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1lgswd6s1 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The number of points, where the curve $$f(x)=\mathrm{e}^{8 x}-\mathrm{e}^{6 x}-3 \mathrm{e}^{4 x}-\mathrm{e}^{2 x}+1, x \in \mathbb{R}$$ cuts $$x$$-axis, is equal to _________.</p> | [] | null | 2 | Firstly, we know that the given function <br/><br/>$f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1$ intersects the x-axis <br/><br/>where $f(x) = 0$. Setting $f(x)$ equal to zero gives us :
<br/><br/>$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0.$
<br/><br/>Let $t = e^{2x}$. The equation now becomes :
<br/><br/>$t^4 - t^3 - 3t^2 - t + 1 = 0.$
<br/><br/>Dividing by $t^2$ and rearranging the equation gives :
<br/><br/>$t^2 - t - 3 - \frac{1}{t} + \frac{1}{t^2} = 0.$
<br/><br/>If we let $y = t + \frac{1}{t}$, we get :
<br/><br/>$y^2 - y - 5 = 0.$
<br/><br/>This quadratic equation in y can be solved using the quadratic formula to give two roots :
<br/><br/>$y = \frac{1 \pm \sqrt{21}}{2}.$
<br/><br/>Since $y = t + \frac{1}{t}$, and $t > 0$ (as $t = e^{2x}$), we must choose the root where $y > 2$. Thus, we take $y = \frac{1 + \sqrt{21}}{2}$.
<br/><br/>So, we have :
<br/><br/>$t + \frac{1}{t} = \frac{1 + \sqrt{21}}{2}.$
<br/><br/>Solving for $t$ gives us :
<br/><br/>$t^2 - yt + 1 = 0,$
<br/><br/>or
<br/><br/>$t = \frac{y \pm \sqrt{y^2 - 4}}{2}.$
<br/><br/>Substituting $y = \frac{1 + \sqrt{21}}{2}$ into the formula gives :
<br/><br/>$t = \frac{\frac{1 + \sqrt{21}}{2} \pm \sqrt{\left(\frac{1 + \sqrt{21}}{2}\right)^2 - 4}}{2}.$
<br/><br/>This quadratic formula for $t$ yields two possible values (t = 2.37 and t = 0.42). Both of them are positive, thus both are acceptable values for $t = e^{2x}$.
<br/><br/>Finally, to get $x$, take the natural log of both sides and divide by 2 :
<br/><br/>$x = \frac{1}{2}\ln(t).$
<br/><br/>This gives us two solutions for $x$, corresponding to the two positive solutions for $t$. So, the number of points where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$ intersects the x-axis is 2. | integer | jee-main-2023-online-11th-april-evening-shift |
1lguwygfx | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If $$a$$ and $$b$$ are the roots of the equation $$x^{2}-7 x-1=0$$, then the value of $$\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$$ is equal to _____________.</p> | [] | null | 51 | We have, $a$ and $b$ are the roots of the equation
<br/><br/>$$
\begin{aligned}
& x^2-7 x-1=0 \\\\
& \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a .........(i)
\end{aligned}
$$
<br/><br/>On squaring both sides, we get $a^4+1=51 a^2$
<br/><br/>Similarly, $b^4+1=51 b^2$ ...........(ii)
<br/><br/>$$
\text { Now, } \frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}}
$$
<br/><br/>$$
\begin{aligned}
& =\frac{a^{17}\left(51 a^2\right)+b^{17}\left(51 b^2\right)}{a^{19}+b^{19}} \quad[\because \text { From Eq. (i) and (ii) }] \\\\
& =\frac{51\left[a^{19}+b^{19}\right]}{a^{19}+b^{19}}=51
\end{aligned}
$$ | integer | jee-main-2023-online-11th-april-morning-shift |
1lgzzo8pe | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta, \gamma$$ be the three roots of the equation $$x^{3}+b x+c=0$$. If $$\beta \gamma=1=-\alpha$$, then $$b^{3}+2 c^{3}-3 \alpha^{3}-6 \beta^{3}-8 \gamma^{3}$$ is equal to :</p> | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "19"}, {"identifier": "C", "content": "$$\\frac{169}{8}$$"}, {"identifier": "D", "content": "$$\\frac{155}{8}$$"}] | ["B"] | null | Given cubic equation is :
<br/><br/>$$
x^3+b x+c=0
$$
<br/><br/>$\because \alpha, \beta, \gamma$ are the roots of above equation.
<br/><br/>And $\beta \gamma=1=-\alpha$
<br/><br/>$$
\begin{aligned}
& \text { So, product of roots }=-c \\\\
& \Rightarrow \alpha \beta \gamma=-c \\\\
& \Rightarrow(-1)(1)=-c \\\\
& \Rightarrow c=1
\end{aligned}
$$
<br/><br/>Since, $\alpha=-1$ is the root. So,
<br/><br/>$$
\begin{aligned}
& \Rightarrow-1-b+c=0 \\\\
& \Rightarrow c-b=1 \\\\
& \Rightarrow 1-b=1 \Rightarrow b=0
\end{aligned}
$$
<br/><br/>The given equation becomes $x^3+1=0$
<br/><br/>So, roots are $-1,-\omega,-\omega^2$
<br/><br/>$$
\begin{aligned}
& \therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\\\
& =0+2-3(-1)^3-6(-\omega)^3-8\left(-\omega^2\right)^3 \\\\
& =2+3+6 \omega^3+8 \omega^6 \\\\
& =5+6+8=19
\end{aligned}
$$
<br/><br/><b>Concept :</b>
<br/><br/>For a cubic equation, $a x^3+b x^2+c x+d=0$
<br/><br/>Sum of roots $=\frac{-b}{a}$
<br/><br/>Product of roots taken two at a time $=\frac{c}{a}$
<br/><br/>Product of roots $=\frac{-d}{a}$ | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh23bm1u | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The sum of all the roots of the equation $$\left|x^{2}-8 x+15\right|-2 x+7=0$$ is :</p> | [{"identifier": "A", "content": "$$11+\\sqrt{3}$$"}, {"identifier": "B", "content": "$$9+\\sqrt{3}$$"}, {"identifier": "C", "content": "$$9-\\sqrt{3}$$"}, {"identifier": "D", "content": "$$11-\\sqrt{3}$$"}] | ["B"] | null | $$
\begin{aligned}
& \text { We have, }\left|x^2-8 x+15\right|-2 x+7=0 \\\\
& \Rightarrow |(x-3)(x-5)|-2 x+7=0
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkn5295/e47cf7ce-efd4-4752-9bbb-34637d3af9f9/c36b1990-6798-11ee-97fe-41fa1903ca4f/file-6y3zli1lnkn5296.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnkn5295/e47cf7ce-efd4-4752-9bbb-34637d3af9f9/c36b1990-6798-11ee-97fe-41fa1903ca4f/file-6y3zli1lnkn5296.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Morning Shift Mathematics - Quadratic Equation and Inequalities Question 19 English Explanation">
<br><br>Now, when $x \leq 3$ or $x \geq 5$, then
<br><br>$$
\begin{array}{ccrl}
& x^2-8 x+15-2 x+7 =0 \\\\
&\Rightarrow x^2-10 x+22=0 \\\\
&\Rightarrow x^2-10 x+25-3=0 \\\\
&\Rightarrow (x-5)^2-3=0 \\\\
&\Rightarrow (x-5)= \pm \sqrt{3} \\\\
&\Rightarrow x=5+\sqrt{3}, 5-\sqrt{3}
\end{array}
$$
<br><br>Since, $x \leq 3$ or $x \geq 5$
<br><br>$$
\therefore x=5+\sqrt{3}
$$
<br><br>When, $3 < x < 5$, then
<br><br>$$
\begin{array}{rlrl}
& -\left(x^2-8 x+15\right)-2 x+7 =0 \\\\
& \Rightarrow -x^2+8 x-15-2 x+7 =0 \\\\
& \Rightarrow -x^2+6 x-8 =0 \\\\
& \Rightarrow x^2-6 x+8 =0 \\\\
& \Rightarrow (x-4)(x-2) =0 \Rightarrow x=2,4
\end{array}
$$
<br><br>Since, $3 < x < 5$
<br><br>$$
\therefore x=4
$$
<br><br>$$
\therefore \text { Sum of roots }=(5+\sqrt{3})+4=9+\sqrt{3}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
lsamarl9 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $\alpha$ and $\beta$ be the roots of the equation $p x^2+q x-r=0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$, then the value of $(\alpha-\beta)^2$ is : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "$\\frac{20}{3}$"}, {"identifier": "D", "content": "$\\frac{80}{9}$"}] | ["D"] | null | Given : $p x^2+q x-r=0$
<br/><br/>Let $p=\frac{a}{r_1}, q=a, r=a r_1$
<br/><br/>$\begin{aligned} & \text { and } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4} \\\\ & \Rightarrow \frac{q}{r}=\frac{3}{4} \\\\ & \Rightarrow \frac{1}{r_1}=\frac{3}{4} \\\\ & \Rightarrow r_1=\frac{4}{3}\end{aligned}$
<br/><br/>$\begin{aligned}(\alpha-\beta)^2 & =(\alpha+\beta)^2-4 \alpha \beta \\\\ & =\left(\frac{-q}{p}\right)^2-4\left(\frac{-r}{p}\right) \\\\ & =\frac{q^2}{p^2}+\frac{4 r}{p} \\\\ & =r_1^2+4 r_1^2=5 r_1^2 \\\\ & =5\left(\frac{4}{3}\right)^2=\frac{80}{9}\end{aligned}$ | mcq | jee-main-2024-online-1st-february-evening-shift |
lsaolop2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $\mathbf{S}=\left\{x \in \mathbf{R}:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>Notice that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are reciprocals of each other because :</p>
<ul>
<li>$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$</li>
</ul>
<br/><strong>Using the Reciprocal Property :</strong>
<br/><ul>
<br/><li>This means $(\sqrt{3} - \sqrt{2})^x = \frac{1}{(\sqrt{3} + \sqrt{2})^x} $</li>
</ul>
<br/><strong>Let</strong> $a = (\sqrt{3} + \sqrt{2})^x $. The equation given in the problem becomes :
<ul>
<br/><li> $a + \frac{1}{a} = 10$</li>
</ul>
<br/><strong>Simplifying</strong>
<p>Multiplying both sides by *a*, we get a quadratic equation :</p>
<ul>
<li> $a^2 + 1 = 10a$</li><br>
<li> $a^2 - 10a + 1 = 0$</li>
</ul>
<br/><strong>Solving the Quadratic</strong>
<p>Using the quadratic formula, we find :</p>
<ul>
<li> $a = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$</li>
</ul>
<br/><strong>Possible values of x</strong>
<p>Since $a = (\sqrt{3} + \sqrt{2})^x $, we have two cases :</p>
<ol>
<li>$(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$ = $(\sqrt{3} + \sqrt{2})^2$. There is one real solution for x in this case which is x = 2.</li><br>
<li>$(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$ = $(\sqrt{3} - \sqrt{2})^2= \frac{1}{(\sqrt{3} + \sqrt{2})^2} $ = $(\sqrt{3} + \sqrt{2})^{-2}$. There is one real solution for x in this case which is x = - 2.</li>
</ol>
<br/><strong>Conclusion</strong>
<p>There are <strong>two</strong> real solutions for *x*. Therefore, the number of elements in the set S is <strong>2</strong>. This corresponds with option <strong>(C)</strong>.</p>
<p></p> | mcq | jee-main-2024-online-1st-february-morning-shift |
1lsgcitu2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta \in \mathbf{N}$$ be roots of the equation $$x^2-70 x+\lambda=0$$, where $$\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N}$$. If $$\lambda$$ assumes the minimum possible value, then $$\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$$ is equal to :</p> | [] | null | 60 | <p>$$\begin{aligned}
& x^2-70 x+\lambda=0 \\
& \alpha+\beta=70 \\
& \alpha \beta=\lambda \\
& \therefore \alpha(70-\alpha)=\lambda
\end{aligned}$$</p>
<p>Since, 2 and 3 does not divide $$\lambda$$</p>
<p>$$\therefore \alpha=5, \beta=65, \lambda=325$$</p>
<p>By putting value of $$\alpha, \beta, \lambda$$ we get the required value $$60$$.</p> | integer | jee-main-2024-online-30th-january-morning-shift |
luxwdj41 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta ; \alpha>\beta$$, be the roots of the equation $$x^2-\sqrt{2} x-\sqrt{3}=0$$. Let $$\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$$. Then $$(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$$ is equal to</p> | [{"identifier": "A", "content": "$$10 \\sqrt{3} \\mathrm{P}_9$$\n"}, {"identifier": "B", "content": "$$11 \\sqrt{3} \\mathrm{P}_9$$\n"}, {"identifier": "C", "content": "$$11 \\sqrt{2} \\mathrm{P}_9$$\n"}, {"identifier": "D", "content": "$$10 \\sqrt{2} \\mathrm{P}_9$$"}] | ["A"] | null | <p>$$\begin{aligned}
& x^2-\sqrt{2} x-\sqrt{3}=0 \\
& P_n=\alpha^n-\beta^n
\end{aligned}$$</p>
<p>$$\alpha$$ and $$\beta$$ are the roots of the equation</p>
<p>Using Newton's theorem</p>
<p>$$\begin{aligned}
& P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\
& \text { Put } n=10 \\
& P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0
\end{aligned}$$</p>
<p>$$P_{12}=\sqrt{2} P_{11}+\sqrt{3} P_{10}$$</p>
<p>Put $$n=9$$</p>
<p>$$\begin{aligned}
& P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\
& P_{11}=\sqrt{2} P_{10}+\sqrt{3} P_9 \\
& (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12}
\end{aligned}$$</p>
<p>Put the value of $$P_{12}$$ & $$P_{12}$$ in above equation.</p>
<p>$$\begin{aligned}
= & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10)\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right)-11\left(\sqrt{2} P_{11}+\sqrt{3} P_{10}\right) \\
= & 11 \sqrt{3} P_{10}-10 \sqrt{2} P_{10}+22 P_{10}+10 \sqrt{2} P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2} P_{11}-11 \sqrt{3} P_{10} \\
= & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2}\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right) \\
= & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-22 P_{10}-11 \sqrt{6} P_9 \\
= & 10 \sqrt{3} P_9
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luy6z57u | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be the roots of the equation $$x^2+2 \sqrt{2} x-1=0$$. The quadratic equation, whose roots are $$\alpha^4+\beta^4$$ and $$\frac{1}{10}(\alpha^6+\beta^6)$$, is:</p> | [{"identifier": "A", "content": "$$x^2-180 x+9506=0$$\n"}, {"identifier": "B", "content": "$$x^2-195 x+9506=0$$\n"}, {"identifier": "C", "content": "$$x^2-190 x+9466=0$$\n"}, {"identifier": "D", "content": "$$x^2-195 x+9466=0$$"}] | ["B"] | null | <p>$$\begin{aligned}
& x^2+2 \sqrt{2 x}-1=0 \\
& \alpha+\beta=-2 \sqrt{2} \text { and } \alpha \beta=-1 \\
& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\
& =8+2=10 \\
& \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2 \\
& =100-2=98 \\
& \alpha^6+\beta^6=\left(\alpha^2+\beta^2\right)^3-3 \alpha^2 \beta^2\left(\alpha^2+\beta^2\right) \\
& =1000-3(10) \\
& =970 \\
& \therefore \quad \frac{1}{10}\left(\alpha^6+\beta^6\right)=97
\end{aligned}$$</p>
<p>Equation whose roots are $$\alpha^4+\beta^4$$ and $$\frac{1}{10}\left(\alpha^6+\beta^6\right)$$ is</p>
<p>$$\begin{aligned}
& x^2-(98+97) x+98 \times 97=0 \\
& x^2-195 x+9506=0
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxd39 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If 2 and 6 are the roots of the equation $$a x^2+b x+1=0$$, then the quadratic equation, whose roots are $$\frac{1}{2 a+b}$$ and $$\frac{1}{6 a+b}$$, is :</p> | [{"identifier": "A", "content": "$$x^2+8 x+12=0$$\n"}, {"identifier": "B", "content": "$$2 x^2+11 x+12=0$$\n"}, {"identifier": "C", "content": "$$4 x^2+14 x+12=0$$\n"}, {"identifier": "D", "content": "$$x^2+10 x+16=0$$"}] | ["A"] | null | <p>Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots.</p>
<p>The given quadratic equation is:</p>
<p>$$a x^2 + b x + 1 = 0$$</p>
<p>By Vieta's formulas, the sum of the roots is:</p>
<p>$$2 + 6 = -\frac{b}{a}$$</p>
<p>So:</p>
<p>$$8 = -\frac{b}{a} \Rightarrow b = -8a$$</p>
<p>And the product of the roots is:</p>
<p>$$2 \times 6 = \frac{1}{a} \Rightarrow 12 = \frac{1}{a} \Rightarrow a = \frac{1}{12}$$</p>
<p>Therefore, $b = -8a = -8 \left( \frac{1}{12} \right) = -\frac{2}{3}$.</p>
<p>Given the roots of the new quadratic equation are:</p>
<p>$$\frac{1}{2a+b}$$ and $$\frac{1}{6a+b}$$</p>
