question_id string | subject string | chapter string | topic string | question string | options string | correct_option string | answer string | explanation string | question_type string | paper_id string |
|---|---|---|---|---|---|---|---|---|---|---|
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 =... | 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/>$$
\text { Now, } \frac{a... | integer | jee-main-2023-online-11th-april-morning-shift |
1lgzzo8pe | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\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 \\\\
& \Rightar... | 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.p... | 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}=\... | 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} ... | 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 requ... | 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>$$\begin{aligned}
& P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\
& \text { Put } n=10 \\
& P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0
... | mcq | jee-main-2024-online-9th-april-evening-shift |
luy6z57u | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\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... | 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>$$2 + 6... | mcq | jee-main-2024-online-4th-april-morning-shift |
lv5gst25 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The sum of all the solutions of the equation $$(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$$. To find the roots of this quadratic equation, we can use the quadratic... | mcq | jee-main-2024-online-8th-april-morning-shift |
lvb2952b | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\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}$$ is equal to ________.</p> | [] | null | 4 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwapcytl/545aa1a6-edf5-4e15-9137-baf3d0f62d19/8d6a2390-144f-11ef-860c-d121cbcdd1fc/file-1lwapcytm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwapcytl/545aa1a6-edf5-4e15-9137-baf3d0f62d19/8d6a2390-144f-11ef-860c-d121cbcdd1fc... | integer | jee-main-2024-online-6th-april-evening-shift |
lvc57aw6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\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$$
<br><br... | mcq | jee-main-2015-offline |
UvIf51uHbb6DvoeXkhC7K | maths | sequences-and-series | am,-gm-and-hm | Let x, y, z be positive real numbers such that x + y + z = 12 and x<sup>3</sup>y<sup>4</sup>z<sup>5</sup> = (0.1) (600)<sup>3</sup>. Then x<sup>3</sup> + y<sup>3</sup> + z<sup>3</sup>is equal to : | [{"identifier": "A", "content": "270"}, {"identifier": "B", "content": "258"}, {"identifier": "C", "content": "342"}, {"identifier": "D", "content": "216"}] | ["D"] | null | As we know
<br><br>AM $$ \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... | 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>$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}$$ = $${y \... | mcq | jee-main-2016-online-10th-april-morning-slot |
wuYGlr4ZcEF2WUJlcQDCt | maths | sequences-and-series | am,-gm-and-hm | If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric mean, then $${{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 }}$$
<br><br>Use componendo and Dividendo
<br><br... | mcq | jee-main-2017-online-8th-april-morning-slot |
GNsf4xF6cvZXlPv0FTnFO | maths | sequences-and-series | am,-gm-and-hm | Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression $${{{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 $$
<br><br>Give... | mcq | jee-main-2019-online-12th-january-evening-slot |
PS85QIo4S7Avsvyt1bjgy2xukf49qm01 | maths | sequences-and-series | am,-gm-and-hm | If m arithmetic means (A.Ms) and three
geometric means (G.Ms) are inserted between
3 and 243 such that 4<sup>th</sup> A.M. is equal to 2<sup>nd</sup>
G.M., then m is equal to _________ . | [] | null | 39 | Given m arithmetic means (A.Ms) present between 3 and 243<br><br>$$ \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 ... | 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>$$ \Rightarrow \min ({2^{\sin x}} + {2^{\cos x}}) = {2^{1 - {1 \over {\... | mcq | jee-main-2020-online-4th-september-evening-slot |
FLfk7pqpuOUDRypvuS1klt80kz2 | maths | sequences-and-series | am,-gm-and-hm | The minimum value of $$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}}$$<br>... | integer | jee-main-2021-online-26th-february-evening-slot |
1l55h7hrj | maths | sequences-and-series | am,-gm-and-hm | <p>If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is :</p> | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "23"}, {"identifier": "D", "content": "24"}] | ["C"] | null | <p>a, A<sub>1</sub>, A<sub>2</sub> ........... A<sub>n</sub>, 100</p>
<p>Let d be the common difference of above A.P. then</p>
<p>$${{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 ... | 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 $$... | mcq | jee-main-2022-online-24th-june-evening-shift |
1l6gglr6d | maths | sequences-and-series | am,-gm-and-hm | <p>Consider two G.Ps. 2, 2<sup>2</sup>, 2<sup>3</sup>, ..... and 4, 4<sup>2</sup>, 4<sup>3</sup>, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is $${(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... | 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_{... | ["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 commo... | 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/>$$\frac{\frac... | mcq | jee-main-2023-online-11th-april-evening-shift |
