$ a + b = $${c \over 3} + {{5c} \over 3}$$ = $${{6c} \over 3}$$ = 2c
<br><br>$$ \therefore $$ b, c, a are in A.P. | mcq | jee-main-2017-offline |
cNdhbmgFjjDuKylw4LXXW | maths | sequences-and-series | summation-of-series | Let
<br/><br/>S<sub>n</sub> = $${1 \over {{1^3}}}$$$$ + {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.$$
<br/><br/>If 100 S<sub>n</sub> = n, then n is equal to : | [{"identifier": "A", "content": "199"}, {"identifier": "B", "content": "99"}, {"identifier": "C", "content": "200"}, {"identifier": "D", "content": "19"}] | ["A"] | null | n<sup>th</sup> term, T<sub>n</sub> = $${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$$
<br><br>T<sub>n</sub> = $${{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ T<sub>n</sub> = $${2 \over {n\left( {n + 1} \right)}}$$ = $$2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$$
<br><br>$$ \therefore $$ S<sub>n</sub> = $$\sum {{T_n}} $$
<br><br>= $$2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} $$
<br><br>= $$2\left( {1 - {1 \over n}} \right)$$
<br><br>= $${{{2n} \over {n + 1}}}$$
<br><br>Given that,
<br><br>100 S<sub>n</sub> = n
<br><br>$$ \Rightarrow $$ 100 $$ \times $$ $${{{2n} \over {n + 1}}}$$ = n
<br><br>$$ \Rightarrow $$ n + 1 = 200
<br><br>$$ \Rightarrow $$ n = 199 | mcq | jee-main-2017-online-9th-april-morning-slot |
OME7QaClOHVLgwxtyldwp | maths | sequences-and-series | summation-of-series | If the sum of the first n terms of the series $$\,\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$$ is $$435\sqrt 3 ,$$ then n equals : | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "29"}] | ["B"] | null | Given,
<br><br>$$\sqrt 3 $$ + $$\sqrt {75} $$ + $$\sqrt {243} $$ + $$\sqrt {507} $$ + . . . . . .+ n terms
<br><br>= $$\sqrt 3 $$ + $$\sqrt {25 \times 3} $$ + $$\sqrt {81 \times 3} $$ + $$\sqrt {169 \times 3} $$ + . . . . . .+ n terms
<br><br>= $$\sqrt 3 $$ + 5$$\sqrt 3 $$ + 9$$\sqrt 3 $$ + 13$$\sqrt 3 $$ + . . . . . .+ n terms
<br><br>= $$\sqrt 3 $$ [ 1 + 5 + 9 + 13 + . . . . .+ n terms]
<br><br>= $$\sqrt 3 $$ $$\left[ {{n \over 2}\left( {2.1 + \left( {n - 1} \right)4} \right)} \right]$$
<br><br>= $$\sqrt 3 $$ $$\left[ {{n \over 2}\left( {2 + 4n - 4} \right)} \right]$$
<br><br>= $$\sqrt 3 $$ $$\left[ {{n \over 2}\left( {4n - 2} \right)} \right]$$
<br><br>= $$\sqrt 3 $$ [n (2n $$-$$ 1)]
<br><br>According to question,
<br><br>$$\sqrt 3 $$ [n (2n $$-$$ 1)] = 435$$\sqrt 3 $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2n<sup>2</sup> $$-$$ n = 435
<br><br>$$\therefore\,\,\,$$ n = $${{1 \pm \sqrt {1 + 4 \times 2 \times 435} } \over 4}$$ = $${{1 \pm 59} \over 4}$$
<br><br>$$\therefore\,\,\,$$ n = $${{1 + 59} \over 4}$$ = 15 or $${{1 - 59} \over 4}$$ = $$-$$ 14.5
<br><br>$$\therefore\,\,\,$$ n = 15 (as n can't be $$-$$ve) | mcq | jee-main-2017-online-8th-april-morning-slot |
OJSjRBTgKunyFAcA | maths | sequences-and-series | summation-of-series | Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series
<br/>1<sup>2</sup> + 2.2<sup>2</sup> + 3<sup>2</sup> + 2.4<sup>2</sup> + 5<sup>2</sup> + 2.6<sup>2</sup> ...........
<br/>If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to | [{"identifier": "A", "content": "496"}, {"identifier": "B", "content": "232"}, {"identifier": "C", "content": "248"}, {"identifier": "D", "content": "464"}] | ["C"] | null | <b><u>Note</u> : </b>
<br><br>Sum of square of first n odd terms
<br><br>1<sup>2</sup> + 3<sup>2</sup> + 5<sup>2</sup> + . . . . .+ n<sup>2</sup> = $${{n\left( {2n - 1} \right)\left( {2n + 1} \right)} \over 3}$$
<br><br>Given,
<br><br>1<sup>2</sup> + 2. 2<sup>2</sup> + 3<sup>2</sup> + 2.4<sup>2</sup> + 5<sup>2</sup> + 2.6<sup>2</sup> + . . . . . .