<p>We know $a = \frac{1}{12}$ and $b = -\frac{2}{3}$, so:</p>
<p>$$2a + b = 2 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = \frac{1 - 4}{6} = -\frac{3}{6} = -\frac{1}{2}$$</p>
<p>and:</p>
<p>$$6a + b = 6 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = -\frac{1}{6}$$</p>
<p>Thus, the roots of the new quadratic equation are:</p>
<p>$$\frac{1}{-\frac{1}{2}} = -2$$</p>
<p>and:</p>
<p>$$\frac{1}{-\frac{1}{6}} = -6$$</p>
<p>The new quadratic equation with roots $-2$ and $-6$ can be formulated as:</p>
<p>$$x^2 - (\text{sum of roots}) x + (\text{product of roots}) = 0$$</p>
<p>The sum of the roots is:</p>
<p>$$-2 + (-6) = -8$$</p>
<p>The product of the roots is:</p>
<p>$$(-2) \times (-6) = 12$$</p>
<p>Thus, the quadratic equation becomes:</p>
<p>$$x^2 - (-8)x + 12 = x^2 + 8x + 12 = 0$$</p>
<p>Hence, the correct option is:</p>
<p>Option A</p>
<p>$$x^2 + 8 x + 12 = 0$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv5gst25 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The sum of all the solutions of the equation $$(8)^{2 x}-16 \cdot(8)^x+48=0$$ is :</p> | [{"identifier": "A", "content": "$$1+\\log _8(6)$$\n"}, {"identifier": "B", "content": "$$1+\\log _6(8)$$\n"}, {"identifier": "C", "content": "$$\\log _8(6)$$\n"}, {"identifier": "D", "content": "$$\\log _8(4)$$"}] | ["A"] | null | <p>First, let's start by substituting $$y = (8)^x$$ in the given equation. By substituting, the equation $$8^{2x} - 16 \cdot 8^x + 48 = 0$$ will be transformed into</p>
<p>$$ y^2 - 16y + 48 = 0 $$</p>
<p>Now, we have a quadratic equation in $$y$$. To find the roots of this quadratic equation, we can use the quadratic formula:</p>
<p>$$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$</p>
<p>In this case, $$a = 1$$, $$b = -16$$, and $$c = 48$$. Substituting these values into the formula, we get:</p>
<p>$$ y = \frac{16 \pm \sqrt{256 - 192}}{2} $$</p>
<p>$$ y = \frac{16 \pm \sqrt{64}}{2} $$</p>
<p>$$ y = \frac{16 \pm 8}{2} $$</p>
<p>Solving for the two possible values of $$y$$, we have:</p>
<p>$$ y = \frac{16 + 8}{2} = 12 $$</p>
<p>$$ y = \frac{16 - 8}{2} = 4 $$</p>
<p>Now, recall that we substituted $$y = (8)^x$$. So, we need to solve for $$x$$ when $$y = 12$$ and $$y = 4$$:</p>
<p>$$ 8^x = 12 $$</p>
<p>$$ x = \log_8(12) $$</p>
<p>$$ 8^x = 4 $$</p>
<p>$$ x = \log_8(4) $$</p>
<p>Therefore, the solutions for $$x$$ are $$\log_8(12)$$ and $$\log_8(4)$$. The sum of these solutions is:</p>
<p>$$ \log_8(12) + \log_8(4) $$</p>
<p>Using the logarithmic property that $$\log_b(m) + \log_b(n) = \log_b(m \cdot n)$$, we get:</p>
<p>$$ \log_8(12) + \log_8(4) = \log_8(12 \cdot 4) $$</p>
<p>$$ = \log_8(48) $$</p>
<p>Now, we note that:</p>
<p>$$ 48 = 8 \cdot 6 $$</p>
<p>Thus,</p>
<p>$$ \log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6) $$</p>
<p>Therefore, the sum of all the solutions of the equation is:</p>
<p>Option A $$1 + \log_8(6)$$.</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lvb2952b | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be roots of $$x^2+\sqrt{2} x-8=0$$. If $$\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$$, then $$\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$$ is equal to ________.</p> | [] | null | 4 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwapcytl/545aa1a6-edf5-4e15-9137-baf3d0f62d19/8d6a2390-144f-11ef-860c-d121cbcdd1fc/file-1lwapcytm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwapcytl/545aa1a6-edf5-4e15-9137-baf3d0f62d19/8d6a2390-144f-11ef-860c-d121cbcdd1fc/file-1lwapcytm.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 - Quadratic Equation and Inequalities Question 3 English Explanation"></p>
<p>$$\Rightarrow \alpha^2+\sqrt{2} \alpha=8$$</p>
<p>$$\begin{aligned}
\alpha+\beta=\sqrt{2}, \quad \alpha \beta=-8, & \Rightarrow \alpha+\sqrt{2}=\frac{8}{\alpha} \\
& \Rightarrow \beta+\sqrt{2}=\frac{8}{\beta}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{U_{10}+\sqrt{2} U_9}{2 U_8}=\frac{\alpha^{10}+\beta^{10}+\sqrt{2} \alpha^9+\sqrt{2} \beta^9}{2 \alpha^8+2 \beta^8} \\
& =\frac{\alpha^9(\alpha+\sqrt{2})+\beta^9(\beta+\sqrt{2})}{2\left(\alpha^8+\beta^8\right)} \\
& =\frac{\alpha^9 \cdot\left(\frac{8}{\alpha}\right)+\beta^9\left(\frac{8}{\beta}\right)}{2\left(\alpha^8+\beta^8\right)}=\frac{8}{2}=4
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-evening-shift |
lvc57aw6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be the distinct roots of the equation $$x^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R}$$ and $$a_n=\alpha^n+\beta^n$$. Then the minimum value of $$\frac{a_{2023}+a_{2025}}{a_{2024}}$$ is</p> | [{"identifier": "A", "content": "$$-1 / 2$$\n"}, {"identifier": "B", "content": "$$-1 / 4$$\n"}, {"identifier": "C", "content": "$$1 / 4$$\n"}, {"identifier": "D", "content": "$$1 / 2$$"}] | ["B"] | null | <p>$$\begin{aligned}
& x^2-\left(t^2-5 t+6\right) x+1=0 \\
& \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \\
& \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \\
& =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \\
& \text { Minimum value }=\frac{-1}{4}
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
YBEbaVOfFKiWxDON | maths | sequences-and-series | am,-gm-and-hm | If m is the A.M. of two distinct real numbers l and n $$(l,n > 1)$$ and $${G_1},{G_2}$$ and $${G_3}$$ are three geometric means between $$l$$ and n, then $$G_1^4\, + 2G_2^4\, + G_3^4$$ equals: | [{"identifier": "A", "content": "$$4\\,lm{n^2}$$ "}, {"identifier": "B", "content": "$$4\\,{l^2}{m^2}{n^2}$$ "}, {"identifier": "C", "content": "$$4\\,{l^2}m\\,n$$ "}, {"identifier": "D", "content": "$$4\\,l\\,{m^2}n$$ "}] | ["D"] | null | $$m = {{l + n} \over 2}$$ and common ratio of
<br><br>$$G.P.$$ $$ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$
<br><br>$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$
<br><br>$$G_1^4 + 2G_2^4 + G_3^4$$
<br><br>$$ = {l^3}n + 2{l^2}{n^2} + {\ln ^3}$$
<br><br>$$ = \ln {\left( {1 + n} \right)^2}$$
<br><br>$$ = \ln \times 2{m^2}$$
<br><br>$$ = 4l{m^2}n$$ | mcq | jee-main-2015-offline |
UvIf51uHbb6DvoeXkhC7K | maths | sequences-and-series | am,-gm-and-hm | Let x, y, z be positive real numbers such that x + y + z = 12 and x<sup>3</sup>y<sup>4</sup>z<sup>5</sup> = (0.1) (600)<sup>3</sup>. Then x<sup>3</sup> + y<sup>3</sup> + z<sup>3</sup>is equal to : | [{"identifier": "A", "content": "270"}, {"identifier": "B", "content": "258"}, {"identifier": "C", "content": "342"}, {"identifier": "D", "content": "216"}] | ["D"] | null | As we know
<br><br>AM $$ \ge $$ GM
<br><br>$$ \Rightarrow $$ $${{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}$$ $$ \ge $$ $${\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}$$
<br><br>$$ \Rightarrow $$ 1 $$ \ge $$ $${{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}$$
<br><br>$$ \Rightarrow $$ x<sup>3</sup> y<sup>4</sup> z<sup>5</sup> $$ \le $$ 3<sup>3</sup> . 4<sup>4</sup> . 5<sup>5</sup>
<br><br>$$ \Rightarrow $$ x<sup>3</sup> y<sup>4</sup> z<sup>5</sup> $$ \le $$ (0.1)(600)<sup>3</sup>
<br><br>but given that,
<br><br>x<sup>3</sup> y<sup>4</sup> z<sup>5</sup> = (0.1) (600)<sup>3</sup>
<br><br>$$ \therefore $$ AM $$=$$ GM
<br><br>$$ \Rightarrow $$ All the number are equal.
<br><br>$$ \therefore $$ $${x \over 3} = {y \over 4} = {z \over 5} = k$$
<br><br>$$ \Rightarrow $$ x $$=$$ 3k, y = 4k, z = 5k
<br><br>given that,
<br><br> x + y + z $$=$$ 12
<br><br>$$ \Rightarrow $$ 3k + 4k + 5k $$=$$ 12
<br><br>$$ \Rightarrow $$ 12k $$=$$ 12
<br><br>$$ \Rightarrow $$ k = 1
<br><br>$$ \therefore $$ x $$=$$ 3, y $$=$$ 4, z $$=$$ 5
<br><br>So, x<sup>3</sup> + y<sup>3</sup> + z<sup>3</sup>
<br><br>$$=$$ 3<sup>3</sup> + 4<sup>3</sup> + 5<sup>3</sup>
<br><br>$$=$$ 216 | mcq | jee-main-2016-online-9th-april-morning-slot |
DhoQKCuInbj9DsKERyHLW | maths | sequences-and-series | am,-gm-and-hm | If A > 0, B > 0 and A + B = $${\pi \over 6}$$, <br/><br/>then the minimum value of tanA + tanB is : | [{"identifier": "A", "content": "$$\\sqrt 3 - \\sqrt 2 $$ "}, {"identifier": "B", "content": "$$2 - \\sqrt 3 $$"}, {"identifier": "C", "content": "$$4 - 2\\sqrt 3 $$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$ "}] | ["C"] | null | Given,
<br><br>A + B = $${\pi \over 6}$$
<br><br>$$ \therefore $$ tan(A + B) = tan$$\left( {{\pi \over 6}} \right)$$ = $${1 \over {\sqrt 3 }}$$
<br><br>We know,
<br><br>tan(A + B) = $${{\tan A + \tan B} \over {1 - \tan A\tan B}}$$
<br><br>$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}$$ = $${y \over {1 - \tan A\tan B}}$$
<br><br>where y = tan A + tan B
<br><br>$$ \Rightarrow $$ tanA tanB = 1 $$-$$ $$\sqrt 3 $$ y
<br><br>Also AM $$ \ge $$ GM
<br><br>$$ \Rightarrow $$ $${{\tan A + \tan B} \over 2} \ge \sqrt {\tan A\tan B} $$
<br><br>$$ \Rightarrow $$ y $$ \ge $$ 2$$\sqrt {1 - \sqrt 3 y} $$
<br><br>$$ \Rightarrow $$ y<sup>2</sup> $$ \ge $$ 4 $$-$$ 4$${\sqrt 3 y}$$
<br><br>$$ \Rightarrow $$ y<sup>2</sup> + 4$${\sqrt 3 y}$$ $$-$$ 4 $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ y $$ \le $$ $$-$$ 2$$\sqrt 3 $$ $$-$$ 4
<br><br>or y $$ \ge $$ $$-$$ 2$$\sqrt 3 $$ + 4
<br><br>(y $$ \le $$ $$-$$ 2$$\sqrt 3 $$ $$-$$ 4 is not possible as tan B > 0) | mcq | jee-main-2016-online-10th-april-morning-slot |
wuYGlr4ZcEF2WUJlcQDCt | maths | sequences-and-series | am,-gm-and-hm | If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric mean, then $${{a + b} \over {a - b}}$$ is equal to : | [{"identifier": "A", "content": "$${{\\sqrt 6 } \\over 2}$$ "}, {"identifier": "B", "content": "$${{3\\sqrt 2 } \\over 4}$$"}, {"identifier": "C", "content": "$${{7\\sqrt 3 } \\over {12}}$$"}, {"identifier": "D", "content": "$${{5\\sqrt 6 } \\over {12}}$$ "}] | ["D"] | null | A.T.Q.,
<br><br>A.M. = 5G.M.
<br><br>$${{a + b} \over 2} = 5\sqrt {ab} $$
<br><br>$${{a + b} \over {\sqrt {ab} }}$$ $$ = 10$$
<br><br>$$ \therefore $$ $${a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}$$
<br><br>Use componendo and Dividendo
<br><br>$${{a + b} \over {a - b}} = {{20} \over {8\sqrt 6 }} = {5 \over {2\sqrt 6 }} = {{5\sqrt 6 } \over {12}}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
GNsf4xF6cvZXlPv0FTnFO | maths | sequences-and-series | am,-gm-and-hm | Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression $${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}}$$ is :
| [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$${{m + n} \\over {6mn}}$$"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}} = {1 \over {\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right)}} \le {1 \over 4}$$
<br><br>using AM $$ \ge $$ GM | mcq | jee-main-2019-online-11th-january-evening-slot |
TQWG3sOpx0jCpXORb97nm | maths | sequences-and-series | am,-gm-and-hm | If sin<sup>4</sup>$$\alpha $$ + 4 cos<sup>4</sup>$$\beta $$ + 2 = 4$$\sqrt 2 $$ sin $$\alpha $$ cos $$\beta $$; $$\alpha $$, $$\beta $$ $$ \in $$ [0, $$\pi $$],
<br/>then cos($$\alpha $$ + $$\beta $$) $$-$$ cos($$\alpha $$ $$-$$ $$\beta $$) is equal to : | [{"identifier": "A", "content": "$$ - \\sqrt 2 $$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$-$$ 1"}, {"identifier": "D", "content": "$$\\sqrt 2 $$ "}] | ["A"] | null | A.M. $$ \ge $$ G.M.
<br><br>$${{{{\sin }^4}\alpha + 4{{\cos }^4}\beta + 1 + 1} \over 4} \ge {\left( {{{\sin }^4}\alpha .4{{\cos }^4}\beta .1.1} \right)^{{1 \over 4}}}$$
<br><br>sin<sup>4</sup><sup></sup>$$\alpha $$ + 4 cos<sup>2</sup>$$\beta $$ + 2 $$ \ge $$ 4 $$\sqrt 2 $$ sin $$\alpha $$ cos $$\beta $$
<br><br>Given that sin<sup>4</sup>$$\alpha $$ + 4cos<sup>4</sup>$$\beta $$ + 2 = 4$$\sqrt 2 $$ sin$$\alpha $$ cos$$\beta $$
<br><br>$$ \Rightarrow $$ A.M.=G.M. $$ \Rightarrow $$ sin<sup>4</sup><sup></sup><sup></sup>$$\alpha $$ = 1 = 4 cos<sup>4</sup> $$\beta $$
<br><br>sin $$\alpha $$ = 1, cos $$\beta $$ = $$ \pm $$ $${1 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow $$ sin$$\beta $$ = $${1 \over {\sqrt 2 }}$$ as $$\beta $$ $$ \in $$ [0, $$\pi $$]
<br><br>cos($$\alpha $$ + $$\beta $$) $$-$$ cos ($$\alpha $$ $$-$$ $$\beta $$) = $$-$$ 2 sin $$\alpha $$ $$\beta $$
<br><br>= $$ - \sqrt 2 $$ | mcq | jee-main-2019-online-12th-january-evening-slot |
PS85QIo4S7Avsvyt1bjgy2xukf49qm01 | maths | sequences-and-series | am,-gm-and-hm | If m arithmetic means (A.Ms) and three
geometric means (G.Ms) are inserted between
3 and 243 such that 4<sup>th</sup> A.M. is equal to 2<sup>nd</sup>
G.M., then m is equal to _________ . | [] | null | 39 | Given m arithmetic means (A.Ms) present between 3 and 243<br><br>$$ \therefore $$ Common difference, $$d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$$<br><br>$$ \therefore $$ 4th A.M. = a + 4d<br><br>= 3 + 4 $$ \times $$ $${{240} \over {m + 1}}$$<br><br>Also there are 3 G.M between 3 and 243<br><br>$$ \therefore $$ Common ratio (r) = $${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$$<br><br>where n = number of G.M inserted.<br><br>$$ \therefore $$ r = $${\left( {{{243} \over 3}} \right)^{{1 \over {3 + 1}}}} = 3$$<br><br>Given, <br><br>4<sup>th</sup> A.M = 2<sup>nd</sup> G.M<br><br>$$ \Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$$<br><br>$$ \Rightarrow {{960} \over {m + 1}} = 24$$<br><br>$$ \Rightarrow m = 39$$ | integer | jee-main-2020-online-3rd-september-evening-slot |
kAHFV06ZK5AUlNQmT8jgy2xukfahb1zf | maths | sequences-and-series | am,-gm-and-hm | The minimum value of 2<sup>sinx</sup> + 2<sup>cosx</sup> is : | [{"identifier": "A", "content": "$${2^{-1 + \\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${2^{1 - {1 \\over {\\sqrt 2 }}}}$$"}, {"identifier": "C", "content": "$${2^{1 - \\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${2^{-1 + {1 \\over {\\sqrt 2 }}}}$$"}] | ["B"] | null | Using AM $$ \ge $$ GM<br><br>$$ \Rightarrow {{{2^{\sin \,x}} + {2^{\cos \,x}}} \over 2} \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $$<br><br>$$ \Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \left( {{{\sin x + \cos x} \over 2}} \right)}}$$<br><br>$$ \Rightarrow \min ({2^{\sin x}} + {2^{\cos x}}) = {2^{1 - {1 \over {\sqrt 2 }}}}$$
<br><br>As we know range of sin x + cos x is :
<br><br>$$ - \sqrt 2 $$ $$ \le $$ sin x + cos x $$ \le $$ $$\sqrt 2 $$.