<br><br>A = Sum of first 20 terms
<br><br>$$\therefore\,\,\,$$A = 1<sup>2</sup> + 2.2<sup>2</sup> + 3<sup>2</sup> + 24<sup>2</sup> + 5<sup>2</sup> + 2.6<sup>2</sup> + . . . . . .20 terms
<br><br>Arrange those terms this way,
<br><br>A = [1<sup>2</sup> + 3<sup>2</sup> + 5<sup>2</sup> + . . . . . 10 terms] + [ 2.2<sup>2</sup> + 2.4<sup>2</sup> + 2.6<sup>2</sup> + . . . . 10 terms]
<br><br>A = [ 1<sup>2</sup> + 3<sup>2</sup> + 5<sup>2</sup> + . . . . 10 terms ] + 2.<sup>2</sup> [ 1<sup>2</sup> + 2<sup>2</sup> + 3<sup>2</sup> + . . . .10 terms ]
<br><br>A = $$ {{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]$$
<br><br>A = $$ {{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}$$
<br><br>A =70 $$ \times $$ 19 + 70 $$ \times $$ 44
<br><br>A = 70 $$ \times $$ 63
<br><br>B = Sum of first 40 terms
<br><br>Arrange those terms this way.
<br><br>B = [1<sup>2</sup>+ 3<sup>2</sup> + 5<sup>2</sup> +. . . . 20 terms ] + [2.2<sup>2</sup> + 2.4<sup>2</sup> +. . . . . 20 terms ]
<br><br>B = [1<sup>2</sup> + 3<sup>2</sup> + 5<sup>2</sup> + . . . . 20 terms] + 2.2<sup>2</sup> [1<sup>2</sup> + 2<sup>2</sup> + . . . 20 terms ]
<br><br>B = $${{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}$$
<br><br>B = 260 $$ \times $$ 41 + 560 $$ \times $$ 41
<br><br>B = 41 $$ \times \,\,\,820$$
<br><br>$$\therefore\,\,\,$$ B $$-$$ 2A = 41 $$ \times \,$$ 820 $$-$$ 2 $$ \times \,$$ 70 $$ \times \,$$ 63 = 24800
<br><br>Given that B $$-$$ 2A = 100 $$\lambda $$
<br><br>$$\therefore\,\,\,$$ 100 $$\lambda $$ = 24800
<br><br>$$ \Rightarrow \,\,\,\lambda $$ = 248 | mcq | jee-main-2018-offline |
QtfYNrtuTFProPJ2lotH6 | maths | sequences-and-series | summation-of-series | The sum of the first 20 terms of the series
<br/><br/>$$1 + {3 \over 2} + {7 \over 4} + {{15} \over 8} + {{31} \over {16}} + ...,$$ is : | [{"identifier": "A", "content": "$$38 + {1 \\over {{2^{19}}}}$$"}, {"identifier": "B", "content": "$$38 + {1 \\over {{2^{20}}}}$$"}, {"identifier": "C", "content": "$$39 + {1 \\over {{2^{20}}}}$$"}, {"identifier": "D", "content": "$$39 + {1 \\over {{2^{19}}}}$$"}] | ["A"] | null | 1 + $${3 \over 2}$$ + $${7 \over 4}$$ + $${15 \over 8}$$ + $${31 \over 16}$$ + . . . .
<br><br>= (2 $$-$$ 1) + (2 $$-$$ $${1 \over 2}$$ ) + (2 $$-$$ $${1 \over 4}$$) + (2 $$-$$ $${1 \over 8}$$) + . . . . .+ 20 terms
<br><br>= (2 + 2 + . . . . . 20 terms) $$-$$ (1 + $${1 \over 2}$$ + $${1 \over 4}$$ + . . . . . 20 terms)
<br><br>= 2 $$ \times $$ 20 $$-$$ $$\left( {{{1 - {{\left( {{1 \over 2}} \right)}^{20}}} \over {1 - {1 \over 2}}}} \right)$$
<br><br>= 40 $$-$$ 2 + 2 $${\left( {{1 \over 2}} \right)^{20}}$$
<br><br>= 38 + $${1 \over {{2^{19}}}}$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
BLiwgey2PSssP0DBRJxwz | maths | sequences-and-series | summation-of-series | Let A<sub>n</sub> = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3}$$ $$-$$. . . . . + ($$-$$1)<sup>n-1</sup> $${\left( {{3 \over 4}} \right)^n}$$ and B<sub>n</sub> = 1 $$-$$ A<sub>n</sub>.