<br><br>So Minimum value of sin x + cos x = $$ - \sqrt 2 $$ | mcq | jee-main-2020-online-4th-september-evening-slot |
FLfk7pqpuOUDRypvuS1klt80kz2 | maths | sequences-and-series | am,-gm-and-hm | The minimum value of $$f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$$, where a, $$x \in R$$ and a > 0, is equal to : | [{"identifier": "A", "content": "$$a + {1 \\over a}$$"}, {"identifier": "B", "content": "2a"}, {"identifier": "C", "content": "a + 1"}, {"identifier": "D", "content": "$$2\\sqrt a $$"}] | ["D"] | null | We know, $$AM \ge GM$$<br><br>$$ \therefore $$ $${{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}} $$
<br><br>$$\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a $$ | mcq | jee-main-2021-online-25th-february-evening-slot |
MYzMY4FrIZlEIa93ir1kluynti6 | maths | sequences-and-series | am,-gm-and-hm | If the arithmetic mean and geometric mean of the p<sup>th</sup> and q<sup>th</sup> terms of the <br/>sequence $$-$$16, 8, $$-$$4, 2, ...... satisfy the equation<br/> 4x<sup>2</sup> $$-$$ 9x + 5 = 0, then p + q is equal to __________. | [] | null | 10 | Given, $$4{x^2} - 9x + 5 = 0$$<br><br>$$ \Rightarrow (x - 1)(4x - 5) = 0$$<br><br>$$ \Rightarrow $$ A. M. $$ = {5 \over 4}$$, G. M. = 1 (As A. M. $$ \ge $$ G. M)<br><br>Again, for the series<br><br>$$-$$16, 8, $$-$$4, 2 ..........<br><br>$${p^{th}}$$ term $${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$$<br><br>$${q^{th}}$$ term $${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$$<br><br>Now, A. M. = $${{{t_p} + {t_q}} \over 2} = {5 \over 4}$$ & G. M. = $$\sqrt {{t_p}{t_q}} = 1$$<br><br>$$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$$<br><br>$$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$$<br><br>$$ \Rightarrow p + q = 10$$ | integer | jee-main-2021-online-26th-february-evening-slot |
1l55h7hrj | maths | sequences-and-series | am,-gm-and-hm | <p>If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is :</p> | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "23"}, {"identifier": "D", "content": "24"}] | ["C"] | null | <p>a, A<sub>1</sub>, A<sub>2</sub> ........... A<sub>n</sub>, 100</p>
<p>Let d be the common difference of above A.P. then</p>
<p>$${{a + d} \over {100 - d}} = {1 \over 7}$$</p>
<p>$$ \Rightarrow 7a + 8d = 100$$ ...... (i)</p>
<p>and $$a + n = 33$$ ..... (ii)</p>
<p>and $$100 = a + (n + 1)d$$</p>
<p>$$ \Rightarrow 100 = a + (34 - a){{(100 - 7a)} \over 8}$$</p>
<p>$$ \Rightarrow 800 = 8a + 7{a^2} - 338a + 3400$$</p>
<p>$$ \Rightarrow 7{a^2} - 330a + 2600 = 0$$</p>
<p>$$ \Rightarrow a = 10,\,{{260} \over 7},$$ but $$a \ne {{260} \over 7}$$</p>
<p>$$\therefore$$ $$n = 23$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l5b7vzyn | maths | sequences-and-series | am,-gm-and-hm | <p>Let x, y > 0. If x<sup>3</sup>y<sup>2</sup> = 2<sup>15</sup>, then the least value of 3x + 2y is</p> | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "40"}] | ["D"] | null | <p>x, y > 0 and x<sup>3</sup>y<sup>2</sup> = 2<sup>15</sup></p>
<p>Now, 3x + 2y = (x + x + x) + (y + y)</p>
<p>So, by A.M $$\ge$$ G.M inequality</p>
<p>$${{3x + 2y} \over 5} \ge \root 5 \of {{x^3}\,.\,{y^2}} $$</p>
<p>$$\therefore$$ $$3x + 2y \ge 5\root 5 \of {{2^{15}}} \ge 40$$</p>
<p>$$\therefore$$ Least value of $$3x + 4y = 40$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l6gglr6d | maths | sequences-and-series | am,-gm-and-hm | <p>Consider two G.Ps. 2, 2<sup>2</sup>, 2<sup>3</sup>, ..... and 4, 4<sup>2</sup>, 4<sup>3</sup>, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is $${(2)^{{{225} \over 8}}}$$, then $$\sum\limits_{k = 1}^n {k(n - k)} $$ is equal to :</p> | [{"identifier": "A", "content": "560"}, {"identifier": "B", "content": "1540"}, {"identifier": "C", "content": "1330"}, {"identifier": "D", "content": "2600"}] | ["C"] | null | <p>Given G.P's 2, 2<sup>2</sup>, 2<sup>3</sup>, .... 60 terms</p>
<p>4, 4<sup>2</sup>, .... n terms</p>
<p>Now, G.M $$ = {2^{{{225} \over 8}}}$$</p>
<p>$${\left( {{{2.2}^2}...\,{{4.4}^2}...} \right)^{{1 \over {60 + n}}}} = {2^{{{225} \over 8}}}$$</p>
<p>$$\left( {{2^{{{{n^2} + n + 1830} \over {60 + n}}}}} \right) = {2^{{{225} \over 8}}}$$</p>
<p>$$ \Rightarrow {{{n^2} + n + 1830} \over {60 + n}} = {{225} \over 8}$$</p>
<p>$$ \Rightarrow 8{n^2} - 217n + 1140 = 0$$</p>
<p>$$n = {{57} \over 8},\,20,\,$$ so $$n = 20$$</p>
<p>$$\therefore$$ $$\sum\limits_{k = 1}^{20} {k(20 - k) = 20 \times {{20 \times 21} \over 2} - {{20 \times 21 \times 41} \over 6}} $$</p>
<p>$$ = {{20 \times 21} \over 2}\left[ {20 - {{41} \over 3}} \right] = 1330$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
lgnybkbk | maths | sequences-and-series | am,-gm-and-hm | Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}, G_{3}$ be three geometric<br/><br/> means of two distinct positive numbers. Then $G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2} G_{3}^{2}$ is equal to : | [{"identifier": "A", "content": "$\\left(A_{1}+A_{2}\\right)^{2} G_{1} G_{3}$"}, {"identifier": "B", "content": "$\\left(A_{1}+A_{2}\\right) G_{1}^{2} G_{3}^{2}$"}, {"identifier": "C", "content": "$2\\left(A_{1}+A_{2}\\right) G_{1}^{2} G_{3}^{2}$"}, {"identifier": "D", "content": "$2\\left(A_{1}+A_{2}\\right) G_{1} G_{3}$"}] | ["A"] | null | <p>Now, we have the following relations :</p>
<p>Arithmetic progression :</p>
<p>Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$, we can say that $a$, $A_1$, $A_2$, and $b$ are in an arithmetic progression. This means there are three equal intervals between $a$ and $b$, which are represented by the common difference $d$.</p>
<p>To find the value of $d$, we can use the following equation :</p>
<p>$$
b - a = 3d
$$</p>
<p>From this equation, we can find the value of $d$ :
</p>
<p>$$
d = \frac{b - a}{3}
$$</p>
<p>$$
A_1 = a + \frac{b - a}{3} = \frac{2a + b}{3}
$$</p>
<p>$$
A_2 = \frac{a + 2b}{3}
$$</p>
<p>$$
A_1 + A_2 = a + b
$$</p>
<p>Geometric progression :</p>
<p>$$
a, G_1, G_2, G_3, b \text{ are in G.P. }
$$</p>
<p>$$
r = \left(\frac{b}{a}\right)^{\frac{1}{4}}
$$</p>
<p>$$
G_1 = \left(a^3b\right)^{\frac{1}{4}}
$$</p>
<p>$$
G_2 = \left(a^2b^2\right)^{\frac{1}{4}}
$$</p>
<p>$$
G_3 = \left(ab^3\right)^{\frac{1}{4}}
$$</p>
<p>We have the expression :</p>
<p>$$
G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^3b + a^2b^2 + ab^3 + \left(a^3b\right)^{\frac{1}{2}}\cdot\left(ab^3\right)^{\frac{1}{2}}
$$</p>
<p>Simplify the expression :</p>
<p>$$
a^3b + a^2b^2 + ab^3 + ab(a^2b^2)
$$</p>
<p>Factor out $ab$:</p>
<p>$$
ab(a^2 + ab + b^2 + a^2b^2)
$$</p>
<p>Combine the terms :</p>
<p>$$
ab(a^2 + 2ab + b^2)
$$</p>
<p>Rewrite the expression using the sum of squares :</p>
<p>$$
ab(a + b)^2
$$</p>
<p>Now, recall that $A_1 + A_2 = a + b$. Substitute this into the expression :</p>
<p>$$
G_1 \cdot G_3 \cdot (A_1 + A_2)^2
$$</p> | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgsuajhg | maths | sequences-and-series | am,-gm-and-hm | <p>Let $$a, b, c$$ and $$d$$ be positive real numbers such that $$a+b+c+d=11$$. If the maximum value of $$a^{5} b^{3} c^{2} d$$ is $$3750 \beta$$, then the value of $$\beta$$ is</p> | [{"identifier": "A", "content": "110"}, {"identifier": "B", "content": "108"}, {"identifier": "C", "content": "90"}, {"identifier": "D", "content": "55"}] | ["C"] | null | Given that $$a+b+c+d=11$$ and the maximum value of $$a^5 b^3 c^2 d$$ is $$3750\beta$$, you assumed the numbers to be $$\frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d$$.
<br/><br/>Applying the AM-GM inequality:
<br/><br/>$$\frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+d}{11} \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}$$
<br/><br/>Since $$a+b+c+d=11$$, we have:
<br/><br/>$$1 \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}$$
<br/><br/>Now, raising both sides to the power of 11:
<br/><br/>$$1^{11} \geq \frac{a^5 b^3 c^2 d}{5^5 3^3 2^2 1}$$
<br/><br/>From the given information, we know that $$a^5 b^3 c^2 d \geq 3750\beta$$:
<br/><br/>$$5^5 3^3 2^2 \geq 3750\beta$$
<br/><br/>Now, we can solve for $$\beta$$:
<br/><br/>$$\beta \leq \frac{1}{3750} \cdot 5^5 3^3 2^2$$
<br/><br/>Since we are looking for the maximum value of $$\beta$$, we take the equality case:
<br/><br/>$$\beta = \frac{1}{3750} \cdot 5^5 3^3 2^2$$
<br/><br/>Calculating the value, we find that:
<br/><br/>$$\beta = 90$$
<br/><br/>So, the value of $$\beta$$ is 90. | mcq | jee-main-2023-online-11th-april-evening-shift |
lsappvw8 | maths | sequences-and-series | am,-gm-and-hm | Let $3, a, b, c$ be in A.P. and $3, a-1, b+1, c+9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is : | [{"identifier": "A", "content": "-4"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "11"}] | ["D"] | null | <p>Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:</p>
<p>$$ d = a - 3 $$</p>
<p>The nth term of an A.P. is given by the formula:</p>
<p>$$ T_n = a + (n-1)d $$</p>
<p>So, using this formula, we can express $b$ and $c$ in terms of $a$ and $d$:</p>
<p>$$ b = a + d $$</p>
<p>$$ c = a + 2d $$</p>
<p>Substituting $d = a - 3$ into these expressions:</p>
<p>$$ b = a + (a - 3) $$</p>
<p>$$ c = a + 2(a - 3) $$</p>
<p>Therefore:</p>
<p>$$ b = 2a - 3 $$</p>
<p>$$ c = 3a - 6 $$</p>
<p>Now, let's consider that $3, a-1, b+1, c+9$ are in geometric progression (G.P.). For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant. So:</p>
<p>$$ \frac{a - 1}{3} = \frac{b + 1}{a - 1} = \frac{c + 9}{b + 1} $$</p>
<p>Now, we will establish the relation between the terms using the property of G.P.:</p>
<p>$$ \frac{a - 1}{3} = \frac{b + 1}{a - 1} $$</p>
<p>$$ (a - 1)^2 = 3(b + 1) $$</p>
<p>$$ a^2 - 2a + 1 = 3b + 3 $$</p>
<p>Substituting $b = 2a - 3$, we get:</p>
<p>$$ a^2 - 2a + 1 = 3(2a - 3) + 3 $$</p>
<p>$$ a^2 - 2a + 1 = 6a - 9 + 3 $$</p>
<p>$$ a^2 - 8a + 7 = 0 $$</p>
<p>Solving this quadratic equation:</p>
<p>$$ (a - 7)(a - 1) = 0 $$</p>
<p>Hence, $a = 7$ or $a = 1$. However, if $a = 1$, the terms $3, a-1, b+1, c+9$ cannot form a G.P. as it would involve division by zero. Therefore, $a = 7$. We use this value to find $b$ and $c$:</p>
<p>$$ b = 2a - 3 = 2(7) - 3 = 14 - 3 = 11 $$</p>
<p>$$ c = 3a - 6 = 3(7) - 6 = 21 - 6 = 15 $$</p>
<p>Now we can find the arithmetic mean ($A$) of $a$, $b$, and $c$:</p>
<p>$$ A = \frac{a + b + c}{3} $$</p>
<p>$$ A = \frac{7 + 11 + 15}{3} $$</p>
<p>$$ A = \frac{33}{3} $$</p>
<p>$$ A = 11 $$</p>
<p>Hence, the arithmetic mean of $a$, $b$, and $c$ is $11$, which corresponds to Option D.</p> | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lse508pa | maths | sequences-and-series | am,-gm-and-hm | <p>For $$0 < c < b < a$$, let $$(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$$ and $$\alpha \neq 1$$ be one of its root. Then, among the two statements</p>
<p>(I) If $$\alpha \in(-1,0)$$, then $$b$$ cannot be the geometric mean of $a$ and $$c$$</p>
<p>(II) If $$\alpha \in(0,1)$$, then $$b$$ may be the geometric mean of $$a$$ and $$c$$</p> | [{"identifier": "A", "content": "only (II) is true\n"}, {"identifier": "B", "content": "Both (I) and (II) are true\n"}, {"identifier": "C", "content": "only (I) is true\n"}, {"identifier": "D", "content": "Neither (I) nor (II) is true"}] | ["B"] | null | <p>$$\begin{aligned}
& f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\
& f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\
& f(1)=0 \\
& \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\
& \alpha=\frac{c+a-2 b}{a+b-2 c} \\
& \text { If, }-1<\alpha<0 \\
& -1<\frac{c+a-2 b}{a+b-2 c}<0 \\
& b+c<2 a \text { and } b>\frac{a+c}{2}
\end{aligned}$$</p>
<p>therefore, b cannot be G.M. between a and c.</p>
<p>$$\begin{aligned}
& \text { If, } 0<\alpha<1 \\
& 0<\frac{c+a-2 b}{a+b-2 c}<1 \\
& b>c \text { and } b<\frac{a+c}{2}
\end{aligned}$$</p>
<p>Therefore, $$\mathrm{b}$$ may be the G.M. between $$\mathrm{a}$$ and $$\mathrm{c}$$.</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
lv2erzp0 | maths | sequences-and-series | am,-gm-and-hm | <p>Let three real numbers $$a, b, c$$ be in arithmetic progression and $$a+1, b, c+3$$ be in geometric progression. If $$a>10$$ and the arithmetic mean of $$a, b$$ and $$c$$ is 8, then the cube of the geometric mean of $$a, b$$ and $$c$$ is</p> | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "316"}, {"identifier": "C", "content": "312"}, {"identifier": "D", "content": "128"}] | ["A"] | null | <p>$$\begin{aligned}
& 2 b=a+c \quad \text{.... (1)}\\
& b^2=(a+1)(c+3) \quad \text{.... (2)}\\
& \frac{a+b+c}{3}=8 \quad \text{.... (3)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow & \frac{3 b}{3}=8 \\
& b=8 \\
\Rightarrow \quad & a c+3 a+c+3=64
\end{aligned}$$</p>
<p>$$\begin{aligned}
& 3 a+c+a c=61 \quad \text{... (4)}\\
& a+c=16 \\
& c=16-a
\end{aligned}$$</p>
<p>from equation (4)</p>
<p>$$\begin{aligned}
& 3 a+16-a+a(16-a)=61 \\
& \Rightarrow \quad(a-15)(a-3)=0 \\
& \quad a=15(a>10) \\
& \Rightarrow \quad a=15, b=8, c=1 \\
& \left((a \cdot b \cdot c)^{\frac{1}{3}}\right)^3=15 \times 8 \times 1=120
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
vlG8kHqWShZMhOkT | maths | sequences-and-series | arithmetic-progression-(a.p) | If 1, $${\log _9}\,\,({3^{1 - x}} + 2),\,\,{\log _3}\,\,({4.3^x} - 1)$$ are in A.P. then x equals | [{"identifier": "A", "content": "$${\\log _3}\\,4\\,\\,\\,$$ "}, {"identifier": "B", "content": "$$1 - \\,{\\log _3}\\,4\\,$$ "}, {"identifier": "C", "content": "$$1 - \\,{\\log _4}\\,3$$ "}, {"identifier": "D", "content": "$${\\log _4}\\,3$$ "}] | ["B"] | null | $$1,\,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log _3}\left( {{{4.3}^x} - 1} \right)$$ are in $$A.P.$$
<br><br>$$ \Rightarrow 2{\log _9}\left( {{3^{1 - x}} + 2} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,$$ $$ = 1 + {\log _3}\left( {{{4.3}^x} - 1} \right)$$
<br><br>$$ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,$$ $$ = {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right)$$
<br><br>$$ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,$$ $$ = {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right]$$
<br><br>$$ \Rightarrow {3^{1 - x}} + 2 = 3\,\left( {{{4.3}^x} - 1} \right)$$
<br><br>$$ \Rightarrow {3.3^{ - x}} + 2 = {12.3^x} - 3.$$
<br><br>Put $${3^x} = t$$
<br><br>$$ \Rightarrow {3 \over t} + 2 = 12t - 3$$
<br><br>or $$12{t^2} - 5t - 3 = 0;$$
<br><br>Hence $$t = - {1 \over 3},{3 \over 4} \Rightarrow {3^x} = {3 \over 4}$$
<br><br>(as $${3^x}\,\, \ne \,\, - ve$$ )
<br><br>$$ \Rightarrow x = {\log _3}\left( {{3 \over 4}} \right)$$
<br><br>or $$x = {\log _3}3 - {\log _3}4$$
<br><br>$$ \Rightarrow x = 1 - {\log _3}4$$ | mcq | aieee-2002 |
wcwhgSoE2D9djeuf | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${{T_r}}$$ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, $$m \ne n,\,\,{T_m} = {1 \over n}\,\,and\,{T_n} = {1 \over m},\,$$ then a - d equals | [{"identifier": "A", "content": "$${1 \\over m} + {1 \\over n}$$ "}, {"identifier": "B", "content": "1 "}, {"identifier": "C", "content": "$${1 \\over {m\\,n}}$$ "}, {"identifier": "D", "content": "0 "}] | ["D"] | null | $${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$$
<br><br>$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$
<br><br>$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$
<br><br>$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$$
<br><br>From $$\left( 1 \right)$$ $$a = {1 \over {mn}} \Rightarrow a - d = 0$$ | mcq | aieee-2004 |
Xm2Y0pl2ITjxWaRb | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${a_1}$$, $${a_2}$$, $${a_3}$$.....be terms on A.P. If $${{{a_1} + {a_2} + .....{a_p}} \over {{a_1} + {a_2} + .....{a_q}}} = {{{p^2}} \over {{q^2}}},\,p \ne q,\,then\,{{{a_6}} \over {{a_{21}}}}\,$$ equals | [{"identifier": "A", "content": "$${{41} \\over {11}}$$ "}, {"identifier": "B", "content": "$${7 \\over 2}$$ "}, {"identifier": "C", "content": "$${2 \\over 7}$$ "}, {"identifier": "D", "content": "$${{11} \\over {41}}$$ "}] | ["D"] | null | $${{{p \over 2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]} \over {{q \over 2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = {{{p^2}} \over {{q^2}}}$$
<br><br>$$ \Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}$$
<br><br>$${{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}$$
<br><br>For $${{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41$$
<br><br>$$ \Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}$$ | mcq | aieee-2006 |
AdurnkJA7QNZXqLq | maths | sequences-and-series | arithmetic-progression-(a.p) | A person is to count 4500 currency notes. Let $${a_n}$$ denote the number of notes he counts in the $${n^{th}}$$ minute. If $${a_1}$$ = $${a_2}$$ = ....= $${a_{10}}$$= 150 and $${a_{10}}$$, $${a_{11}}$$,.... are in an AP with common difference - 2, then the time taken by him to count all notes is | [{"identifier": "A", "content": "34 minutes"}, {"identifier": "B", "content": "125 minutes"}, {"identifier": "C", "content": "135 minutes "}, {"identifier": "D", "content": "24 minutes "}] | ["A"] | null | Till $$10$$<sup>th</sup> minute number of counted notes $$ = 1500$$
<br><br>$$3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = n\left[ {148 - n + 1} \right]$$
<br><br>$$ \Rightarrow $$$${n^2} - 149n + 3000 = 0$$
<br><br>$$ \Rightarrow n = 125,24$$
<br><br>But $$n=125$$ is not possible
<br><br>$$\therefore$$ total time $$ = 24 + 10 = 34$$ minutes. | mcq | aieee-2010 |
GjqY94gBeyZFKgR8 | maths | sequences-and-series | arithmetic-progression-(a.p) | A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after | [{"identifier": "A", "content": "19 months"}, {"identifier": "B", "content": "20 months"}, {"identifier": "C", "content": "21 months "}, {"identifier": "D", "content": "18 months "}] | ["C"] | null | Let required number of months $$=n$$
<br><br>$$\therefore$$ $$200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$$
<br><br>$$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$$
<br><br>$$ \Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600$$
<br><br>$$ \Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440$$
<br><br>$$ \Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440$$
<br><br>$$ \Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522$$
<br><br>$$ \Rightarrow {n^2} + 5n - 546 = 0$$
<br><br>$$ \Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0$$
<br><br>$$\therefore$$ $$n = 21$$ | mcq | aieee-2011 |
a2Ydbym8jWeSPitXaOUmI | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, . . . . . . . , a<sub>n</sub>, . . . . . be in A.P.
<br/><br/>If a<sub>3</sub> + a<sub>7</sub> + a<sub>11</sub> + a<sub>15</sub> = 72,
<br/><br/>then the sum of its first 17 terms is equal to : | [{"identifier": "A", "content": "306"}, {"identifier": "B", "content": "153"}, {"identifier": "C", "content": "612"}, {"identifier": "D", "content": "204"}] | ["A"] | null | As a<sub>1</sub> a<sub>2</sub> . . . . . a<sub>n</sub> . . . . . are in A.P.
<br><br>$$ \therefore $$ a<sub>3</sub> + a<sub>15</sub> = a<sub>7</sub> + a<sub>11</sub> = a<sub>1</sub> + a<sub>17</sub>
<br><br>Given,
<br><br>a<sub>3</sub> + a<sub>7</sub> + a<sub>11</sub> + a<sub>15</sub> + a<sub>15</sub> = 72
<br><br>$$ \Rightarrow $$ (a<sub>3</sub> + a<sub>15</sub>) + (a<sub>7</sub> + a<sub>11</sub>) = 72
<br><br>$$ \Rightarrow $$ 2(a<sub>1</sub> + a<sub>17</sub>) = 72
<br><br>$$ \Rightarrow $$ (a<sub>1</sub> + a<sub>17</sub>) = 36
<br><br>$$ \therefore $$ Sum of first 17 terms
<br><br>= $${{17} \over 2}$$ (a<sub>1</sub> + a<sub>17</sub>)
<br><br>= $${{17} \over 2}$$ $$ \times $$ 36
<br><br>= 306 | mcq | jee-main-2016-online-10th-april-morning-slot |
9Gv7MfBVcVWqAbtn3Bhao | maths | sequences-and-series | arithmetic-progression-(a.p) | If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4$${^{{1 \\over 3}}}$$"}, {"identifier": "C", "content": "4$${^{{2 \\over 3}}}$$"}, {"identifier": "D", "content": "4"}] | ["A"] | null | a, b and c are in AP.
<br><br>$$ \therefore $$ a + c = 2b
<br><br>As, abc = 8
<br><br> $$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8
<br><br>$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4
<br><br>$$ \therefore $$ ac = 4 and a + c = 4
<br><br>Then,
<br><br>b = $$\left( {{{a + c} \over 2}} \right)$$ = $${4 \over 2}$$ = 2 | mcq | jee-main-2017-online-9th-april-morning-slot |
0pRQnw9nIKu0GgXK | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that
<br/><br/>$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.
<br/><br/>$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to | [{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "66"}, {"identifier": "C", "content": "68"}, {"identifier": "D", "content": "34"}] | ["D"] | null | a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> . . . a<sub>43</sub> are in AP
<br><br>So, a<sub>2</sub> = a<sub>1</sub> + d
<br><br>a<sub>3</sub> = a<sub>1</sub> + 2d
<br><br>.
<br><br>.
<br><br>.
<br><br>a<sub>49</sub> =a<sub>1</sub> + 48d
<br><br>Now given, $${a_9} + {a_{43}} = 66$$
<br><br>$$ \Rightarrow \,\,\,\,$$ a<sub>1</sub> + 8d + a<sub>1</sub> + 42d = 66
<br><br>$$ \Rightarrow \,\,\,\,$$ 2a<sub>1</sub> + 50d = 66
<br><br>$$ \Rightarrow \,\,\,\,$$ a<sub>1</sub> + 25d = 33 . . . . . (1)
<br><br>$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} $$ = 416
<br><br>$$ \Rightarrow \,\,\,\,$$ a<sub>1</sub> + a<sub>5</sub> + a<sub>9</sub> + a<sub>13</sub> +. . . . . 13 items = 416
<br><br>$$ \Rightarrow \,\,\,\,$$ a<sub>1</sub> + a<sub>1</sub> + 4d + a<sub>1</sub> + 8d + . . . . a<sub>1</sub> + 48d = 416
<br><br>$$ \Rightarrow \,\,\,\,$$ 13a<sub>1</sub> + 4d +8d + 12d + . . . . . 48d = 416
<br><br>$$ \Rightarrow \,\,\,\,$$ 13a<sub>1</sub> + 4 (1+ 2 + 3 + . . . + 12) d = 416
<br><br>$$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $$d = 416
<br><br>$$ \Rightarrow \,\,\,\,$$ 13a<sub>1</sub> + 24 $$ \times$$ 13d = 416
<br><br>$$ \Rightarrow \,\,\,\,$$ a<sub>1</sub> + 24 d =32 . . . .(2)
<br><br>Solving (1) and (2) we get,
<br><br>d = 1
<br><br>and $${a_1} = 8$$
<br><br>$$\therefore\,\,\,$$ a<sub>1</sub> = 8
<br><br>a<sub>2</sub> = 8 + 1 = 9
<br><br>a<sub>3</sub> = 8 + 2 = 10
<br><br>.
<br><br>.
<br><br>.
<br><br>a<sub>17</sub> = 8 + 16 = 24
<br><br>Now, $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$$
<br><br>$$ \Rightarrow \,\,\,\,$$ $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$$
<br><br>$$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$$
<br><br>We can write above series like this,
<br><br> $$ \Rightarrow \,\,\,\,\,$$ (1<sup>2</sup> +2<sup>2</sup> + . . . . +24<sup>2</sup>) $$-$$ (1<sup>2</sup> + 2<sup>2</sup> + . . . . .+ 7<sup>2</sup>) = 140 m
<br><br>$$ \Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m$$
<br><br>$$ \Rightarrow \,\,\,\,\,$$ 490 $$-$$ 140 = 140 m
<br><br>$$ \Rightarrow \,\,\,\,\,$$4760 = 140 m
<br><br>$$ \Rightarrow \,\,\,\,\,$$ m = 34 | mcq | jee-main-2018-offline |
woTqZt9YlsJk9DcM8ygkw | maths | sequences-and-series | arithmetic-progression-(a.p) | If x<sub>1</sub>, x<sub>2</sub>, . . ., x<sub>n</sub> and $${1 \over {{h_1}}}$$, $${1 \over {{h_2}}}$$, . . . , $${1 \over {{h_n}}}$$ are two A.P..s such that x<sub>3</sub> = h<sub>2</sub> = 8 and x<sub>8</sub> = h<sub>7</sub> = 20, then x<sub>5</sub>.h<sub>10</sub> equals : | [{"identifier": "A", "content": "2560"}, {"identifier": "B", "content": "2650"}, {"identifier": "C", "content": "3200"}, {"identifier": "D", "content": "1600"}] | ["A"] | null | Assume d<sub>1</sub> is the common difference of A.P x<sub>1</sub>,x<sub>2</sub> ..... x<sub>n</sub><br><br>
Given x<sub>3</sub> = 8 and x<sub>8</sub> = 20<br><br>
$$ \therefore $$ x<sub>1</sub> + 2d<sub>1</sub> = 8 ..... <b>(i)</b><br>
and x<sub>1</sub> + 7d<sub>1</sub> = 20 .....<b> (ii)</b><br><br>
Solving <b>(i)</b> and <b>(ii)</b> we get x<sub>1</sub> = $$16 \over {15}$$ and d<sub>1</sub> = $$12 \over {5}$$<br><br>
Now let $$1 \over d_2$$ is the common difference of A.P $$1 \over h_1$$, $$1 \over h_2$$ ..... $$1 \over h_n$$<br><br>
Given that,<br>
h<sub>2</sub> = 8 and h<sub>7</sub> = 20<br><br>
$$ \therefore $$ $$1 \over h_2$$ = $$1 \over 8$$<br><br>
$$ \Rightarrow $$ $$1 \over h_1$$ + $$1 \over d_2$$ = $$1 \over 8$$ .... <b>(iii)</b><br><br>
and $$1 \over h_7$$ = $$1 \over 20$$<br><br>
$$ \Rightarrow $$ $$1 \over h_1$$ + $$6 \over d_2$$ = $$1 \over 20$$ ... <b>(iv)</b><br><br>
Solving <b>(iii)</b> and <b>(iv)</b> we get <br>
$$1 \over h_1$$ = $$28 \over 200$$ and $$1 \over d_2$$ = $$- {3 \over 200}$$<br>
So, x<sub>5</sub> = x<sub>1</sub> + 4d<sub>1</sub><br> = $$16 \over 5$$ + $$48 \over 5$$= $$64 \over 5$$ and <br>$$1 \over h_{10}$$ = $$1 \over h_1$$ + $$9 \over d_2$$ <br>= $$28 \over 200$$ - $$27 \over 200$$ = $$1\over 200$$<br><br>
$$ \therefore $$ x<sub>5</sub> $$\times$$ h<sub>10</sub> = $${64 \over 5} \times 200$$ = 2560 | mcq | jee-main-2018-online-15th-april-morning-slot |
6FiYbaaf6MERQvv8ffrYX | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,$$ (x<sub>i</sub> $$ \ne $$ 0 for i = 1, 2, ..., n) be in A.P. such that x<sub>1</sub>=4 and x<sub>21</sub> = 20. If n is the least positive integer for which $${x_n} > 50,$$ then $$\sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)} $$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$${{13} \\over 8}$$"}, {"identifier": "D", "content": "$${{13} \\over 4}$$"}] | ["D"] | null | $$ \because $$$$\,\,\,$$ $${1 \over {{x_1}}},{1 \over {{x_2}}},{1 \over {{x_3}}},.....,{1 \over {{x_n}}}$$ are in A.P.
<br><br>x<sub>1</sub> = 4 and x<sub>21</sub> = 20
<br><br>Let 'd' be the common difference of this A.P.
<br><br>$$\therefore\,\,\,$$ its 21<sup>st</sup> term = $${1 \over {{x_{21}}}} = {1 \over {{x_1}}} + \left[ {\left( {21 - 1} \right) \times d} \right]$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ d = $${1 \over {20}}$$ $$ \times $$ $$\left( {{1 \over {20}} - {1 \over 4}} \right)$$ $$ \Rightarrow $$ d = $$-$$ $${1 \over {100}}$$
<br><br>Also x<sub>n</sub> > 50(given).
<br><br>$$\therefore\,\,\,$$ $${1 \over {{x_n}}} = {1 \over {{x_1}}} + \left[ {\left( {n - 1} \right) \times d} \right]$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x<sub>n</sub> = $${{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}}$$
<br><br>$$\therefore\,\,\,$$ $${{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}} > 50$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${4 \over {1 + \left( {n - 1} \right) \times \left( { - {1 \over {100}}} \right) \times 4}} > 50$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 1 + (n $$-$$ 1) $$ \times $$ ($$-$$ $${1 \over {100}}$$) $$ \times $$ 4 < $${4 \over {50}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$-$$ $${1 \over {100}}$$(n $$-$$ 1) < $$-$$ $${{23} \over {100}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ n $$-$$ > 23 $$ \Rightarrow $$ n > 24
<br><br>Therefore$$\,\,\,$$ n = 25.
<br><br>$$ \Rightarrow $$$$\,\,\,$$$$\sum\limits_{i = 1}^{25} {{1 \over {{x_i}}}} $$ = $${{25} \over 2}\left[ {\left( {2 \times {1 \over 4}} \right) + \left( {25 - 1} \right) \times \left( { - {1 \over {100}}} \right)} \right]$$ = $${{13} \over 4}$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
d4VJ7d65BnAyLfgZ7j3rsa0w2w9jwy1gult | maths | sequences-and-series | arithmetic-progression-(a.p) | If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ............... a<sub>n</sub> are in A.P. and a<sub>1</sub> + a<sub>4</sub> + a<sub>7</sub> + ........... + a<sub>16</sub> = 114, then a<sub>1</sub> + a<sub>6</sub> + a<sub>11</sub> + a<sub>16</sub> is equal to : | [{"identifier": "A", "content": "38"}, {"identifier": "B", "content": "98"}, {"identifier": "C", "content": "76"}, {"identifier": "D", "content": "64"}] | ["C"] | null | 3(a<sub>1</sub> + a<sub>16</sub>) = 114<br><br>
$${a_1} + {a_{16}} = 38$$<br><br>
Now a<sub>1</sub> + a<sub>6</sub> + a<sub>11</sub> + a<sub>16</sub> = 2(a<sub>1</sub> + a<sub>16</sub>)<br><br>
= 2 × 38 = 76 | mcq | jee-main-2019-online-10th-april-morning-slot |
kAbLnyXnMBfNTGXpwC3rsa0w2w9jxb4m3oi | maths | sequences-and-series | arithmetic-progression-(a.p) | If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... are in A.P. such that a<sub>1</sub> + a<sub>7</sub> + a<sub>16</sub> = 40, then the sum of the first 15 terms of this A.P. is : | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "150"}, {"identifier": "D", "content": "280"}] | ["B"] | null | a<sub>1</sub> + a<sub>7</sub> + a<sub>16</sub> = 40<br><br>
$${a_1} + \left( {{a_1} + 6d} \right) + ({a_1} + 15d) = 40$$<br><br>
$$ \Rightarrow 3{a_1} + 21d = 40$$<br><br>
$$ \Rightarrow {a_1} + 7d = {{40} \over 3}$$<br><br>
$$ \Rightarrow {a_1} + {a_2}....... + {a_{15}} = {{15} \over 2}[{a_1} + {a_{15}}]$$<br><br>
$$ \Rightarrow {{15} \over 2}[{a_1} + {a_1} + 14d] \Rightarrow 15({a_1} + 7d) = 15 \times {{40} \over 3} = 200$$
| mcq | jee-main-2019-online-12th-april-evening-slot |
MAuiUu0xJFaSUmPVeZ3rsa0w2w9jx6g3geq | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>n</sub> denote the sum of the first n terms of an A.P. If S<sub>4</sub> = 16 and S<sub>6</sub>= – 48, then S<sub>10</sub> is equal to : | [{"identifier": "A", "content": "- 320"}, {"identifier": "B", "content": "- 380"}, {"identifier": "C", "content": "- 460"}, {"identifier": "D", "content": "- 210"}] | ["A"] | null | S<sub>4</sub> = $${4 \over 2}\left( {2a + 3d} \right) = 16$$<br><br>
$$ \Rightarrow 2a + 3d = 8$$<br><br>
S<sub>4</sub> = $${6 \over 2}\left( {2a + 5d} \right) = -48$$<br><br>
$$ \Rightarrow 2a + 5d = -16$$<br><br>
$$ \therefore $$ d = -12 and a = 22, Now S<sub>10</sub> = $${{10} \over 2}\left( {44 - 108} \right) = - 320$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
mkSTbBbv8X594mlprT3rsa0w2w9jx2g4463 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>,......be an A.P. with a<sub>6</sub> = 2. Then the common difference of this A.P., which maximises the
product a<sub>1</sub>a<sub>4</sub>a<sub>5</sub>, is :
| [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${6 \\over 5}$$"}, {"identifier": "C", "content": "$${8 \\over 5}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["C"] | null | first term = a, Common difference = d
<br><br>
$$ \therefore $$ a + 5d = 2<br><br>
a<sub>1</sub>. a<sub>4</sub>. a<sub>5</sub> = a(a + 3d) (a + 4d)<br><br>
f(d) = (2 – 5d) (2 – 2d) (2 – d)<br><br>
$$ \Rightarrow $$ $$f'(d) = 0 \Rightarrow d = {2 \over 3},{8 \over 5}$$<br><br>
$$ \Rightarrow $$ $$f''(d) < 0\,at\,d = {8 \over 5}$$<br><br>
$$\, \Rightarrow d = {8 \over 5}$$ | mcq | jee-main-2019-online-10th-april-evening-slot |
dYLoLXwINfh5DdmFfC18hoxe66ijvwubvnl | maths | sequences-and-series | arithmetic-progression-(a.p) | If the sum and product of the first three term in
an A.P. are 33 and 1155, respectively, then a value
of its 11<sup>th</sup> term is :- | [{"identifier": "A", "content": "\u201325"}, {"identifier": "B", "content": "\u201336"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "\u201335"}] | ["A"] | null | Let the three terms are a - d, a, a + d
<br><br>Given a - d + a + a + d = 33
<br><br>$$ \Rightarrow $$ 3a = 33
<br><br>$$ \Rightarrow $$ a = 11
<br><br>Also given,
<br><br>(a - d)a(a + d) = 1155
<br><br>$$ \Rightarrow $$ (a<sup>2</sup> - d<sup>2</sup>)a = 1155
<br><br>$$ \Rightarrow $$ (11<sup>2</sup> - d<sup>2</sup>)11 = 1155
<br><br>$$ \Rightarrow $$ (11<sup>2</sup> - d<sup>2</sup>) = 105
<br><br>$$ \Rightarrow $$ d = $$ \pm $$ 4
<br><br>When d = 4 and a = 11 then series is
<br><br>7, 11, 15, ....
<br><br>$$ \therefore $$ T<sub>11</sub> = a + 10d = 7 + 10$$ \times $$ 4 = 47
<br><br>When d = -4 and a = 11 then series is
<br><br>15, 11, 7, ....
<br><br>$$ \therefore $$ T<sub>11</sub> = a + 10d = 15 + 10$$ \times $$ -4 = -25 | mcq | jee-main-2019-online-9th-april-evening-slot |
9YvLd5bxYwk2gXlduoBxc | maths | sequences-and-series | arithmetic-progression-(a.p) | Let the sum of the first n terms of a non-constant
A.P., a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... be $$50n + {{n(n - 7)} \over 2}A$$, where
A is a constant. If d is the common difference of
this A.P., then the ordered pair (d, a<sub>50</sub>) is equal to | [{"identifier": "A", "content": "(A, 50+45A)"}, {"identifier": "B", "content": "(50, 50+45A)"}, {"identifier": "C", "content": "(A, 50+46A)"}, {"identifier": "D", "content": "(50, 50+46A)"}] | ["C"] | null | S<sub>n</sub> = $$50n + {{n(n - 7)} \over 2}A$$
<br><br>We know, n<sup>th</sup> tem
<br><br>T<sub>n</sub> = S<sub>n</sub> - S<sub>n - 1</sub>
<br><br>= $$50n + {{n(n - 7)} \over 2}A$$ - $$50\left( {n - 1} \right) - {{\left( {n - 1} \right)\left( {n - 8} \right)} \over 2}A$$
<br><br>= 50 + $${A \over 2}\left[ {{n^2} - 7n - {n^2} + 9n - 8} \right]$$
<br><br>= 50 + A(n - 4)
<br><br>We also know, common difference
<br><br>d = T<sub>n</sub> - T<sub>n - 1</sub>
<br><br>= 50 + A(n - 4) - 50 - A(n - 5)
<br><br>= A
<br><br>And T<sub>50</sub> = 50 + A(50 - 4)
<br><br>= 50 + 46A
<br><br>$$ \therefore $$ (d, a<sub>50</sub>) = (A, 50+46A) | mcq | jee-main-2019-online-9th-april-morning-slot |
VbSju1U91PISJVdiGZyBw | maths | sequences-and-series | arithmetic-progression-(a.p) | If <sup>n</sup>C<sub>4</sub>, <sup>n</sup>C<sub>5</sub> and <sup>n</sup>C<sub>6</sub> are in A.P., then n can be : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "14"}] | ["D"] | null | 2.<sup>n</sup>C<sub>5</sub> = <sup>n</sup>C<sub>4</sub> + <sup>n</sup><sup></sup>C<sub>6</sub>
<br><br>2.$${n \over {\left| 5 \right|n - 5}} = {n \over {\left| 4 \right|n - 4}} + {n \over {\left| 6 \right|n - 6}}$$
<br><br>$${2 \over 5}.{1 \over {n - 5}} = {1 \over {\left( {n - 4} \right)\left( {n - 5} \right)}} + {1 \over {30}}$$
<br><br>$$n = 14$$ satisfying equation. | mcq | jee-main-2019-online-12th-january-evening-slot |
jMQGhZPXti3X7SerSWxl2 | maths | sequences-and-series | arithmetic-progression-(a.p) | If 19<sup>th</sup> term of a non-zero A.P. is zero, then its (49<sup>th</sup> term) : (29<sup>th</sup> term) is : | [{"identifier": "A", "content": "2 : 1"}, {"identifier": "B", "content": "4 : 1"}, {"identifier": "C", "content": "1 : 3"}, {"identifier": "D", "content": "3 : 1"}] | ["D"] | null | a + 18d = 0 . . . . .(1)
<br><br>$${{a + 48d} \over {a + 28d}} = {{ - 18d + 48d} \over { - 18d + 28d}} = {3 \over 1}$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
XiEmOz8m70Df51MsaRno1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is - | [{"identifier": "A", "content": "1356"}, {"identifier": "B", "content": "1256"}, {"identifier": "C", "content": "1365"}, {"identifier": "D", "content": "1465"}] | ["A"] | null | $$\sum\limits_{r = 2}^{13} {(7r + 2) = 7.{{2 + 13} \over 2}} \times 6 + 2 \times 12$$
<br><br>= 7 $$ \times $$90 + 24 = 654
<br><br>$$\sum\limits_{r = 1}^{13} {(7r + 5) = 7\left( {{{1 + 13} \over 2}} \right)} \times 13 + 5 \times 13 = 702$$
<br><br>Total = 654 + 702 = 1356 | mcq | jee-main-2019-online-10th-january-morning-slot |
kQCDkroRzgx0A6RuISTkL | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${a_1},{a_2},.......,{a_{30}}$$ be an A.P.,
<br/><br/>$$S = \sum\limits_{i = 1}^{30} {{a_i}} $$ and $$T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}} $$.