<br/>Then, the least dd natural numbr p, so that B<sub>n</sub> > A<sub>n</sub> , for all n$$ \ge $$ p, is : | [{"identifier": "A", "content": "9 "}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "5"}] | ["B"] | null | A<sub>n</sub> = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3} - .... + {\left( { - 1} \right)^{n - 1}}{\left( {{3 \over 4}} \right)^n}$$
<br><br>Which in a G.P. with a = $${{3 \over 4}}$$, r = $${{{ - 3} \over 4}}$$ and number of terms = n
<br><br>$$\therefore\,\,\,$$ A<sub>n</sub> = $${{{3 \over 4}\left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {1 - \left( {{{ - 3} \over 4}} \right)}} = {{{3 \over 4} \times \left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {{7 \over 4}}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$A<sub>n</sub> = $${{3 \over 7}}$$$$\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right]$$ $$\,\,\,\,\,\,\,\,\,$$ . . . . . . . . . .(1)
<br><br>As, B<sub>n</sub> = 1 $$-$$ A<sub>n</sub>
<br><br>For least odd natural number p, such that B<sub>n</sub> > A<sub>n</sub>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 1 $$-$$ A<sub>n</sub> > A<sub>n</sub> $$ \Rightarrow $$ 1 > 2 $$ \times $$ A<sub>n</sub> $$ \Rightarrow $$ A<sub>n</sub> < $${{1 \over 2}}$$
<br><br>From eqn. (1), we get
<br><br>$${{3 \over 7}}$$$$ \times $$ $$\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right] < {1 \over 2}$$ $$ \Rightarrow $$ 1 $$-$$ $${\left( {{{ - 3} \over 4}} \right)^n} < {7 \over 6}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 1 $$-$$ $${7 \over 6}$$ < $${\left( {{{ - 3} \over 4}} \right)^n}$$ $$ \Rightarrow $$ $${{ - 1} \over 6} < {\left( {{{ - 3} \over 4}} \right)^n}$$
<br><br>As n is odd, then $${\left( {{{ - 3} \over 4}} \right)^n}$$ = $$-$$ $${{{{3^n}} \over 4}}$$
<br><br>So $${{ - 1} \over 6}$$ < $$-$$ $${\left( {{3 \over 4}} \right)^n}$$ $$ \Rightarrow $$ $${1 \over 6}$$ > $${\left( {{3 \over 4}} \right)^n}$$
<br><br>log$$\left( {{1 \over 6}} \right)$$ = n log$$\left( {{3 \over 4}} \right)$$ $$ \Rightarrow $$ 6.228 < n
<br><br>Hence, n should be 7. | mcq | jee-main-2018-online-15th-april-evening-slot |
VB4N8NG5lJI3jk4QxM3rsa0w2w9jx5edf2b | maths | sequences-and-series | summation-of-series | For x $$\varepsilon $$ R, let [x] denote the greatest integer $$ \le $$ x, then the sum of the series
$$\left[ { - {1 \over 3}} \right] + \left[ { - {1 \over 3} - {1 \over {100}}} \right] + \left[ { - {1 \over 3} - {2 \over {100}}} \right] + .... + \left[ { - {1 \over 3} - {{99} \over {100}}} \right]$$ is : | [{"identifier": "A", "content": "- 153"}, {"identifier": "B", "content": "- 135"}, {"identifier": "C", "content": "- 133"}, {"identifier": "D", "content": "- 131"}] | ["C"] | null | $$\left[ {{{ - 1} \over 3}} \right] + \left[ {{{ - 1} \over 3} - {1 \over {100}}} \right] + \left[ {{{ - 1} \over 3} - {2 \over {100}}} \right] +$$<br><br>
$$ ....... + \left[ {{{ - 1} \over 3} - {{99} \over {100}}} \right]$$<br><br>
$$ \Rightarrow ( - 1 - 1 - 1 - .....67\,times) + ( - 2 - 2 - 2 - .....33\,times)$$<br><br>
$$ \Rightarrow $$ -133
| mcq | jee-main-2019-online-12th-april-morning-slot |
JcmjVvCtosFouYLOtS3rsa0w2w9jx1zojez | maths | sequences-and-series | summation-of-series | The sum
<br/>$$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$$ is equal to : | [{"identifier": "A", "content": "620"}, {"identifier": "B", "content": "1240"}, {"identifier": "C", "content": "1860"}, {"identifier": "D", "content": "660"}] | ["A"] | null | $$Sum = \sum\limits_{n = 1}^{15} {{{{1^3} + {2^3} + .... + {n^3}} \over {1 + 2 + .... + n}}} - {1 \over 2}{{15 \times 16} \over 2}$$<br><br>
$$ = \sum\limits_{n = 1}^{15} {{{n(n + 1)} \over 2}} - 60$$<br><br>
$$ = {1 \over 2}\sum\limits_{n = 1}^{15} {{n^2}} + {1 \over 2}\sum\limits_{n = 1}^{15} {n - 60} $$<br><br>
$$ = {1 \over 2} \times {{15 \times 16 \times 31} \over 6} + {1 \over 2} \times {{15 \times 16} \over 2} - 60$$<br><br>
= 620 | mcq | jee-main-2019-online-10th-april-evening-slot |
HCoDSSB4LAVjlrMorg3rsa0w2w9jwxuogih | maths | sequences-and-series | summation-of-series | The sum