<br/><br/>If $$a_5$$ = 27 and S - 2T = 75, then $$a_{10}$$ is equal to : | [{"identifier": "A", "content": "47"}, {"identifier": "B", "content": "42"}, {"identifier": "C", "content": "52"}, {"identifier": "D", "content": "57"}] | ["C"] | null | Let the common difference = d
<br><br>S = $$\sum\limits_{i = 1}^{30} {{a_i}} $$
<br><br>= $$a$$<sub>1</sub> + $$a$$<sub>2</sub> + . . . . . + $$a$$<sub>30</sub>
<br><br>$$ \therefore $$ S = $${{30} \over 2}\left[ {{a_1} + {a_{30}}} \right]$$
<br><br>= 15 [$$a$$<sub>1</sub> + $$a$$<sub>1</sub> + 29d]
<br><br>= 15 (2$$a$$<sub>1</sub> + 29d)
<br><br>T = $$\sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}} $$
<br><br>= $$a$$<sub>1</sub> + $$a$$<sub>3</sub> + . . . . . . + $$a$$<sub>29</sub>
<br><br>= $${{15} \over 2}\left[ {a{}_1 + {a_{29}}} \right]$$
<br><br>= $${{15} \over 2}\left[ {a{}_1 + {a_1} + 28d} \right]$$
<br><br>= $${{15} \over 2}\left[ {2a{}_1 + 28d} \right]$$
<br><br>= 15 ($$a$$<sub>1</sub> + 14d)
<br><br>Given,
<br><br>S $$-$$ 2T = 75
<br><br>$$ \Rightarrow $$ 15(2$$a$$<sub>1</sub> + 29d) $$-$$ 2 $$ \times $$ 15 ($$a$$<sub>1</sub> + 14d) = 75
<br><br>$$ \Rightarrow $$ 30$$a$$<sub>1</sub> + 15 $$ \times $$ 29d $$-$$ 30 $$a$$<sub>1</sub> $$-$$ 420d = 75
<br><br>$$ \Rightarrow $$ 435d $$-$$ 420d = 75
<br><br>$$ \Rightarrow $$ 15d = 75
<br><br>$$ \Rightarrow $$ d = 5
<br><br>Given that,
<br><br>$$a$$<sub>5</sub> = 27
<br><br>$$ \Rightarrow $$ $$a$$<sub>1</sub> + 4d = 27
<br><br>$$ \Rightarrow $$ $$a$$<sub>1</sub> + 20 = 27
<br><br>$$ \Rightarrow $$ $$a$$<sub>1</sub> = 7
<br><br>$$ \therefore $$ $$a$$<sub>10</sub> = $$a$$<sub>1</sub> + 9d
<br><br>= 7 + 45
<br><br>= 52 | mcq | jee-main-2019-online-9th-january-morning-slot |
BFKROwAabU9SIEAGMs1jw | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a, b and c be the 7<sup>th</sup>, 11<sup>th</sup> and 13<sup>th</sup> terms respectively of a non-constant A.P. If these are also three consecutive terms of a G.P., then $${a \over c}$$ equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 13}$$"}, {"identifier": "D", "content": "4"}] | ["D"] | null | T<sub>7</sub> = A + 6d = a; T<sub>11</sub> = A + 10d = b; T<sub>13</sub> = A + 12d = c
<br><br>Now a, b, c are in G.P.
<br><br>$$ \therefore $$ b<sup>2</sup> = ac
<br><br>$$ \Rightarrow $$ (A + 10d)<sup>2</sup> = (A + 6d) (A + 12d)
<br><br>$$ \Rightarrow $$ A<sup>2</sup> + 100d<sup>2</sup> + 20Ad = A<sup>2</sup> + 18Ad + 72d<sup>2</sup>
<br><br>$$ \Rightarrow $$ A + 14d = 0, A = $$-$$ 14d
<br><br>$${a \over c} = {{A + 6d} \over {A + 12d}} = {{ - 8d} \over { - 2d}} = 4$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
r4WC0aKydDZkqNCmrD5bM | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all natural numbers 'n' such that
100 < n < 200 and H.C.F. (91, n) > 1 is : | [{"identifier": "A", "content": "3221"}, {"identifier": "B", "content": "3121"}, {"identifier": "C", "content": "3203"}, {"identifier": "D", "content": "3303"}] | ["B"] | null | $$ \because $$ 91 = 13 $$ \times $$ 7
<br><br>So the required numbers are either divisible by
7 or 13.
<br><br>S<sub>A</sub>
= sum of numbers between 100 and 200 which are divisible by 7.
<br><br>$$ \Rightarrow $$ S<sub>A</sub> = 105 + 112 + ..... + 196
<br><br>S<sub>A</sub> = $${{14} \over 2}\left[ {105 + 196} \right]$$ = 2107
<br><br>S<sub>B</sub>
= Sum of numbers between 100 and 200 which are divisible by 13.
<br><br>S<sub>B</sub> = 104 + 117 + ....... + 195
<br><br>S<sub>B</sub> = $${8 \over 2}\left[ {104 + 195} \right]$$ = 1196
<br><br>S<sub>C</sub> = Sum of numbers between 100 and 200 which are divisible by 7 and 13.
<br><br>S<sub>C</sub> = 182
<br><br>Sum of numbers divisible by
7 or 13 = Sum of no. divisible by
7 + sum of the no. divisible by 13 – Sum of the
numbers divisible by 7 and 13
<br><br> = 2107 + 1196 - 182
<br><br>= 3121 | mcq | jee-main-2019-online-8th-april-morning-slot |
WbC8tVOvR5LmL8klSE7k9k2k5khzuf1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of terms common to the two A.P.'s
3, 7, 11, ....., 407 and 2, 9, 16, ....., 709 is ______. | [] | null | 14 | First A.P. is 3, 7, 11, 15, 19, 23, ..... 407
<br><br>d<sub>1</sub> = 4
<br><br>Second A.P. is 2, 9, 16, 23, ..... 709
<br><br>d<sub>2</sub> = 7
<br><br>First common term = 23
<br><br>Common difference of new A.P using the common terms of the two given A.P's is d = L.C.M. (4, 7) = 28
<br><br>Last term $$ \le $$ 407
<br><br>$$ \Rightarrow $$ 23 + (n – 1) (28) $$ \le $$ 407
<br><br>$$ \Rightarrow $$ n $$ \le $$ 14.7
<br><br>$$ \therefore $$ n = 14 | integer | jee-main-2020-online-9th-january-evening-slot |
MHxZwe5tFw0FI9dtSrjgy2xukg0cxml1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The common difference of the A.P. <br/>b<sub>1</sub>, b<sub>2</sub>, … , b<sub>m</sub>
is 2 more than the common<br/> difference of A.P. a<sub>1</sub>, a<sub>2</sub>, …, a<sub>n</sub>. If<br/> a<sub>40 </sub> = –159, a<sub>100</sub> = –399 and
b<sub>100</sub> = a<sub>70</sub>, then b<sub>1</sub>
is equal to : | [{"identifier": "A", "content": "127"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "\u2013127"}, {"identifier": "D", "content": "-81"}] | ["D"] | null | Let common difference of series
<br>a<sub>1</sub>
, a<sub>2</sub>
, a<sub>3</sub>
,..., a<sub>n</sub>
be d.
<br><br>$$ \because $$ a<sub>40</sub> = a<sub>1</sub> + 39d == –159 ...(i)
<br><br>and a<sub>100</sub> = a<sub>1</sub> + 99d = –399 ...(ii)
<br><br>From eqn. (ii) and (i)
<br>d = –4 and a<sub>1</sub>
= –3.
<br><br>The common difference of the A.P. <br>b<sub>1</sub>, b<sub>2</sub>, … , b<sub>m</sub>
is 2 more than the common<br> difference of A.P. a<sub>1</sub>, a<sub>2</sub>, …, a<sub>n</sub>.
<br><br>$$ \therefore $$ Common difference of b<sub>1</sub>
, b<sub>2</sub>
, b<sub>3</sub>
, ..., be (–2).
<br><br>$$ \because $$ b<sub>100</sub> = a<sub>70</sub>
<br><br>$$ \therefore $$ b<sub>1</sub>
+ 99(–2) = (–3) + 69(–4)
<br><br>$$ \therefore $$ b<sub>1</sub>
= 198 – 279
<br><br>$$ \therefore $$ b<sub>1</sub>
= – 81
| mcq | jee-main-2020-online-6th-september-evening-slot |
qMKL8WjZUIAD5hRjSpjgy2xukfg6jne3 | maths | sequences-and-series | arithmetic-progression-(a.p) | If $${3^{2\sin 2\alpha - 1}}$$, 14 and $${3^{4 - 2\sin 2\alpha }}$$ are the first three terms of an A.P. for some $$\alpha $$, then the sixth
terms of this A.P. is: | [{"identifier": "A", "content": "66"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "65"}, {"identifier": "D", "content": "78"}] | ["A"] | null | Given that<br><br>$${3^{4 - \sin 2\alpha }} + {3^{2\sin 2\alpha - 1}} = 28$$<br><br>Let $${3^{2\sin 2\alpha }}$$ = t<br><br>$$ \Rightarrow $$ $${{81} \over t} + {t \over 3} = 28$$<br><br>$$ \Rightarrow $$t = 81, 3<br><br>$$ \therefore $$ $${3^{2\sin 2\alpha }}$$ = 3<sup>1</sup>, 3<sup>4</sup><br><br>$$\sin 2\alpha = {1 \over 2}$$, 2 (rejected)<br><br>First term a = $${3^{2\sin 2\alpha -1}}$$ = 3<sup>0</sup>
<br><br>$$ \Rightarrow $$ a = 1<br><br>Given Second term = 14<br><br>$$ \therefore $$ Common difference d = 13<br><br>$${T_6} = a + 5d$$<br><br>$${T_6} = 1 + 5 \times 13$$<br><br>$${T_6} = 66$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
cFq6eDhgpo0Oel3P33jgy2xukfakjhs2 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, ..., an be a given A.P. whose<br/> common difference is an integer and <br/>S<sub>n</sub> = a<sub>1</sub> + a<sub>2</sub> + .... + a<sub>n</sub>. If a<sub>1</sub> = 1, a<sub>n</sub> = 300 and 15 $$ \le $$ n $$ \le $$ 50, then <br/>the ordered pair (S<sub>n-4</sub>, a<sub>n–4</sub>) is equal to:
| [{"identifier": "A", "content": "(2480, 249) "}, {"identifier": "B", "content": "(2480, 248)"}, {"identifier": "C", "content": "(2490, 248)"}, {"identifier": "D", "content": "(2490, 249)"}] | ["C"] | null | $${a_n} = {a_1} + (n - 1)d$$<br><br>$$ \Rightarrow 300 = 1 + (n - 1)d$$<br><br>$$ \Rightarrow (n - 1)d = 299 = 13 \times 23$$<br><br>since, n $$ \in $$[15, 50]<br><br>$$ \therefore $$ n = 24 and d = 13<br><br>$${a_{n - 4}} = {a_{20}} = 1 + 19 \times 13 = 248$$<br><br>$$ \Rightarrow {a_{n - 4}} = 248$$<br><br>$${S_{n - 4}} = {{20} \over 2}\{ 1 + 248\} = 2490$$ | mcq | jee-main-2020-online-4th-september-evening-slot |
p6030E7IzHd7wg2E0Jjgy2xukf0p91yb | maths | sequences-and-series | arithmetic-progression-(a.p) | If the first term of an A.P. is 3 and the sum of
its first 25 terms is equal to the sum of its next
15 terms, then the common difference of this
A.P. is : | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 5}$$"}, {"identifier": "C", "content": "$${1 \\over 7}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}] | ["D"] | null | First 25 terms = a, a + d, .......,a + 24d
<br><br>Next 15 terms = a + 25d, a + 26d, ......, a + 39d
<br><br>$$ \therefore $$ $${{25} \over 2}\left[ {2a + 24d} \right] = {{15} \over 2}\left[ {2\left( {a + 25d} \right) + 14d} \right]$$
<br><br>$$ \Rightarrow $$ 50a + 600d = 15 [2a + 50d + 14d]
<br><br>$$ \Rightarrow $$ 20a + 600d = 960d
<br><br>$$ \Rightarrow $$ 60 = 360d
<br><br>$$ \Rightarrow $$ d = $${1 \over 6}$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
7cFUZoeafMOtnxiXOLjgy2xukez7017n | maths | sequences-and-series | arithmetic-progression-(a.p) | If the sum of first 11 terms of an A.P.,
<br/>a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ....
is 0 (a $$ \ne $$ 0), then the sum of the A.P.,
<br/>a<sub>1</sub>
, a<sub>3</sub>
, a<sub>5</sub>
,....., a<sub>23</sub> is ka<sub>1</sub>
, where k is equal to :
| [{"identifier": "A", "content": "$${{121} \\over {10}}$$"}, {"identifier": "B", "content": "-$${{121} \\over {10}}$$"}, {"identifier": "C", "content": "$${{72} \\over 5}$$"}, {"identifier": "D", "content": "-$${{72} \\over 5}$$"}] | ["D"] | null | Let common difference be d.
<br><br>$$ \because $$ a<sub>1</sub>
+ a<sub>2</sub>
+ a<sub>3</sub>
+ ... + a<sub>11</sub> = 0
<br><br>$$ \therefore $$ $${{11} \over 2}\left[ {2{a_1} + 10d} \right]$$ = 0
<br><br>$$ \Rightarrow $$ a<sub>1</sub>
+ 5d = 0
<br><br>$$ \Rightarrow $$ d = $${ - {{{a_1}} \over 5}}$$ .....(1)
<br><br>Now a<sub>1</sub>
+ a<sub>3</sub>
+ a<sub>5</sub>
+ ... + a<sub>23</sub>
<br><br>= (a<sub>1</sub> + a<sub>23</sub>) $$ \times $$ $${{12} \over 2}$$
<br><br>= (a<sub>1</sub> + a<sub>1</sub> + 22d) × 6
<br><br>= $$\left[ {2{a_1} + 22\left( { - {{{a_1}} \over 5}} \right)} \right]$$ $$ \times $$ 6
<br><br>= $$ - {{72} \over 2}{a_1}$$
<br><br>$$ \therefore $$ k = $$ - {{72} \over 2}$$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
m38rP9Tkzz02doHgLE7k9k2k5hj1bjo | maths | sequences-and-series | arithmetic-progression-(a.p) | If the 10<sup>th</sup> term of an A.P. is $${1 \over {20}}$$ and its 20<sup>th</sup> term
is $${1 \over {10}}$$, then the sum of its first 200 terms is | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "$$100{1 \\over 2}$$"}, {"identifier": "C", "content": "$$50{1 \\over 4}$$"}, {"identifier": "D", "content": "50"}] | ["B"] | null | T<sub>10</sub> = a + 9d = $${1 \over {20}}$$ ....(1)
<br><br>T<sub>20</sub> = a + 19d = $${1 \over {10}}$$ .....(2)
<br><br>Equation (2) – (1)
<br><br>10d = $${1 \over {10}}$$ - $${1 \over {20}}$$
<br><br>$$ \Rightarrow $$ d = $${1 \over {200}}$$
<br><br>a + $${9 \over {200}}$$ = $${1 \over {20}}$$
<br><br>$$ \Rightarrow $$ a = $${1 \over {200}}$$
<br><br>S<sub>200</sub> = $${{200} \over 2}\left[ {{2 \over {200}} + \left( {200 - 1} \right) \times {1 \over {200}}} \right]$$
<br><br>= $$100\left[ {{2 \over {200}} + {{199} \over {200}}} \right]$$
<br><br>= $${{201} \over 2}$$ = $$100{1 \over 2}$$ | mcq | jee-main-2020-online-8th-january-evening-slot |
MeExSco81dydzJRFtt7k9k2k5gpayn9 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let ƒ : <b>R</b> $$ \to $$ <b>R</b> be such that for all
x $$ \in $$ R <br/>(2<sup>1+x</sup> + 2<sup>1–x</sup>), ƒ(x) and (3<sup>x</sup> + 3<sup>–x</sup>) are in
A.P., <br/>then the minimum value of ƒ(x) is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["C"] | null | f(x) = $${{2\left( {{2^x} + {2^{ - x}}} \right) + \left( {{3^x} + {3^{ - x}}} \right)} \over 2} \ge 3$$
<br><br>As we know, A.M > G.M | mcq | jee-main-2020-online-8th-january-morning-slot |
QcvKqniiG7DVwDMmdz7k9k2k5e4k7ko | maths | sequences-and-series | arithmetic-progression-(a.p) | Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -$${1 \over 2}$$ , then the greatest number amongst them is:
| [{"identifier": "A", "content": "$${{21} \\over 2}$$"}, {"identifier": "B", "content": "27"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "16"}] | ["D"] | null | Let the A.P is
<br>a - 2d, a - d, a, a + d, a + 2d
<br>$$ \because $$ sum = 25 <br>$$ \Rightarrow $$ 5a = 25 $$ \Rightarrow $$ a = 5
<br><br>Also given,
<br><br> product (a<sup>2</sup> – 4d<sup>2</sup>) (a<sup>2</sup> – d<sup>2</sup>).a = 2520
<br><br>$$ \Rightarrow $$ (25 – 4d<sup>2</sup>) (25 –d<sup>2</sup>)5 = 2520
<br><br>$$ \Rightarrow $$ 4d<sup>4</sup>
– 121d<sup>2</sup>
– 4d<sup>2</sup>
+ 121 = 0
<br><br>$$ \Rightarrow $$ (d<sup>2</sup>
– 1) (4d<sup>2</sup> – 121) = 0
<br><br>$$ \Rightarrow $$ d = $$ \pm $$1, d = $$ \pm {{11} \over 2}$$
<br><br>When d = $$ \pm $$1 we can't get any fraction term like -$${1 \over 2}$$.
<br><br>$$ \therefore $$ d = $$ \pm {{11} \over 2}$$
<br><br>And when d = $${{11} \over 2}$$
<br><br>we get largest term = 5 + 2d = 5 + 11 = 16 | mcq | jee-main-2020-online-7th-january-morning-slot |
UXHgHEDPLWdzlEEwzB1kluy1yat | maths | sequences-and-series | arithmetic-progression-(a.p) | The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _________. | [] | null | 1000 | Let N be the four digit number<br><br>gcd(N, 18) = 3<br><br>Hence N is an odd integer which is divisible by 3 but not by 9.<br><br>4 digit odd multiples of 3<br><br>1005, 1011, ..........., 9999 $$ \to $$ 1500<br><br>4 digit odd multiples of 9<br><br>1017, 1035, ..........., 9999 $$ \to $$ 500<br><br>Hence number of such N = 1000 | integer | jee-main-2021-online-26th-february-evening-slot |
s8bWRo7c7px8aS6V1D1kmm3p42s | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>1</sub> be the sum of first 2n terms of an arithmetic progression. Let S<sub>2</sub> be the sum of first 4n terms of the same arithmetic progression. If (S<sub>2</sub> $$-$$ S<sub>1</sub>) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to : | [{"identifier": "A", "content": "7000"}, {"identifier": "B", "content": "1000"}, {"identifier": "C", "content": "3000"}, {"identifier": "D", "content": "5000"}] | ["C"] | null | S<sub>1</sub> = $${{2n} \over 2}$$[2a + (2n $$-$$ 1)d]<br><br>S<sub>2</sub> = $${{4n} \over 2}$$[2a + (4n $$-$$ 1)d]<br><br>(where a = T<sub>1</sub> and d is common difference)<br><br>S<sub>2</sub> $$-$$ S<sub>1</sub>$$ \Rightarrow $$ 2n[2a + (4n $$-$$ 1)d] $$-$$ n[2a + (2n $$-$$ 1)d] = 1000<br><br>$$ \Rightarrow $$ n[2a + d(8n $$-$$ 2 $$-$$ 2n + 1)] = 1000<br><br>$$ \Rightarrow $$ n[2a + (6n $$-$$ 1)d] = 1000<br><br>S<sub>6</sub> = $${{6n} \over 2}$$[2a + (6n $$-$$ 1)d] = 3(S<sub>2</sub> $$-$$ S<sub>1</sub>) = 3000 | mcq | jee-main-2021-online-18th-march-evening-shift |
1krtbqfk7 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>n</sub> denote the sum of first n-terms of an arithmetic progression. If S<sub>10</sub> = 530, S<sub>5</sub> = 140, then S<sub>20</sub> $$-$$ S<sub>6</sub> is equal to: | [{"identifier": "A", "content": "1862"}, {"identifier": "B", "content": "1842"}, {"identifier": "C", "content": "1852"}, {"identifier": "D", "content": "1872"}] | ["A"] | null | Let first term of A.P. be a and common difference is d.<br><br>$$\therefore$$ $${S_{10}} = {{10} \over 2}\{ 2a + 9d\} = 530$$<br><br>$$\therefore$$ $$2a + 9d = 106$$ ..... (i)<br><br>$${S_5} = {5 \over 2}\{ 2a + 4d\} = 140$$<br><br>$$a + 2d = 28$$ ...... (ii)<br><br>From equation (i) and (ii), a = 8, d = 10<br><br>$$\therefore$$ $${S_{20}} - {S_6} = {{20} \over 2}\{ 2 \times 8 + 19 \times 10\} - {6 \over 2}\{ 2 \times 8 + 5 \times 10\} $$<br><br>$$ = 2060 - 198$$<br><br>$$ = 1862$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krub79cd | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all the elements in the set {n$$\in$$ {1, 2, ....., 100} | H.C.F. of n and 2040 is 1} is equal to _____________. | [] | null | 1251 | 2040 = 2<sup>3</sup> $$\times$$ 3 $$\times$$ 5 $$\times$$ 17<br><br>n should not be multiple of 2, 3, 5 and 17.<br><br>Sum of all n = (1 + 3 + 5 + ...... + 99) $$-$$ (3 + 9 + 15 + 21 + ...... + 99) $$-$$ (5 + 25 + 35 + 55 + 65 + 85 + 95) $$-$$ (17)<br><br>= 2500 $$-$$ $${{17} \over 2}$$(3 + 99) $$-$$ 365 $$-$$ 17<br><br>2500 $$-$$ 867 $$-$$ 365 $$-$$ 17<br><br>= 1251 | integer | jee-main-2021-online-22th-july-evening-shift |
1krvs2cgo | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>n</sub> be the sum of the first n terms of an arithmetic progression. If S<sub>3n</sub> = 3S<sub>2n</sub>, then the value of $${{{S_{4n}}} \over {{S_{2n}}}}$$ is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "8"}] | ["A"] | null | Let a be first term and d be common diff. of this A.P.<br><br>Given, S<sub>3n</sub> = 3S<sub>2n</sub><br><br>$$ \Rightarrow {{3n} \over 2}[2a + (3n - 1)d] = 3{{2n} \over 2}[2a + (2n - 1)d]$$<br><br>$$ \Rightarrow 2a + (3n - 1)d = 4a + (4n - 2)d$$<br><br>$$ \Rightarrow 2a + (n - 1)d = 0$$<br><br>Now, $${{{S_{4n}}} \over {{S_{2n}}}} = {{{{4n} \over 2}[2a + (4n - 1)d]} \over {{{2n} \over 2}[2a + (2n - 1)d]}} = {{2\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + 3nd} \right]} \over {\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + nd} \right]}}$$<br><br>$$ = {{6nd} \over {nd}} = 6$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1ks0bytdw | maths | sequences-and-series | arithmetic-progression-(a.p) | If $${\log _3}2,{\log _3}({2^x} - 5),{\log _3}\left( {{2^x} - {7 \over 2}} \right)$$ are in an arithmetic progression, then the value of x is equal to _____________. | [] | null | 3 | $$2{\log _3}({2^x} - 5) = {\log _2} + {\log _3}\left( {{2^x} - {7 \over 2}} \right)$$<br><br>Let $${2^x} = t$$<br><br>$${\log _3}{(t - 5)^2} = {\log _3}2\left( {t - {7 \over 2}} \right)$$<br><br>$${(t - 5)^2} = 2t - 7$$<br><br>$${t^2} - 12t + 32 = 0$$<br><br>$$(t - 4)(t - 8) = 0$$<br><br>$$\Rightarrow$$ 2<sup>x</sup> = 4 or 2<sup>x</sup> = 8<br><br>x = 2 (Rejected)<br><br>Or x = 3 | integer | jee-main-2021-online-27th-july-morning-shift |
1ktd2u4n1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is _____________. | [] | null | 7744 | 209, 220, 231, ..........., 495<br><br>Sum = $${{27} \over 2}$$(209 + 495) = 9504<br><br>Number containing 1 at unit place $$\matrix{
{\underline 2 } & {\underline 3 } & {\underline 1 } \cr
{\underline 3 } & {\underline 4 } & {\underline 1 } \cr
{\underline 4 } & {\underline 5 } & {\underline 1 } \cr
} $$<br><br>Number containing 1 at 10<sup>th</sup> place $$\matrix{
{\underline 3 } & {\underline 1 } & {\underline 9 } \cr
{\underline 4 } & {\underline 1 } & {\underline 8 } \cr
} $$<br><br>Required = 9504 $$-$$ (231 + 341 + 451 + 319 + 418)<br><br>= 7744 | integer | jee-main-2021-online-26th-august-evening-shift |
1ktk9gc86 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... be an A.P. If $${{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}}$$, p $$\ne$$ 10, then $${{{a_{11}}} \over {{a_{10}}}}$$ is equal to : | [{"identifier": "A", "content": "$${{19} \\over {21}}$$"}, {"identifier": "B", "content": "$${{100} \\over {121}}$$"}, {"identifier": "C", "content": "$${{21} \\over {19}}$$"}, {"identifier": "D", "content": "$${{121} \\over {100}}$$"}] | ["C"] | null | $${{{{10} \over 2}(2{a_1} + 9d)} \over {{p \over 2}(2{a_1} + (p - 1)d)}} = {{100} \over {{p^2}}}$$<br><br>$$(2{a_1} + 9d)p = 10(2{a_1} + (p - 1)d)$$<br><br>$$9dp = 20{a_1} - 2p{a_1} + 10d(p - 1)$$<br><br>$$9p = (20 - 2p){{{a_1}} \over d} + 10(p - 1)$$<br><br>$${{{a_1}} \over d} = {{(10 - p)} \over {2(10 - p)}} = {1 \over 2}$$<br><br>$$\therefore$$ $${{{a_{11}}} \over {{a_{10}}}} = {{{a_1} + 10d} \over {{a_1} + 9d}} = {{{1 \over 2} + 10} \over {{1 \over 2} + 9}} = {{21} \over {19}}$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1ktkdo86s | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________. | [] | null | 5143 | A = 4-digit numbers divisible by 3<br><br>A = 1002, 1005, ....., 9999.<br><br>9999 = 1002 + (n $$-$$ 1)3<br><br>$$\Rightarrow$$ (n $$-$$ 1)3 = 8997 $$\Rightarrow$$ n = 3000<br><br>B = 4-digit numbers divisible by 7<br><br>B = 1001, 1008, ......., 9996<br><br>$$\Rightarrow$$ 9996 = 1001 + (n $$-$$ 1)7<br><br>$$\Rightarrow$$ n = 1286<br><br>A $$\cap$$ B = 1008, 1029, ....., 9996<br><br>9996 = 1008 + (n $$-$$ 1)21<br><br>$$\Rightarrow$$ n = 429<br><br>So, no divisible by either 3 or 7<br><br>= 3000 + 1286 $$-$$ 429 = 3857<br><br>total 4-digits numbers = 9000<br><br>required numbers = 9000 $$-$$ 3857 = 5143 | integer | jee-main-2021-online-31st-august-evening-shift |
1kto9lnec | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, ..........., a<sub>21</sub> be an AP such that $$\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = {4 \over 9}} $$. If the sum of this AP is 189, then a<sub>6</sub>a<sub>16</sub> is equal to : | [{"identifier": "A", "content": "57"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "48"}, {"identifier": "D", "content": "36"}] | ["B"] | null | $$\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} } $$<br><br>$$ = {1 \over d}\sum\limits_{n = 1}^{20} {\left( {{1 \over {{a_n}}} - {1 \over {{a_n} + d}}} \right)} $$<br><br>$$ \Rightarrow {1 \over d}\left( {{1 \over {{a_1}}} - {1 \over {{a_{21}}}}} \right) = {4 \over 9}$$ (Given)<br><br>$$ \Rightarrow {1 \over d}\left( {{{{a_{21}} - {a_1}} \over {{a_1}{a_{21}}}}} \right) = {4 \over 9}$$<br><br>$$ \Rightarrow {1 \over d}\left( {{{{a_1} + 20d - {a_1}} \over {{a_1}{a_2}}}} \right) = {4 \over 9} \Rightarrow {a_1}{a_2} = 45$$ .... (1)<br><br>Now sum of first 21 terms = $${{21} \over 2}(2{a_1} + 20d) = 189$$<br><br>$$\Rightarrow$$ a<sub>1</sub> + 10d = 9 ..... (2)<br><br>For equation (1) & (2) we get<br><br>a<sub>1</sub> = 3 & d = $${3 \over 5}$$<br><br>or a<sub>1</sub> = 15 & d = $$ - {3 \over 5}$$<br><br>So, a<sub>6</sub> . a<sub>16</sub> = (a<sub>1</sub> + 5d) (a<sub>1</sub> + 15d)<br><br>$$\Rightarrow$$ a<sub>6</sub>a<sub>16</sub> = 72<br><br>Option (b) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l54tsnl8 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let 3, 6, 9, 12, ....... upto 78 terms and 5, 9, 13, 17, ...... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ________.</p> | [] | null | 2223 | <p>1st AP :</p>
<p>3, 6, 9, 12, ....... upto 78 terms</p>
<p>t<sub>78</sub> = 3 + (78 $$-$$ 1)3</p>
<p>= 3 + 77 $$\times$$ 3</p>
<p>= 234</p>
<p>2nd AP :</p>
<p>5, 9, 13, 17, ...... upto 59 terms</p>
<p>t<sub>59</sub> = 5 + (59 $$-$$ 1)4</p>
<p>= 5 + 58 $$\times$$ 4</p>
<p>= 237</p>
<p>Common term's AP :</p>
<p>First term = 9</p>
<p>Common difference of first AP = 3</p>
<p>And common difference of second AP = 4</p>
<p>$$\therefore$$ Common difference of common terms</p>
<p>AP = LCM (3, 4) = 12</p>
<p>$$\therefore$$ New AP = 9, 21, 33, .......</p>
<p>t<sub>n</sub> = 9 + (n $$-$$ 1)12 $$\le$$ 234</p>
<p>$$ \Rightarrow n \le {{237} \over {12}}$$</p>
<p>$$ \Rightarrow n = 19$$</p>
<p>$$\therefore$$ $${S_{19}} = {{19} \over 2}\left[ {2.9 + (19 - 1)12} \right]$$</p>
<p>$$ = 19(9 + 108)$$</p>
<p>$$ = 2223$$</p> | integer | jee-main-2022-online-29th-june-evening-shift |
1l567rodu | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let A = {1, a<sub>1</sub>, a<sub>2</sub> ....... a<sub>18</sub>, 77} be a set of integers with 1 < a<sub>1</sub> < a<sub>2</sub> < ....... < a<sub>18</sub> < 77. <br/><br/>Let the set A + A = {x + y : x, y $$\in$$ A} contain exactly 39 elements. Then, the value of a<sub>1</sub> + a<sub>2</sub> + ...... + a<sub>18</sub> is equal to _____________.</p> | [] | null | 702 | If we write the elements of $A+A$, we can certainly find 39 distinct elements as $1+1,1+a_{1}, 1+a_{2}, \ldots .1$ $+a_{18}, 1+77, a_{1}+77, a_{2}+77, \ldots \ldots a_{18}+77,77+77$.<br/><br/> It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.
<br/><br/>
Let the common difference be '$d$'.
<br/><br/>
$77=1+19 \mathrm{~d} \Rightarrow d=4$
<br/><br/>
So, $\sum\limits_{i=1}^{18} a_{1}=\frac{18}{2}\left[2 a_{1}+17 d\right]=9[10+68]=702$ | integer | jee-main-2022-online-28th-june-morning-shift |
1l56q8kl9 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> ...... and b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub> ....... are A.P., and a<sub>1</sub> = 2, a<sub>10</sub> = 3, a<sub>1</sub>b<sub>1</sub> = 1 = a<sub>10</sub>b<sub>10</sub>, then a<sub>4</sub> b<sub>4</sub> is equal to -</p> | [{"identifier": "A", "content": "$${{35} \\over {27}}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${{27} \\over {28}}$$"}, {"identifier": "D", "content": "$${{28} \\over {27}}$$"}] | ["D"] | null | <p>a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> .... are in A.P. (Let common difference is d<sub>1</sub>)</p>
<p>b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub> .... are in A.P. (Let common difference is d<sub>2</sub>)</p>
<p>and a<sub>1</sub> = 2, a<sub>10</sub> = 3, a<sub>1</sub>b<sub>1</sub> = 1 = a<sub>10</sub>b<sub>10</sub></p>
<p>$$\because$$ a<sub>1</sub>b<sub>1</sub> = 1</p>
<p>$$\therefore$$ b<sub>1</sub> = $${1 \over 2}$$</p>
<p>a<sub>10</sub>b<sub>10</sub> = 1</p>
<p>$$\therefore$$ b<sub>10</sub> = $${1 \over 3}$$</p>
<p>Now, a<sub>10</sub> = a<sub>1</sub> + 9d<sub>1</sub> $$\Rightarrow$$ d<sub>1</sub> = $${1 \over 9}$$</p>
<p>b<sub>10</sub> = b<sub>1</sub> + 9d<sub>2</sub> $$\Rightarrow$$ d<sub>2</sub> = $${1 \over 9}$$$$\left[ {{1 \over 3} - {1 \over 2}} \right]$$ = $$-$$ $${{1 \over {54}}}$$</p>
<p>Now, a<sub>4</sub> = 2 + $${{3 \over {9}}}$$ = $${{7 \over {3}}}$$</p>
<p>b<sub>4</sub> = $${{1 \over {2}}}$$ $$-%$$ $${{3 \over {54}}}$$ = $${{4 \over {9}}}$$</p>
<p>$$\therefore$$ a<sub>4</sub>b<sub>4</sub> = $${{28 \over {27}}}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l5c1fk8r | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>If $$\{ {a_i}\} _{i = 1}^n$$, where n is an even integer, is an arithmetic progression with common difference 1, and $$\sum\limits_{i = 1}^n {{a_i} = 192} ,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}} = 120} $$, then n is equal to :</p> | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "96"}, {"identifier": "C", "content": "92"}, {"identifier": "D", "content": "104"}] | ["B"] | null | <p>$$\sum\limits_{i = 1}^n {{a_i} = 192} $$</p>
<p>$$\Rightarrow$$ a<sub>1</sub> + a<sub>2</sub> + a<sub>3</sub> + ...... + a<sub>n</sub> = 192</p>
<p>$$ \Rightarrow {n \over 2}[{a_1} + {a_n}] = 192$$</p>
<p>$$ \Rightarrow {a_1} + {a_n} = {{384} \over n}$$ ..... (1)</p>
<p>Now, $$\sum\limits_{i = 1}^{{n \over 2}} {{a_{2i}} = 120} $$</p>
<p>$$\Rightarrow$$ a<sub>2</sub> + a<sub>4</sub> + a<sub>6</sub> + ...... + a<sub>n</sub> = 120</p>
<p>Here total $${n \over 2}$$ terms present.</p>
<p>$$\therefore$$ $${{{n \over 2}} \over 2}[{a_2} + {a_n}] = 120$$</p>
<p>$$ \Rightarrow {n \over 4}[{a_1} + 1 + {a_n}] = 120$$</p>
<p>$$ \Rightarrow {a_1} + {a_n} + 1 = {{480} \over n}$$ ..... (2)</p>
<p>Subtracting (1) from (2), we get</p>
<p>$$1 = {{480} \over n} - {{384} \over n}$$</p>
<p>$$ \Rightarrow 1 = {{96} \over n}$$</p>
<p>$$\Rightarrow$$ n = 96</p> | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6dx827f | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a, b$$ be two non-zero real numbers. If $$p$$ and $$r$$ are the roots of the equation $$x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0$$ and $$\mathrm{q}$$ and s are the roots of the equation $$x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0$$, such that $$\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}}$$ are in A.P., then $$\mathrm{a}^{-1}-\mathrm{b}^{-1}$$ is equal to _____________.</p> | [] | null | 38 | $\because$ Roots of $2 a x^{2}-8 a x+1=0$ are $\frac{1}{p}$ and $\frac{1}{r}$ and roots of $6 b x^{2}+12 b x+1=0$ are $\frac{1}{q}$ and $\frac{1}{s}$.