<br/>$${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$$ upto 10 terms is: | [{"identifier": "A", "content": "600"}, {"identifier": "B", "content": "660"}, {"identifier": "C", "content": "680"}, {"identifier": "D", "content": "620"}] | ["B"] | null | $${T_r} = {{(2r + 1)({1^3} + {2^3} + {3^3} + ...... + {r^3})} \over {{1^2} + {2^2} + {3^2} + ...... + {r^2}}}$$<br><br>
$${T_r} = (2r + 1){\left( {{{r(r + 1)} \over 2}} \right)^2} \times {6 \over {r(r + 1)(2r + 1)}}$$<br><br>
$${T_r} = {{3r(r + 1)} \over 2}$$<br><br>
Now, $$S = \sum\limits_{r = 1}^{10} {{T_r}} = {3 \over 2}\sum\limits_{r = 1}^{10} {({r^2} + r)} $$<br><br>
$$ \Rightarrow {3 \over 2}\left\{ {{{10 \times (10 + 1)(2 \times 10 + 1)} \over 6} + {{10 \times 11} \over 2}} \right\}$$<br><br>
$$ \Rightarrow {3 \over 2}\left\{ {{{10 \times 11 \times 21} \over 6} + 5 \times 11} \right\}$$<br><br>
$$ \Rightarrow {3 \over 2} \times 5 \times 11 \times 8 = 660$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
7PpCk7Hi0qXclaNmDM18hoxe66ijvww22ey | maths | sequences-and-series | summation-of-series | The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term is :- | [{"identifier": "A", "content": "945"}, {"identifier": "B", "content": "916"}, {"identifier": "C", "content": "915"}, {"identifier": "D", "content": "946"}] | ["D"] | null | S = 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term
<br><br>General term T<sub>n</sub> = n (2n - 1)
<br><br>$$ \therefore $$ S<sub>n</sub> = $$\sum {{T_n}} $$
<br><br>$$ \Rightarrow $$ S<sub>n</sub> = $$\sum {\left( {2{n^2} - n} \right)} $$
<br><br>$$ \Rightarrow $$ S<sub>n</sub> = $$2\sum {{n^2} - \sum r } $$
<br><br>$$ \Rightarrow $$ S<sub>n</sub> = $$2\left( {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right) - {{n\left( {n + 1} \right)} \over 2}$$
<br><br>$$ \therefore $$ S<sub>11</sub> = $$2\left( {{{11\left( {12} \right)\left( {23} \right)} \over 6}} \right) - {{11\left( {12} \right)} \over 2}$$
<br><br>$$ \Rightarrow $$ S<sub>11</sub> = 946 | mcq | jee-main-2019-online-9th-april-evening-slot |
50wym4bFg664eLPHoskkF | maths | sequences-and-series | summation-of-series | The sum
$$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $$ is equal to | [{"identifier": "A", "content": "$$2 - {11 \\over {{2^{19}}}}$$"}, {"identifier": "B", "content": "$$2 - {3 \\over {{2^{17}}}}$$"}, {"identifier": "C", "content": "$$1 - {11 \\over {{2^{20}}}}$$"}, {"identifier": "D", "content": "$$2 - {21 \\over {{2^{20}}}}$$"}] | ["A"] | null | Let S = $$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $$
<br><br>$$ \Rightarrow $$ S = $${1 \over 2} + {2 \over {{2^2}}} + {3 \over {{2^3}}} + {4 \over {{2^4}}} + ....... + {{20} \over {{2^{20}}}}$$ .......(1)
<br><br>This is an Arithmetic Geometric Sequence. Here numerator is in A.P and denominator is in G.P.
<br><br>To solve Arithmetic Geometric Sequence, we multiply the series S with the common ratio of G.P. Here common ratio of G.P is $${1 \over 2}$$.
<br><br>By multiplying (1) with $${1 \over 2}$$ we get,
<br><br>$${S \over 2} = {1 \over {{2^2}}} + {2 \over {{2^3}}} + {3 \over {{2^4}}} + {4 \over {{2^5}}} + ....... + {{19} \over {{2^{20}}}} + {{20} \over {{2^{21}}}}$$ ....(2)
<br><br>Subtract (2) from (1),
<br><br>S - $${S \over 2}$$ = $${1 \over 2} + {1 \over {{2^2}}} + {1 \over {{2^3}}} + {1 \over {{2^4}}} + ....... + {1 \over {{2^{20}}}} - {{20} \over {{2^{21}}}}$$
<br><br>$$ \Rightarrow $$ $${S \over 2}$$ = $${{{1 \over 2}\left( {1 - {1 \over {{2^{20}}}}} \right)} \over {1 - {1 \over 2}}}$$ - $${{20} \over {{2^{21}}}}$$
<br><br>$$ \Rightarrow $$ S = $$2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}}$$
<br><br>$$ \Rightarrow $$ S = $$2 - {2 \over {{2^{20}}}} - {{20} \over {{2^{20}}}}$$
<br><br>$$ \Rightarrow $$ S = $$2 - {{22} \over {{2^{20}}}}$$
<br><br>$$ \Rightarrow $$ S = $$2 - {{11} \over {{2^{19}}}}$$
| mcq | jee-main-2019-online-8th-april-evening-slot |
PUNoOi9Ml37qcJFZvBpIU | maths | sequences-and-series | summation-of-series | If the sum of the first 15 terms of the series $${\left( {{3 \over 4}} \right)^3} + {\left( {1{1 \over 2}} \right)^3} + {\left( {2{1 \over 4}} \right)^3} + {3^3} + {\left( {3{3 \over 4}} \right)^3} + ....$$ is equal to 225 k, then k