<br/><br/>
Let $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ as $\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta$
<br/><br/>
So sum of roots $2 \alpha-2 \beta=4$ and $2 \alpha+2 \beta=-2$
<br/><br/>
Clearly $\alpha=\frac{1}{2}$ and $\beta=-\frac{3}{2}$ <br/><br/>Now product of roots, $\frac{1}{p} \cdot \frac{1}{r}=\frac{1}{2 a}=-5 \Rightarrow \frac{1}{a}=-10$<br/><br/> and $\frac{1}{q} \cdot \frac{1}{x}=\frac{1}{6 b}=-8 \Rightarrow \frac{1}{b}=-48$
<br/><br/>
So, $\frac{1}{a}-\frac{1}{b}=38$ | integer | jee-main-2022-online-25th-july-morning-shift |
1l6i041q0 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ___________.</p> | [] | null | 53 | <p>$${d_1} = {{199 - 100} \over 2} \notin I$$</p>
<p>$${d_2} = {{199 - 100} \over 3} = 33$$</p>
<p>$${d_3} = {{199 - 100} \over 4} \notin I$$</p>
<p>$${d_n} = {{199 - 100} \over {i + 1}} \in I$$</p>
<p>$${d_i} = 33 + 11,\,9$$</p>
<p>Sum of CD's $$ = 33 + 11 + 9$$</p>
<p>$$ = 53$$</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6jbbycd | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Suppose $$a_{1}, a_{2}, \ldots, a_{n}$$, .. be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is $$5: 17$$ and , $$110 < {a_{15}} < 120$$, then the sum of the first ten terms of the progression is equal to</p> | [{"identifier": "A", "content": "290"}, {"identifier": "B", "content": "380"}, {"identifier": "C", "content": "460"}, {"identifier": "D", "content": "510"}] | ["B"] | null | <p>$$\because$$ a<sub>1</sub>, a<sub>2</sub>, .... a<sub>n</sub> be an A.P of natural numbers and</p>
<p>$${{{S_5}} \over {{S_9}}} = {5 \over {17}} \Rightarrow {{{5 \over 2}[2{a_1} + 4d]} \over {{9 \over 2}[2{a_1} + 8d]}} = {5 \over {17}}$$</p>
<p>$$ \Rightarrow 34{a_1} + 68d = 18{a_1} + 72d$$</p>
<p>$$ \Rightarrow 16{a_1} = 4d$$</p>
<p>$$\therefore$$ $$d = 4{a_1}$$</p>
<p>And $$110 < {a_{15}} < 120$$</p>
<p>$$\therefore$$ $$110 < {a_1} + 14d < 120 \Rightarrow 110 < 57{a_1} < 120$$</p>
<p>$$\therefore$$ $${a_1} = 2$$ ($$\because$$ $${a_i}\, \in N$$)</p>
<p>$$d = 8$$</p>
<p>$$\therefore$$ $${S_{10}} = 5[4 + 9 \times 8] = 380$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6p3bg7l | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a_{1}, a_{2}, a_{3}, \ldots$$ be an A.P. If $$\sum\limits_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$$, then $$4 a_{2}$$ is equal to _________.</p> | [] | null | 16 | <p>Given</p>
<p>$$S = {{{a_1}} \over 2} + {{{a_2}} \over {{2^2}}} + {{{a_3}} \over {{2^3}}} + {{{a_4}} \over {{2^4}}}\, + \,.....\,\infty $$</p>
<p>$${{{1 \over 2}S = {{{a_1}} \over {{2^2}}} + {{{a_2}} \over {{2^3}}}\, + \,.........\,\infty } \over {{S \over 2} = {{{a_1}} \over 2} + {{({a_2} + {a_1})} \over {{2^2}}} + {{({a_3} + {a_2})} \over {{2^3}}}\, + \,......\,\infty }}$$</p>
<p>$$ \Rightarrow {S \over 2} = {{{a_1}} \over 2} + {d \over 2}$$</p>
<p>$$ \Rightarrow {a_1} + d = {a_2} = 4 \Rightarrow 4{a_2} = 16$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
1ldo7ckhv | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>The sum of the common terms of the following three arithmetic progressions.</p>
<p>$$3,7,11,15, \ldots ., 399$$,</p>
<p>$$2,5,8,11, \ldots ., 359$$ and</p>
<p>$$2,7,12,17, \ldots ., 197$$,</p>
<p>is equal to _____________.</p> | [] | null | 321 | $$
\begin{array}{ll}
3,7,11,15, \ldots \ldots \ldots . .399 : & \mathrm{~d}_1=4 \\\\
2,5,8,11, \ldots \ldots \ldots \ldots, 359 : & \mathrm{~d}_2=3 \\\\
2,7,12,17, \ldots \ldots, 197 : & \mathrm{~d}_3=5 \\\\
\operatorname{LCM}\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=60 &
\end{array}
$$
<br/><br/>Common terms are 47, 107, 167
<br/><br/>Sum $=321$ | integer | jee-main-2023-online-1st-february-evening-shift |
ldo8c1vp | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $a_1, a_2, a_3, \ldots$ be an A.P. If $a_7=3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n !-4 a_{n(n+2)}$ is equal to : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "$\\frac{381}{4}$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "$\\frac{33}{4}$"}] | ["A"] | null | $a_{7}=3 \Rightarrow a+6 d=3 \Rightarrow a=3-6 d$
<br/><br/>$$
\begin{aligned}
& a_{1} \cdot a_{4}=a(a+3 d) \\\\
& \Rightarrow(3-6 d)(3-6 d+3 d) \\\\
& \Rightarrow 3(1-2 d) 3(1-d) \\\\
& \Rightarrow 9\left(2 d^{2}-3 d+1\right)
\end{aligned}
$$
<br/><br/>Let $f(d)=2 d^{2}-3 d+1$
<br/><br/>$f^{\prime}(d)=4 d-3 \Rightarrow d=\frac{3}{4}$
<br/><br/>$\therefore a=3-6 \cdot \frac{3}{4}=3-\frac{9}{2}=-\frac{3}{2}$
<br/><br/>$S_{n}=0$
<br/><br/>$\frac{n}{2}(29+(n-1) d)=0$
<br/><br/>$\Rightarrow 2 \cdot\left(-\frac{3}{2}\right)+(n-1)\left(\frac{3}{4}\right)=0$
<br/><br/>$\Rightarrow \quad 3=\frac{3}{4}(n-1)$
<br/><br/>$\Rightarrow n=5$
<br/><br/>Now, $n !-4 \cdot a_{n(n+2)}$
<br/><br/>$$
\begin{aligned}
& =5 !-4 \cdot a_{35} \\\\
& =120-4\left(-\frac{3}{2}+34 \cdot \frac{3}{4}\right) \\\\
& =120-(-6+102) \\\\
& =120-(96) \\\\
& =24
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldonp1se | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a_{1}=8, a_{2}, a_{3}, \ldots, a_{n}$$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is ___________.</p> | [] | null | 754 | $$
\begin{aligned}
& a_1+a_2+a_3+a_4=50 \\\\
& \Rightarrow 32+6 d=50 \\\\
& \Rightarrow d=3 \\\\
& \text { and, } a_{n-3}+a_{n-2}+a_{n-1}+a_n=170 \\\\
& \Rightarrow 32+(4 n-10) \cdot 3=170 \\\\
& \Rightarrow \mathrm{n}=14 \\\\
& a_7=26, a_8=29 \\\\
& \Rightarrow a_7 \cdot a_8=754
\end{aligned}
$$ | integer | jee-main-2023-online-1st-february-morning-shift |
1ldptn1y5 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a_{1}, a_{2}, \ldots, a_{n}$$ be in A.P. If $$a_{5}=2 a_{7}$$ and $$a_{11}=18$$, then
<br/><br/>$$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$$ is equal to ____________.</p> | [] | null | 8 | $a_{11}=18$
<br/><br/>$$
\begin{aligned}
& a+10 d=18 \\\\
& a_{5}=2 a_{7} \\\\
& a+4 d=2(a+6 d) \\\\
& a=-8 d
\end{aligned}
$$
<br/><br/>(i) and (ii) $\Rightarrow a=-72, d=9$.
<br/><br/>On rationalising the denominator, given expression
<br/><br/>$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-\sqrt{a_{12}}}{-d}+\ldots+\frac{\sqrt{a_{17}}-\sqrt{a_{18}}}{-d}\right]$ <br/><br/>$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{18}}}{-d}\right]$
<br/><br/>$=12\left[\frac{\sqrt{a_{11}-d}-\sqrt{a_{11}+7 d}}{-d}\right]$
<br/><br/>$=12\left[\frac{\sqrt{18-9}-\sqrt{18+63}}{-9}\right]$ $=12 \times \frac{2}{3}=8$ | integer | jee-main-2023-online-31st-january-morning-shift |
1ldybitxm | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>For three positive integers p, q, r, $${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$$ and r = pq + 1 such that 3, 3 log$$_yx$$, 3 log$$_zy$$, 7 log$$_xz$$ are in A.P. with common difference $$\frac{1}{2}$$. Then r-p-q is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "$$-$$6"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $x^{p q^{2}}=y^{q r}=z^{p^{2} r}$
<br/><br/>
$$
3 \log _{y} x=\frac{7}{2}, 3 \log _{z} y=4,7 \log _{x} z=\frac{9}{2}
$$
<br/><br/>
$$
\begin{aligned}
& \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\\\
& y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\\\
& \Rightarrow \frac{7}{6} p q^{2}=q r=\frac{3}{4} p^{2} r \\\\
& \therefore 7 p q=6 r, 4 q=3 p^{2} \\\\
& r=p q+1 \\\\
& r=\frac{6 r}{7}+1 \Rightarrow r=7 \\\\
& p q=6 \\\\
& p\left(\frac{3 p^{2}}{4}\right)=6 \\\\
& p=2, q=3 \\\\
& r-p-q=7-5=2
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift |
1lgpy2jk4 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$s_{1}, s_{2}, s_{3}, \ldots, s_{10}$$ respectively be the sum to 12 terms of 10 A.P. s whose first terms are $$1,2,3, \ldots .10$$ and the common differences are $$1,3,5, \ldots \ldots, 19$$ respectively. Then $$\sum_\limits{i=1}^{10} s_{i}$$ is equal to :</p> | [{"identifier": "A", "content": "7360"}, {"identifier": "B", "content": "7220"}, {"identifier": "C", "content": "7260"}, {"identifier": "D", "content": "7380"}] | ["C"] | null | We have 10 arithmetic progressions (A.P.s) with the first terms $$a_i$$ and the common differences $$d_i$$, where $$i = 1, 2, \ldots, 10$$.
<br/><br/>The first terms are $$a_i = i$$ and the common differences are $$d_i = 2i - 1$$.
<br/><br/>Now, we need to find the sum of the first 12 terms for each A.P. The formula for the sum of the first n terms of an A.P. is:
<br/><br/>$$S_n = n\left(\frac{2a + (n - 1)d}{2}\right)$$
<br/><br/>In this case, we need to find the sum of the first 12 terms for each A.P., so we have:
<br/><br/>$$S_{12} = 12\left(\frac{2a + 11d}{2}\right)$$
<br/><br/>Now, we can compute the sum $$s_i$$ for each A.P.:
<br/><br/>$$s_i = 12\left(\frac{2i + 11(2i - 1)}{2}\right) = 6(2i + 22i - 11) = 6(24i - 11)$$
<br/><br/>Finally, we need to find the sum of all $$s_i$$ for $$i = 1, 2, \ldots, 10$$:
<br/><br/>$$\sum\limits_{i=1}^{10} s_i = 6\sum\limits_{i=1}^{10} (24i - 11) = 6\left(24\sum\limits_{i=1}^{10} i - 11\sum\limits_{i=1}^{10} 1\right)$$
<br/><br/>The sum of the first 10 integers is $$\sum\limits_{i=1}^{10} i = \frac{10(10 + 1)}{2} = 55$$, so we have:
<br/><br/>$$\sum\limits_{i=1}^{10} s_i = 6\left(24\cdot55 - 11\cdot10\right) = 6(1320 - 110) = 6\cdot1210 = 7260$$
<br/><br/>Thus, the sum $$\sum\limits_{i=1}^{10} s_i$$ is equal to 7260. | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgutxklh | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$x_{1}, x_{2}, \ldots, x_{100}$$ be in an arithmetic progression, with $$x_{1}=2$$ and their mean equal to 200 . If $$y_{i}=i\left(x_{i}-i\right), 1 \leq i \leq 100$$, then the mean of $$y_{1}, y_{2}, \ldots, y_{100}$$ is :</p> | [{"identifier": "A", "content": "10051.50"}, {"identifier": "B", "content": "10049.50"}, {"identifier": "C", "content": "10100"}, {"identifier": "D", "content": "10101.50"}] | ["B"] | null | We have, mean of $x_1, x_2 \ldots \ldots x_{100}=200$
<br/><br/>Where, $x_1, x_2 \ldots x_{100}$ are in AP with first term as 2.
<br/><br/>$$
\begin{aligned}
\text { Mean } & =200 \\\\
& =\frac{\sum\limits_{i=1}^{100} x_i}{100}=200
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\frac{100}{2} \times[2 \times 2+99 d] =20000 \\\\
\Rightarrow 4+99 d =400 \\\\
\Rightarrow 99 d =396 \\\\
d =4
\end{aligned}
$$
<br/><br/>Also,
<br/><br/>$$
\begin{aligned}
y_i & =i\left(x_i-i\right) \\\\
& =i[2+(i-1) 4-i] \\\\
& =i[3 i-2] \\\\
& =3 i^2-2 i
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\text { Required mean } & =\frac{\sum\limits_{i=1}^{100} y_i}{100} \\\\
& =\frac{1}{100}\left[\sum_{i=1}^{100}\left(3 i^2-2 i\right)\right] \\\\
& =\frac{1}{100}\left[\frac{3 \times 100 \times 101 \times 201}{6}-2 \times \frac{100 \times 101}{2}\right]\\\\
& =\frac{20301}{2}-101 \\\\
& =10049.50
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgzztm49 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$S_{K}=\frac{1+2+\ldots+K}{K}$$ and $$\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$$, where $$A, B, C, D \in \mathbb{N}$$ and $$A$$ has least value. Then</p> | [{"identifier": "A", "content": "$$A+B+C+D$$ is divisible by 5"}, {"identifier": "B", "content": "$$A+C+D$$ is not divisible by $$B$$"}, {"identifier": "C", "content": "$$A+B=5(D-C)$$"}, {"identifier": "D", "content": "$$A+B$$ is divisible by $$\\mathrm{D}$$"}] | ["D"] | null | $$
\begin{aligned}
& \because S_k=\frac{1+2+\ldots+k}{k} \\\\
& =\frac{k(k+1)}{2 k}=\frac{k+1}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\\\
& \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\right]
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+\frac{2 n(n+1)}{2}\right] \\\\
& =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\\\
& =\frac{n}{24}\left[2 n^2+3 n+1+6+6 n+6\right] \\\\
& =\frac{n}{24}\left[2 n^2+9 n+13\right]
\end{aligned}
$$
<br/><br/>On comparing, we get
<br/><br/>$$
\mathrm{A}=24, \mathrm{~B}=2, \mathrm{C}=9, \mathrm{D}=13
$$
<br/><br/>(A) $A+B+C+D=48$, which is not divisible by 5.
<br/><br/>(B) $\mathrm{A}+\mathrm{C}+\mathrm{D}=46$, which is divisible by 2(B).
<br/><br/>(C) $A+B=26$
<br/><br/>$$
\begin{aligned}
& 5(\mathrm{D}-\mathrm{C})=5(13-9)=20 \\\\
& \therefore 26 \neq 20
\end{aligned}
$$
<br/><br/>(D) $A+B=24+2=26$, divisible by 13. | mcq | jee-main-2023-online-8th-april-morning-shift |
lsamv2vn | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $\mathrm{S}_{15}-\mathrm{S}_5$ is equal to : | [{"identifier": "A", "content": "800"}, {"identifier": "B", "content": "890"}, {"identifier": "C", "content": "790"}, {"identifier": "D", "content": "690"}] | ["C"] | null | <p>To solve this problem, we will start by using the properties of an arithmetic progression (AP).</p>
<p>The sum of the first $n$ terms of an AP can be calculated using the formula:
$$ S_n = \frac{n}{2} (2a + (n-1)d) $$
where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference between the terms.</p>
<p>Given the information:
$$ S_{10} = 390 $$</p>
<p>We can plug $n=10$ into the sum formula to get:</p>
<p>$$ S_{10} = \frac{10}{2} (2a + (10-1)d) $$</p>
<p>$$ 390 = 5(2a + 9d) $$</p>
<p>$$ 390 = 10a + 45d $$</p>
<p>$$ 78 = 2a + 9d \quad .........\text{(1)} $$</p>
<p>Next, we're given the ratio of the tenth term ($T_{10}$) to the fifth term ($T_5$):
$$ \frac{T_{10}}{T_5} = \frac{15}{7} $$</p>
<p>The $n$th term of an AP is given by:
<br/><br/>$$ T_n = a + (n-1)d $$</p>
<p>So, for the tenth term:
$$ T_{10} = a + (10-1)d = a + 9d $$</p>
<p>And for the fifth term:
<br/><br/>$$ T_5 = a + (5-1)d = a + 4d $$</p>
<p>Now we can write the ratio as:</p>
<p>$$ \frac{a + 9d}{a + 4d} = \frac{15}{7} $$</p>
<p>$$ 7(a + 9d) = 15(a + 4d) $$</p>
<p>$$ 7a + 63d = 15a + 60d $$</p>
<p>$$ 63d - 60d = 15a - 7a $$</p>
<p>$$ 3d = 8a \quad .........\text{(2)} $$</p>
<p>Now we have two equations (1) and (2):</p>
<p>$$ 78 = 2a + 9d \quad \text{(1)} $$</p>
<p>$$ 3d = 8a \quad \text{(2)} $$</p>
<p>We can solve these equations simultaneously.</p>
<p>From equation (2):</p>
<p>$$ d = \frac{8}{3}a $$</p>
<p>Plugging this back into (1):</p>
<p>$$ 78 = 2a + 9\left(\frac{8}{3}a\right) $$</p>
<p>$$ 78 = 2a + 24a $$</p>
<p>$$ 78 = 26a $$</p>
<p>$$ a = 3 $$</p>
<p>Now we can find $d$:</p>
<p>$$ d = \frac{8}{3}a $$</p>
<p>$$ d = \frac{8}{3} \times 3 $$</p>
<p>$$ d = 8 $$</p>
<p>Now we can find $S_{15}$ and $S_5$ using the formula for the sum of an AP.</p>
<p>For $S_{15}$:</p>
<p>$$ S_{15} = \frac{15}{2} (2 \cdot 3 + (15-1) \cdot 8) $$</p>
<p>$$ S_{15} = \frac{15}{2} (6 + 14 \cdot 8) $$</p>
<p>$$ S_{15} = \frac{15}{2} (6 + 112) $$</p>
<p>$$ S_{15} = \frac{15}{2} \cdot 118 $$</p>
<p>$$ S_{15} = 15 \cdot 59 $$</p>
<p>$$ S_{15} = 885 $$</p>
<p>For $S_5$:</p>
<p>$$ S_5 = \frac{5}{2} (2 \cdot 3 + (5-1) \cdot 8) $$</p>
<p>$$ S_5 = \frac{5}{2} (6 + 4 \cdot 8) $$</p>
<p>$$ S_5 = \frac{5}{2} (6 + 32) $$</p>
<p>$$ S_5 = \frac{5}{2} \cdot 38 $$</p>
<p>$$ S_5 = 5 \cdot 19 $$</p>
<p>$$ S_5 = 95 $$</p>
<p>The difference $S_{15} - S_{5}$ is:</p>
<p>$$ S_{15} - S_{5} = 885 - 95 $$</p>
<p>$$ S_{15} - S_{5} = 790 $$</p>
<p>Therefore, the correct answer is Option C, which is 790.</p> | mcq | jee-main-2024-online-1st-february-evening-shift |
lsapy2m8 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $3,7,11,15, \ldots, 403$ and $2,5,8,11, \ldots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ___________. | [] | null | 6699 | <p>To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).