is equal to : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "108"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "54"}] | ["C"] | null | S = $${\left( {{3 \over 4}} \right)^3} + {\left( {{6 \over 4}} \right)^3} + {\left( {{9 \over 4}} \right)^3} + {\left( {{{12} \over 4}} \right)^3} + \,........15$$ term
<br><br>= $${{27} \over {64}}$$ $$\sum\limits_{r = 1}^{15} {{r^3}} $$
<br><br>= $${{27} \over {64.}}{\left[ {{{15\left( {15 + 1} \right)} \over 2}} \right]^2}$$
<br><br>= 225 K (Given in question)
<br><br>K = 27 | mcq | jee-main-2019-online-12th-january-evening-slot |
wUMtiArnVk8GEbYuJ01FI | maths | sequences-and-series | summation-of-series | Let S<sub>k</sub> = $${{1 + 2 + 3 + .... + k} \over k}.$$ If $$S_1^2 + S_2^2 + .....\, + S_{10}^2 = {5 \over {12}}$$A, then A is equal to : | [{"identifier": "A", "content": "283"}, {"identifier": "B", "content": "156"}, {"identifier": "C", "content": "301"}, {"identifier": "D", "content": "303"}] | ["D"] | null | S<sub>k</sub> = $${{K + 1} \over 2}$$
<br><br>$$\sum {S_k^2} = {5 \over {12}}$$ A
<br><br>$$\sum\limits_{K = 1}^{10} {{{\left( {{{K + 1} \over 2}} \right)}^2}} = {{{2^2} + {3^2} + - - + {{11}^2}} \over 4} = {5 \over {12}}$$ A
<br><br>$${{11 \times 12 \times 23} \over 6} - 1 = {5 \over 3}$$ A
<br><br>505 $$ = {5 \over 3}$$ A, A = 303 | mcq | jee-main-2019-online-12th-january-morning-slot |
qOmtyXLtSaFKZemyMR1Mj | maths | sequences-and-series | summation-of-series | The sum of the following series
<br/><br/>$$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9}$$
<br/><br/> $$ + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....$$ up to 15 terms, is : | [{"identifier": "A", "content": "7520"}, {"identifier": "B", "content": "7510"}, {"identifier": "C", "content": "7830"}, {"identifier": "D", "content": "7820"}] | ["D"] | null | $$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9} + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....\,15$$
<br><br>$$ = {{3\left( {{1^2}} \right)} \over 3} + {{6\left( {{1^2} + {2^2}} \right)} \over 5} + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12} \over 9}\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right) + ......$$
<br><br>$${T_r} = {{3r} \over {2r + 1}}\left( {{1^2} + {2^2} + .... + {r^2}} \right)$$
<br><br>$${T_r} = {{3r} \over {2r + 1}}{{r\left( {r + 1} \right)\left( {2r + 1} \right)} \over 6} = {1 \over 2}{r^2}\left( {r + 1} \right)$$
<br><br>Sum of $$n$$ terms $$ = \sum\limits_{r = 1}^n {{T_r}} = {1 \over 2}\sum\limits_{r = 1}^n {\left( {{r^3} + {r^2}} \right)} $$
<br><br>$$ = {1 \over 2}\left[ {{{{n^2}{{\left( {n + 1} \right)}^2}} \over 4} + {{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right]$$
<br><br>Sum upto 15 terms $$ \Rightarrow $$ then put $$n$$ = 15
<br><br>$$ = {1 \over 2}\left( {{{{{\left( {15 \times 16} \right)}^2}} \over 4} + {{15 \times 16 \times 31} \over 6}} \right) = 7820$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
QRFlNrjhUGdaY1JzcO18hoxe66ijvwq8ul6 | maths | sequences-and-series | summation-of-series | Some identical balls are arranged in rows to form
an equilateral triangle. The first row consists of one
ball, the second row consists of two balls and so
on. If 99 more identical balls are addded to the total
number of balls used in forming the equilaterial
triangle, then all these balls can be arranged in a
square whose each side contains exactly 2 balls
less than the number of balls each side of the
triangle contains. Then the number of balls used to
form the equilateral triangle is :- | [{"identifier": "A", "content": "262"}, {"identifier": "B", "content": "190"}, {"identifier": "C", "content": "157"}, {"identifier": "D", "content": "225"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264552/exam_images/ooaanpllsv7wqqhp4zns.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Sequences and Series Question 185 English Explanation 1">
<br><br>As triangle is equilateral so each side of triangle has n-balls.
<br><br>At the top of the triangle there is 1 ball then next line has 2 balls. Similarly last line has n balls.
<br><br>$$ \therefore $$ Total no of balls used to create this triangle
<br><br>= 1 + 2 + 3 + ..............+ n
<br><br>= $${{n\left( {n + 1} \right)} \over 2}$$
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264168/exam_images/piff2idkgob8rasda4s6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Sequences and Series Question 185 English Explanation 2">
<br>Here Number of balls in each side of square is = (n – 2)
<br><br>$$ \therefore $$ Total no of balls used to create this square
<br><br>= (n – 2)<sup>2</sup>
<br><br>According to the question,
<br><br>$${{n\left( {n + 1} \right)} \over 2}$$ + 99 = (n – 2)<sup>2</sup>
<br><br>$$ \Rightarrow $$ n<sup>2</sup> - 9n - 190 = 0
<br><br>$$ \Rightarrow $$ (n - 19) (n + 10) = 0
<br><br>$$ \therefore $$ n = 19
<br><br>So total balls used to form triangle
<br><br>= $${{n\left( {n + 1} \right)} \over 2}$$ = $${{19\left( {20} \right)} \over 2}$$ = 190 | mcq | jee-main-2019-online-9th-april-evening-slot |
U9xxSdQCH7SEff1HPajgy2xukfg6hk2u | maths | sequences-and-series | summation-of-series | If 2<sup>10</sup> + 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> = S - 2<sup>11</sup>, then S is equal to : | [{"identifier": "A", "content": "$${{{3^{11}}} \\over 2} + {2^{10}}$$"}, {"identifier": "B", "content": "3<sup>11</sup> \u2014 2<sup>12</sup>"}, {"identifier": "C", "content": "2.3<sup>11</sup>"}, {"identifier": "D", "content": "3<sup>11</sup>"}] | ["D"] | null | Let S<sub>1</sub> = 2<sup>10</sup> + 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> ....(1)
<br><br>Also $${{3{S_1}} \over 2}$$ = 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> + $${{{3^{11}}} \over 2}$$ ......(2)
<br><br>Performing (1) - (2), we get
<br><br>$$ - {{{S_1}} \over 2} = {2^{10}} - {{{3^{11}}} \over 2}$$
<br><br>$$ \Rightarrow $$ S<sub>1</sub> = 3<sup>11</sup> - 2<sup>11</sup>
<br><br>According to question,
<br><br>3<sup>11</sup> - 2<sup>11</sup> = S - 2<sup>11</sup>
<br><br>$$ \Rightarrow $$ S = 3<sup>11</sup> | mcq | jee-main-2020-online-5th-september-morning-slot |
q6m7BJB7vRxsb0PkbGjgy2xukfqd7bft | maths | sequences-and-series | summation-of-series | If the sum of the first 20 terms of the series
<br/>$${\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...$$ is 460,
<br/>then x is equal to : | [{"identifier": "A", "content": "e<sup>2</sup>"}, {"identifier": "B", "content": "7<sup>1/2</sup>"}, {"identifier": "C", "content": "7<sup>2</sup>"}, {"identifier": "D", "content": "7<sup>46/21</sup>"}] | ["C"] | null | 460 = log<sub>7</sub> <sup>x</sup>·(2 + 3 + 4 + ..... + 20 + 21)
<br><br>$$ \Rightarrow $$ 460 = log<sub>7</sub> <sup>x</sup>. $$\left( {{{21 \times 22} \over 2} - 1} \right)$$
<br><br>$$ \Rightarrow $$ 460 = 230. log<sub>7</sub> <sup>x</sup>
<br><br>$$ \Rightarrow $$ log<sub>7</sub> <sup>x</sup> = 2
<br><br>$$ \Rightarrow $$ x = 49 | mcq | jee-main-2020-online-5th-september-evening-slot |
6CDZ22WJaLTtjOakazjgy2xukf7gbwk0 | maths | sequences-and-series | summation-of-series | If 1+(1–2<sup>2</sup>.1)+(1–4<sup>2</sup>.3)+(1-6<sup>2</sup>.5)+......+(1-20<sup>2</sup>.19)= $$\alpha $$ - 220$$\beta $$, <br/>then an ordered pair $$\left( {\alpha ,\beta } \right)$$ is equal to: | [{"identifier": "A", "content": "(11, 103)"}, {"identifier": "B", "content": "(10, 103)"}, {"identifier": "C", "content": "(10, 97)"}, {"identifier": "D", "content": "(11, 97)"}] | ["A"] | null | $$1 + (1 - {2^2}.1) + (1 - {4^2}.3) + (1 - {6^2}.5) + ....(1 - {20^2}.19)$$<br><br>$$S = 1 + \sum\limits_{r = 1}^{10} {\left[ {1 - {{(2r)}^2}(2r - 1)} \right] = 1 + \sum\limits_{r = 1}^{10} {\left( {1 - 8{r^3} + 4{r^2}} \right)} = 1 + 10 - } \sum\limits_{r = 1}^{10} {\left( {8{r^3} - 4{r^2}} \right)} $$<br><br>$$= 11 - 8{\left( {{{10 \times 11} \over 2}} \right)^2} + 4 \times \left( {{{10 \times 11 \times 21} \over 6}} \right)$$<br><br>$$ = 11 - 2 \times {(110)^2} + 4 \times 55 \times 7$$<br><br>$$ = 11 - 220(110 - 7)$$<br><br>$$ = 11 - 220 \times 103 = \alpha - 220\beta $$
<br><br>$$\Rightarrow \alpha = 11,\beta = 103$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
iitp3b1OAO5bXqqUlkjgy2xukf44u0jj | maths | sequences-and-series | summation-of-series | If the sum of the series
<br/><br/>20 + 19$${3 \over 5}$$ + 19$${1 \over 5}$$ + 18$${4 \over 5}$$ + ...