<p>The first AP is:</p></p>
<p>$$3, 7, 11, 15, \ldots, 403$$</p>
<p>The common difference ($d_1$) for the first AP can be calculated by subtracting the first term from the second term:</p>
<p>$$d_1 = 7 - 3 = 4$$</p>
<p>The second AP is:</p>
<p>$$2, 5, 8, 11, \ldots, 404$$</p>
<p>The common difference ($d_2$) for the second AP is:</p>
<p>$$d_2 = 5 - 2 = 3$$</p>
<p>To find the terms common to both sequences, we need to find a term that appears in both sequences. Any common term must be of the form $3 + 4k$ and $2 + 3l$ for some integers $k$ and $l$. We want to find when these two forms will give us the same number, so we set them equal to each other:</p>
<p>$$3 + 4k = 2 + 3l$$</p>
<p>Rearranging the terms gives us:</p>
<p>$$4k - 3l = 2 - 3$$</p>
<p>This simplifies to:</p>
<p>$$4k - 3l = -1 ....... (1)$$</p>
<p>The solutions to equation $(1)$ will give us the common terms. Notice this is a Diophantine equation (A Diophantine equation is a polynomial equation, usually with two or more variables,) and has an infinite number of solutions. Let's find one such solution. We can see that:</p>
<p>$$k = 1 \quad \text{yields} \quad 4(1) - 3l = -1 \implies 4 - 3l = -1 \implies 3l = 5 \implies l = 1\frac{2}{3}$$</p>
<p>This is not an integer solution for $l$, so $k = 1$ does not work. Trying $k = 2$ gives:</p>
<p>$$4(2) - 3l = -1 \implies 8 - 3l = -1 \implies 3l = 9 \implies l = 3$$</p>
<p>Now we've found integers $k = 2$ and $l = 3$ that satisfy the equation. The corresponding term in both sequences would be:</p>
<p>$$3 + 4(2) = 3 + 8 = 11 \quad \text{and} \quad 2 + 3(3) = 2 + 9 = 11$$</p>
<p>Since $11$ is a common term, we can assert that every common term in both APs will be of the form $11 + m(4 \times 3)$, where $m$ is a non-negative integer, and $4 \times 3 = 12$ is the LCM of the common differences of the two APs. Thus, the general form for the common terms would be:</p>
<p>$$11 + 12m$$</p>
<p>Now we are to find all terms that are common up to $403$ in the first sequence and up to $404$ in the second sequence. Because the first sequence doesn't exceed $403$, we'll use this as our limit:</p>
<p>$$11 + 12m \leq 403$$</p>
<p>To find the largest possible integer value for $m$, we solve the inequality:</p>
<p>$$12m \leq 403 - 11$$</p>
<p>$$12m \leq 392$$</p>
<p>$$m \leq 32\frac{2}{3}$$</p>
<p>Since $m$ has to be an integer, the largest possible value for $m$ is $32$. Therefore, the common terms are generated by $m = 0, 1, 2, \ldots, 32$. There are $32 + 1 = 33$ terms in total.
<p>We will now sum these up. The sum of an AP is given by the formula:</p></p>
<p>$$S = \frac{n}{2}(a_1 + a_n)$$</p>
<p>Where $S$ is the sum, $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. Using the formula:</p>
<p>$$S = \frac{33}{2}(11 + (11 + 12 \times 32))$$</p>
<p>$$S = \frac{33}{2}(11 + 11 + 384)$$</p>
<p>$$S = \frac{33}{2}(11 + 11 + 384)$$</p>
<p>$$S = \frac{33}{2}(406)$$</p>
<p>$$S = 33 \times 203$$</p>
<p>$$S = 6699$$</p>
<p>Therefore, the sum of the common terms in the two arithmetic progressions is 6699.</p> | integer | jee-main-2024-online-1st-february-morning-shift |
lsbl2vr8 | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of common terms in the progressions <br/><br/>$4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and <br/><br/>$3,6,9,12, \ldots \ldots$, up to $37^{\text {th }}$ term is : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p>$$4,9,14,19, \ldots$$, up to $$25^{\text {th }}$$ term</p>
<p>$$\mathrm{T}_{25}=4+(25-1) 5=4+120=124$$</p>
<p>$$3,6,9,12, \ldots$$, up to $$37^{\text {th }}$$ term</p>
<p>$$\mathrm{T}_{37}=3+(37-1) 3=3+108=111$$</p>
<p>Common difference of $$\mathrm{I}^{\text {st }}$$ series $$\mathrm{d}_1=5$$</p>
<p>Common difference of $$\mathrm{II}^{\text {nd }}$$ series $$\mathrm{d}_2=3$$</p>
<p>First common term $$=9$$, and their common difference $$=15\left(\operatorname{LCM}\right.$$ of $$\mathrm{d}_1$$ and $$\left.\mathrm{d}_2\right)$$ then common terms are $$9,24,39,54,69,84,99$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscn8k0e | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>$$\text { The } 20^{\text {th }} \text { term from the end of the progression } 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4} \text { is : }$$</p> | [{"identifier": "A", "content": "$$-115$$"}, {"identifier": "B", "content": "$$-100$$"}, {"identifier": "C", "content": "$$-110$$"}, {"identifier": "D", "content": "$$-118$$"}] | ["A"] | null | <p>$$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}$$</p>
<p>This is A.P. with common difference</p>
<p>$$\begin{aligned}
& d_1=-1+\frac{1}{4}=-\frac{3}{4} \\
& -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20
\end{aligned}$$</p>
<p>This is also A.P. $$\mathrm{a}=-129 \frac{1}{4}$$ and $$\mathrm{d}=\frac{3}{4}$$</p>
<p>Required term $$=$$</p>
<p>$$\begin{aligned}
& -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\
& =-129-\frac{1}{4}+15-\frac{3}{4}=-115
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lseyi8z2 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>In an A.P., the sixth term $$a_6=2$$. If the product $$a_1 a_4 a_5$$ is the greatest, then the common difference of the A.P. is equal to</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{8}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$\n"}, {"identifier": "D", "content": "$$\\frac{8}{5}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& a_6=2 \Rightarrow a+5 d=2 \\
& a_1 a_4 a_5=a(a+3 d)(a+4 d) \\
& =(2-5 d)(2-2 d)(2-d) \\
& f(d)=8-32 d+34 d^2-20 d+30 d^2-10 d^3 \\
& f^{\prime}(d)=-2(5 d-8)(3 d-2)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2xo2gn/47c8e21d-8001-480c-b87e-4bf1dfc70a37/1fd43f80-d4a5-11ee-bdd1-01c80c3e2d9a/file-1lt2xo2go.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2xo2gn/47c8e21d-8001-480c-b87e-4bf1dfc70a37/1fd43f80-d4a5-11ee-bdd1-01c80c3e2d9a/file-1lt2xo2go.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) 29th January Morning Shift Mathematics - Sequences and Series Question 25 English Explanation"></p>
<p>$$\mathrm{d}=\frac{8}{5}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfkmm1m | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>If $$\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c}$$ are in an A.P. and $$\log _e \mathrm{a}-\log _e 2 \mathrm{~b}, \log _e 2 \mathrm{~b}-\log _e 3 \mathrm{c}, \log _e 3 \mathrm{c} -\log _e$$ a are also in an A.P, then $$a: b: c$$ is equal to</p> | [{"identifier": "A", "content": "$$6: 3: 2$$\n"}, {"identifier": "B", "content": "$$9: 6: 4$$\n"}, {"identifier": "C", "content": "$$25: 10: 4$$\n"}, {"identifier": "D", "content": "$$16: 4: 1$$"}] | ["B"] | null | <p>$$\log _{\mathrm{e}} \mathrm{a}, \log _{\mathrm{e}} \mathrm{b}, \log _{\mathrm{e}} \mathrm{c}$$ are in A.P.</p>
<p>$$\therefore \mathrm{b}^2=\mathrm{ac}$$ ..... (i)</p>
<p>Also</p>
<p>$$\begin{aligned}
& \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text { are in A.P. } \\
& \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\
& \frac{b}{c}=\frac{3}{2}
\end{aligned}$$</p>
<p>Putting in eq. (i) $$b^2=a \times \frac{2 b}{3}$$</p>
<p>$$\begin{aligned}
& \frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{2} \\
& \mathrm{a}: \mathrm{b}: \mathrm{c}=9: 6: 4
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsgb2052 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$S_n$$ denote the sum of first $$n$$ terms of an arithmetic progression. If $$S_{20}=790$$ and $$S_{10}=145$$, then $$\mathrm{S}_{15}-\mathrm{S}_5$$ is :</p> | [{"identifier": "A", "content": "405"}, {"identifier": "B", "content": "390"}, {"identifier": "C", "content": "410"}, {"identifier": "D", "content": "395"}] | ["D"] | null | <p>$$\begin{aligned}
&\begin{aligned}
& \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \\
& 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\\
& \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \\
& 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)}
\end{aligned}\\
&\text { From (1) and (2) } a=-8, d=5
\end{aligned}$$</p>
<p>$$\begin{aligned}
& S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\
& =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\
& =405-10 \\
& =395
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
lv3vegci | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>An arithmetic progression is written in the following way</p>
<p><img src="data:image/png;base64,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"/></p>
<p>The sum of all the terms of the 10<sup>th</sup> row is _________.</p> | [] | null | 1505 | <p>First term is each row form pattern</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4ntqyk/94850946-5efa-47dd-aab7-199c5589db35/d9f12fc0-10fc-11ef-aaa0-17ca36a32505/file-1lw4ntqyl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4ntqyk/94850946-5efa-47dd-aab7-199c5589db35/d9f12fc0-10fc-11ef-aaa0-17ca36a32505/file-1lw4ntqyl.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) 8th April Evening Shift Mathematics - Sequences and Series Question 11 English Explanation"></p>
<p>$$\begin{aligned}
& \Rightarrow T_n=a n^2+b n+c \\
& \Rightarrow T_1=a+b+c=2 \\
& \Rightarrow T_2=4 a+2 b+c=5 \\
& \Rightarrow T_3=9 a+3 b+c=11 \\
& \Rightarrow 3 a+b=3 \\
& 5 a+b=6 \\
& \Rightarrow 2 a=3 \Rightarrow a=\frac{3}{2}, \quad b=\frac{-3}{2} \Rightarrow c=2 \\
& \Rightarrow T_n=\frac{3}{2} n^2-\frac{3(n)}{2}+2 \Rightarrow \frac{3 n^2-3 n+4}{2} \\
& T_{10}=\frac{3 \times 100-3 \times 10+4}{2}=\frac{274}{2}=137
\end{aligned}$$</p>
<p>Terms in $$10^{\text {th }}$$ row is 10 with 3 differences</p>
<p>$$\begin{aligned}
& \Rightarrow 137,140,143 \ldots \\
& \Rightarrow \quad S_{10}=\frac{10}{2}(2 \times 137+(10-1) \times 3) \\
& =5(274+27)=5 \times 301=1505
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv7v4g4n | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a_1, a_2, a_3, \ldots$$ be in an arithmetic progression of positive terms.</p>
<p>Let $$A_k=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2 k-1}^2-a_{2 k}^2$$.</p>
<p>If $$\mathrm{A}_3=-153, \mathrm{~A}_5=-435$$ and $$\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=66$$, then $$\mathrm{a}_{17}-\mathrm{A}_7$$ is equal to ________.</p> | [] | null | 910 | <p>Let $$a_n=a+(n-1) d \forall n \in N$$</p>
<p>$$\begin{aligned}
A_k & =\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\ldots a_{2 k-1}^2-a_{2 k}^2 \\
& =(-d)\left(a_1+a_2+\ldots+a_{2 k}\right) \\
& A_k=(-d k)(2 a+(2 k-1) d) \\
\Rightarrow & A_3=(-3 d)(2 a+5 d)=-153 \\
\Rightarrow & d(2 a+5 d)=51 \quad \text{... (i)}\\
& A_5=(-5 d)(2 a+9 d)=-435
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow & d(2 a+9 d)=87 \\
\Rightarrow & 4 d^2=36 \Rightarrow d= \pm 3(d=3 \text { positive terms }) \\
\Rightarrow & 3(2 a+27)=87 \\
\Rightarrow & 2 a=29-27 \\
\Rightarrow & a=1 \\
& a_{17}-A_7=(a+16 d)-(-7 d)(2+13 d) \\
& =49+7 \times 3(2+39) \\
& =49+21 \times 41=910
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s20ci | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>For $$x \geqslant 0$$, the least value of $$\mathrm{K}$$, for which $$4^{1+x}+4^{1-x}, \frac{\mathrm{K}}{2}, 16^x+16^{-x}$$ are three consecutive terms of an A.P., is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "16"}] | ["A"] | null | <p>To determine the least value of $$\mathrm{K}$$ for which the terms $$4^{1+x} + 4^{1-x}, \frac{\mathrm{K}}{2}, 16^x + 16^{-x}$$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.</p>
<p>For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms. Therefore, we can write:</p>
<p>$$ \frac{4^{1+x} + 4^{1-x} + 16^x + 16^{-x}}{2} = \frac{K}{2} $$</p>
<p>First, simplify each term individually:</p>
<p>1. Consider $$4^{1+x} + 4^{1-x}$$:</p>
<p>$$4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x} = 4 \cdot 2^{2x}$$</p>
<p>and</p>
<p>$$4^{1-x} = 4 \cdot 4^{-x} = 4 \cdot (2^2)^{-x} = 4 \cdot 2^{-2x} = 4 \cdot 2^{-2x}$$</p>
<p>Thus,</p>
<p>$$4^{1+x} + 4^{1-x} = 4 \cdot 2^{2x} + 4 \cdot 2^{-2x} = 4(2^{2x} + 2^{-2x})$$</p>
<p>2. Consider $$16^x + 16^{-x}$$:</p>
<p>$$16^x = (2^4)^x = 2^{4x}$$</p>
<p>and</p>
<p>$$16^{-x} = (2^4)^{-x} = 2^{-4x}$$</p>
<p>Thus,</p>
<p>$$16^x + 16^{-x} = 2^{4x} + 2^{-4x}$$</p>
<p>3. Combine the terms and set up the equation:</p>
<p>$$ \frac{4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x}}{2} = \frac{K}{2} $$</p>
<p>Multiply both sides by 2:</p>
<p>$$ 4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x} = K $$</p>
<p>To find the least value of $$\mathrm{K}$$, let's assume $$x = 0$$ (since $$x$$ can range over non-negative values):</p>
<p>For $$x = 0$$:</p>
<p>$$4(2^{2 \cdot 0} + 2^{-2 \cdot 0}) + 2^{4 \cdot 0} + 2^{-4 \cdot 0}$$</p>
<p>This simplifies to:</p>
<p>$$4(2^0 + 2^0) + 2^0 + 2^0$$</p>
<p>$$ = 4(1 + 1) + 1 + 1 $$</p>
<p>$$ = 4 \cdot 2 + 1 + 1 $$</p>
<p>$$ = 8 + 1 + 1 $$</p>
<p>$$ = 10 $$</p>
<p>Therefore, the least value of $$\mathrm{K}$$ that ensures the values form an arithmetic progression is $ 10 $. Hence, the correct option is:</p>
<p>Option A: 10</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294px | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $$\mathrm{m}$$ is equal to:</p> | [{"identifier": "A", "content": "125"}, {"identifier": "B", "content": "160"}, {"identifier": "C", "content": "150"}, {"identifier": "D", "content": "180"}] | ["C"] | null | <p>To determine the value of $$\mathrm{m}$$, we need to formulate the problem using some basic concepts of arithmetic progression and work. Let's first understand the nature of the problem:</p>
<p>Initially, there are $$\mathrm{m}$$ computers, and it is estimated that with these $$\mathrm{m}$$ computers, the assignment can be completed in 17 days.</p>
<p>However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.</p>
<p>To begin with, let's define the total work (W) in terms of the number of computers and days:</p>
<p>The total work (W) is given by:</p>
<p>The amount of work completed each day with $$\mathrm{m}$$ computers for 17 days:</p>
<p>$$W = 17m$$</p>
<p>When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are $$\mathrm{m}$$ computers. On the second day, there are $$\mathrm{m} - 4$$ computers, on the third day, there are $$\mathrm{m} - 8$$ computers, and so on. We need to sum this series until 25 days are completed.</p>
<p>This can be formulated as:</p>
<p>Total work done over 25 days with decrement in the number of computers:</p>
<p>$$W = m + (m - 4) + (m - 8) + \ldots + \left[m - 4 \times (n - 1)\right]$$</p>
<p>where $$n$$ is the number of days. Here, $$n = 25$$.</p>
<p>Notice that we form an arithmetic series where the first term (a) is $$\mathrm{m}$$ and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:</p>
<p>$$S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]$$</p>
<p>Plugging in the values:</p>
<p>$$S_{25} = \frac{25}{2} \left[ 2m + (25 - 1)(-4) \right]$$</p>
<p>$$S_{25} = \frac{25}{2} \left[ 2m - 96 \right]$$</p>
<p>$$S_{25} = \frac{25}{2} \left[ 2m - 96 \right] = 25(m - 48)$$</p>
<p>This work should be equivalent to the work calculated earlier, so:</p>
<p>$$17m = 25(m - 48)$$</p>
<p>Solving for $$\mathrm{m}$$:</p>
<p>$$17m = 25m - 1200$$</p>
<p>$$8m = 1200$$</p>
<p>$$m = 150$$</p>
<p>Thus, the value of $$\mathrm{m}$$ is equal to:</p>
<p>Option C: 150</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
1l5w0vbmu | maths | sequences-and-series | arithmetico-geometric-progression | <p>Let for $$f(x) = {a_0}{x^2} + {a_1}x + {a_2},\,f'(0) = 1$$ and $$f'(1) = 0$$. If a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub> are in an arithmatico-geometric progression, whose corresponding A.P. has common difference 1 and corresponding G.P. has common ratio 2, then f(4) is equal to _____________.</p> | [] | null | 2 | <p>Given,</p>
<p>$$f(x) = {a_0}{x^2} + {a_1}x + {a_2}$$</p>
<p>$$f'(0) = 1$$</p>
<p>$$f'(1) = 0$$</p>
<p>a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub> are in A. G. P</p>
<p>Common difference of $$AP = 1$$</p>
<p>Common ratio of $$GP = 2$$</p>
<p>A.P terms = a, a + 1, a + 2</p>
<p>G.P terms = y, ry, r<sup>2</sup>y</p>
<p>$$\therefore$$ AGP terms = ay, (a+1)ry, (a+2)r<sup>2</sup>y</p>
<p>$$\therefore$$ $${a_0} = ay$$</p>
<p>$${a_1} = (a + 1)ry = (a + 1)2y$$</p>
<p>$${a_2} = (a + 2){r^2}y = (a + 2)4y$$</p>
<p>Now, $$f'(x) = 2x{a_0} + {a_1}$$</p>
<p>$$\therefore$$ $$f'(0) = {a_1} = 1$$</p>
<p>and $$f'(1) = 2{a_0} + {a_1} = 0$$</p>
<p>$$ \Rightarrow 2{a_0} + 1 = 0$$</p>
<p>$$ \Rightarrow {a_0} = - {1 \over 2}$$</p>
<p>$$\therefore$$ $$ay = - {1 \over 2}$$</p>
<p>and $$(a + 1)2y = 1$$</p>
<p>$$ \Rightarrow 2ay + 2y = 1$$</p>
<p>$$ \Rightarrow 2 \times \left( { - {1 \over 2}} \right) + 2y = 1$$</p>
<p>$$ \Rightarrow 2y = + \,2$$</p>
<p>$$ \Rightarrow y = + \,1$$</p>
<p>$$\therefore$$ $$a = - {1 \over 2}$$</p>
<p>$$\therefore$$ $${a_2} = (a + 2)4y$$</p>
<p>$$ = \left( { - {1 \over 2} + 2} \right) \times 4\,.\,1$$</p>
<p>$$ = 6$$</p>
<p>$$\therefore$$ $$f(x) = - {1 \over 2}{x^2} + x + 6$$</p>
<p>$$\therefore$$ $$f(4) = - {1 \over 2}{(4)^2} + 4 + 6$$</p>
<p>$$ = - 8 + 10$$</p>
<p>$$ = 2$$</p> | integer | jee-main-2022-online-30th-june-morning-shift |
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