<br/><br/>upto nth term is 488
and the n<sup>th</sup> term is negative, then : | [{"identifier": "A", "content": "n = 41"}, {"identifier": "B", "content": "n = 60"}, {"identifier": "C", "content": "n<sup>th</sup> term is \u20134"}, {"identifier": "D", "content": "n<sup>th</sup> term is -4$${2 \\over 5}$$"}] | ["C"] | null | $$S = {{100} \over 5} + {{98} \over 5} + {{96} \over 5} + {{94} \over 5} + ...\,n$$<br><br>$$\,{S_n} = {n \over 2}\left( {2 \times {{100} \over 5} + (n - 1)\left( {{{ - 2} \over 5}} \right)} \right) = 188$$<br><br>$$ \Rightarrow $$ $$n(100 - n + 1) = 488 \times 5$$<br><br>$$ \Rightarrow $$ $${n^2} - 101n + 488 \times 5 = 0$$<br><br>$$ \Rightarrow $$ $$n = 61,\,40$$
<br><br>For negative term n = 61
<br><br>$$\,{T_n} = a + (n - 1)d = {{100} \over 5} - {2 \over 5} \times 60$$<br><br>$$ = 20 - 24 = - 4$$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
grhqiUH1FX5lb2FWm3jgy2xukezf3clg | maths | sequences-and-series | summation-of-series | Let S be the sum of the first 9 terms of the
series :
<br/>{x + k$$a$$} + {x<sup>2</sup> + (k + 2)$$a$$} + {x<sup>3</sup> + (k + 4)$$a$$}
<br/>+ {x<sup>4</sup> + (k + 6)$$a$$} + .... where a $$ \ne $$ 0 and x $$ \ne $$ 1.
<br/><br/>If S = $${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$$, then k is equal to : | [{"identifier": "A", "content": "-3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "-5"}, {"identifier": "D", "content": "3"}] | ["A"] | null | S = {x + k$$a$$} + {x<sup>2</sup> + (k + 2)$$a$$} + {x<sup>3</sup> + (k + 4)$$a$$}
<br>+ {x<sup>4</sup> + (k + 6)$$a$$} + ....<br><br>
$$S = \left( {x + {x^2} + {x^3} + ....\,9terms} \right) + $$<br> $$a\left( {k + \left( {k + 2} \right) + (k + 4) + (k + 6) + ....9terms} \right)$$<br><br>
$$S = {{x\left( {{x^9} - 1} \right)} \over {\left( {x - 1} \right)}} + a\left[ {{9 \over 2}\left[ {2k + \left( {9 - 1} \right)2} \right]} \right]$$<br><br>
$$S = {{{x^{10}} - x} \over {x - 1}} + 9a\left( {k + 8} \right)$$<br><br>
$$S = {{{x^{10}} - x + 9a\left( {k + 8} \right)\left( {x - 1} \right)} \over {x - 1}}$$<br><br>
$$S = {{{x^{10}} - x + 9\left( {k + 8} \right)a\left( {x - 1} \right)} \over {x - 1}}$$<br><br>
Compare with given sum, then we get<br><br>
$${9\left( {k + 8} \right) = 45}$$<br><br>
$$ \Rightarrow \left( {k + 8} \right) = 5$$<br><br>
$$ \Rightarrow k = - 3$$
| mcq | jee-main-2020-online-2nd-september-evening-slot |
iYlXL3LrAQdpyM6ikm7k9k2k5is98y3 | maths | sequences-and-series | summation-of-series | The product $${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ... to $$\infty $$ is equal
to : | [{"identifier": "A", "content": "$${2^{{1 \\over 4}}}$$"}, {"identifier": "B", "content": "$${2^{{1 \\over 2}}}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ...
<br><br>= $${2^{{1 \over 4} + {2 \over {16}} + {3 \over {48}} + ...\infty }}$$
<br><br>= $${2^{{1 \over 4} + {1 \over 8} + {1 \over {16}} + ...\infty }}$$
<br><br>= $${2^{\left( {{{{1 \over 4}} \over {1 - {1 \over 2}}}} \right)}}$$
<br><br>= $${2^{{1 \over 2}}}$$ | mcq | jee-main-2020-online-9th-january-morning-slot |
jfd0PqVwAR7IawfFqN7k9k2k5hke3hj | maths | sequences-and-series | summation-of-series | The sum, $$\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $$ is equal to
________. | [] | null | 504 | $$\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $$
<br><br>= $${1 \over 4}\sum\limits_{n = 1}^7 {\left( {2{n^3} + 3{n^2} + n} \right)} $$
<br><br>= $${1 \over 2}\sum\limits_{n = 1}^7 {{n^3}} $$ + $${3 \over 4}\sum\limits_{n = 1}^7 {{n^2}} $$ + $${1 \over 4}\sum\limits_{n = 1}^7 n $$
<br><br>= $${1 \over 2}{\left( {{{7\left( {7 + 1} \right)} \over 2}} \right)^2}$$ + $${3 \over 4}\left( {{{7\left( {7 + 1} \right)\left( {14 + 1} \right)} \over 6}} \right)$$ + $${1 \over 4}{{7\left( 8 \right)} \over 2}$$
<br><br>= (49)(8) + (15$$ \times $$7) + (7)
<br><br>= 392 + 105 + 7 = 504 | integer | jee-main-2020-online-8th-january-evening-slot |
mYMcFJJVdAchWsiX9b7k9k2k5h0a9ty | maths | sequences-and-series | summation-of-series | The sum $$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $$ is : | [] | null | 1540 | $$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $$
<br><br>= $$\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $$
<br><br>= $$\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $$
<br><br>= $${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 \over 2} \times {{20 \times 21} \over 2}$$
<br><br>= 1540 | integer | jee-main-2020-online-8th-january-morning-slot |
vRpvwMQFOTlboi3Zw47k9k2k5fmzvev | maths | sequences-and-series | summation-of-series | If the sum of the first 40 terms of the series, <br/>3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to : | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "25"}] | ["A"] | null | 3 + 4 + 8 + 9 + 13 + 14 +…….upto 40 terms
<br><br>$$ \Rightarrow $$ 7 + 17 + 27 +…….20 terms
<br><br>S = $${{20} \over 2}$$ [2 × 7 + 19 × 10]
<br><br> = 102 × 20 = 102 m
<br><br>$$ \therefore $$ m = 20 | mcq | jee-main-2020-online-7th-january-evening-slot |
tUvK0XGzjI8Hr4plFFjgy2xukewmjblr | maths | sequences-and-series | summation-of-series | If |x| < 1, |y| < 1 and x $$ \ne $$ y, then the sum to infinity
of the following series
<br/><br/>(x + y) + (x<sup>2</sup>+xy+y<sup>2</sup>) + (x<sup>3</sup>+x<sup>2</sup>y + xy<sup>2</sup>+y<sup>3</sup>) + .... | [{"identifier": "A", "content": "$${{x + y - xy} \\over {\\left( {1 + x} \\right)\\left( {1 + y} \\right)}}$$"}, {"identifier": "B", "content": "$${{x + y - xy} \\over {\\left( {1 - x} \\right)\\left( {1 - y} \\right)}}$$"}, {"identifier": "C", "content": "$${{x + y + xy} \\over {\\left( {1 + x} \\right)\\left( {1 + y} \\right)}}$$"}, {"identifier": "D", "content": "$${{x + y + xy} \\over {\\left( {1 - x} \\right)\\left( {1 - y} \\right)}}$$"}] | ["B"] | null | (x + y) + (x<sup>2</sup>+xy+y<sup>2</sup>) + (x<sup>3</sup>+x<sup>2</sup>y + xy<sup>2</sup>+y<sup>3</sup>) + ....
<br><br>By multiplying and dividing x – y :
<br><br>$${{\left( {{x^2} - {y^2}} \right) + \left( {{x^3} - {y^3}} \right) + \left( {{x^4} - {y^4}} \right) + ...} \over {x - y}}$$
<br><br>= $${{\left( {{x^2} + {x^3} + {x^4} + ....} \right) - \left( {{y^2} + {y^3} + {y^4} + ...} \right)} \over {x - y}}$$
<br><br>= $${{{{{x^2}} \over {1 - x}} - {{{y^2}} \over {1 - y}}} \over {x - y}}$$
<br><br>= $${{\left( {{x^2} - {y^2}} \right) - xy\left( {x - y} \right)} \over {\left( {1 - x} \right)\left( {1 - y} \right)\left( {x - y} \right)}}$$
<br><br>= $${{x + y - xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$$ | mcq | jee-main-2020-online-2nd-september-morning-slot |
yuKByka9gb3axlQDX41kls445x1 | maths | sequences-and-series | summation-of-series | If $$0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi } $$ and $$z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi } $$ then : | [{"identifier": "A", "content": "xy $$-$$ z = (x + y)z"}, {"identifier": "B", "content": "xyz = 4"}, {"identifier": "C", "content": "xy + z = (x + y)z"}, {"identifier": "D", "content": "xy + yz + zx = z"}] | ["C"] | null | $$x = 1 + {\cos ^2}\theta + ..........\infty $$<br><br>$$x = {1 \over {1 - {{\cos }^2}\theta }} = {1 \over {{{\sin }^2}\theta }}$$ .......(1)<br><br>$$y = 1 + {\sin ^2}\phi + ........\infty $$<br><br>$$y = {1 \over {1 - {{\sin }^2}\phi }} = {1 \over {{{\cos }^2}\phi }}$$ ....... (2)<br><br>$$z = {1 \over {1 - {{\cos }^2}\theta .{{\sin }^2}\phi }} = {1 \over {1 - \left( {1 - {1 \over x}} \right)\left( {1 - {1 \over y}} \right)}} = {{xy} \over {xy - (x - 1)(y - 1)}}$$<br><br>$$ \Rightarrow $$ $$xz + yz - z = xy$$<br><br>$$ \Rightarrow $$ $$xy + z = (x + y)z$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
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