lsappvw8 | maths | sequences-and-series | am,-gm-and-hm | Let $3, a, b, c$ be in A.P. and $3, a-1, b+1, c+9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is : | [{"identifier": "A", "content": "-4"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "11"}] | ["D"] | null | <p>Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:</p>
<p>$$ 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 $$</p>
<p>So, using this formula, we can express $b$ and $c$ ... | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lse508pa | maths | sequences-and-series | am,-gm-and-hm | <p>For $$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$$ may be the geometric mean ... | [{"identifier": "A", "content": "only (II) is true\n"}, {"identifier": "B", "content": "Both (I) and (II) are true\n"}, {"identifier": "C", "content": "only (I) is true\n"}, {"identifier": "D", "content": "Neither (I) nor (II) is true"}] | ["B"] | null | <p>$$\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{align... | 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>$$\begin{aligned}
& 3 a+c+a c=61 \quad \t... | mcq | jee-main-2024-online-4th-april-evening-shift |
vlG8kHqWShZMhOkT | maths | sequences-and-series | arithmetic-progression-(a.p) | If 1, $${\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)... | 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>$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn... | mcq | aieee-2004 |
Xm2Y0pl2ITjxWaRb | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${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} \o... | 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>$$... | mcq | aieee-2010 |
GjqY94gBeyZFKgR8 | maths | sequences-and-series | arithmetic-progression-(a.p) | A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after | [{"identifier": "A", "content": "19 months"}, {"identifier": "B", "content": "20 months"}, {"identifier": "C", "content": "21 months "}, {"identifier": "D", "content": "18 months "}] | ["C"] | null | Let required number of months $$=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} \ri... | 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</s... | 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 = $$\left( {{{a + c} \over 2}} \right)$$ =... | 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>$$ \Rightarrow \,\,\,\,$$ ... | 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... | 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 equ... | [{"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 = $${1 \over {{x_{21}}}} = {1 \over {{x_1}}... | mcq | jee-main-2018-online-16th-april-morning-slot |
d4VJ7d65BnAyLfgZ7j3rsa0w2w9jwy1gult | maths | sequences-and-series | arithmetic-progression-(a.p) | If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ............... a<sub>n</sub> are in A.P. and a<sub>1</sub> + a<sub>4</sub> + a<sub>7</sub> + ........... + a<sub>16</sub> = 114, then a<sub>1</sub> + a<sub>6</sub> + a<sub>11</sub> + a<sub>16</sub> is equal to : | [{"identifier": "A", "content": "38"}, {"identifier": "B", "content": "98"}, {"identifier": "C", "content": "76"}, {"identifier": "D", "content": "64"}] | ["C"] | null | 3(a<sub>1</sub> + a<sub>16</sub>) = 114<br><br>
$${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}}]$$<br><br>
$$ ... | 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> = $${{10} \over 2}\left( {44 - 108} \right) = - ... | mcq | jee-main-2019-online-12th-april-morning-slot |
mkSTbBbv8X594mlprT3rsa0w2w9jx2g4463 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>,......be an A.P. with a<sub>6</sub> = 2. Then the common difference of this A.P., which maximises the
product a<sub>1</sub>a<sub>4</sub>a<sub>5</sub>, is :
| [{"identifier": "A", "content": "$${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 ... | 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>)1... | 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 + $${A \over 2}\left[ {{n^2} - 7... | mcq | jee-main-2019-online-9th-april-morning-slot |
VbSju1U91PISJVdiGZyBw | maths | sequences-and-series | arithmetic-progression-(a.p) | If <sup>n</sup>C<sub>4</sub>, <sup>n</sup>C<sub>5</sub> and <sup>n</sup>C<sub>6</sub> are in A.P., then n can be : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "14"}] | ["D"] | null | 2.<sup>n</sup>C<sub>5</sub> = <sup>n</sup>C<sub>4</sub> + <sup>n</sup><sup></sup>C<sub>6</sub>
<br><br>2.$${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 ... | 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><b... | 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</... | 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> = $${{14} \over 2}\left[ {105 + 196} \right]... | mcq | jee-main-2019-online-8th-april-morning-slot |
WbC8tVOvR5LmL8klSE7k9k2k5khzuf1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of terms common to the two A.P.'s
3, 7, 11, ....., 407 and 2, 9, 16, ....., 709 is ______. | [] | null | 14 | First A.P. is 3, 7, 11, 15, 19, 23, ..... 407
<br><br>d<sub>1</sub> = 4
<br><br>Second A.P. is 2, 9, 16, 23, ..... 709
<br><br>d<sub>2</sub> = 7
<br><br>First common term = 23
<br><br>Common difference of new A.P using the common terms of the two given A.P's is d = L.C.M. (4, 7) = 28
<br><br>Last term $$ \le $$ 407
<br... | integer | jee-main-2020-online-9th-january-evening-slot |
MHxZwe5tFw0FI9dtSrjgy2xukg0cxml1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The common difference of the A.P. <br/>b<sub>1</sub>, b<sub>2</sub>, … , b<sub>m</sub>
is 2 more than the common<br/> difference of A.P. a<sub>1</sub>, a<sub>2</sub>, …, a<sub>n</sub>. If<br/> a<sub>40 </sub> = –159, a<sub>100</sub> = –399 and
b<sub>100</sub> = a<sub>70</sub>, then b<sub>1</sub>
is equal to : | [{"identifier": "A", "content": "127"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "\u2013127"}, {"identifier": "D", "content": "-81"}] | ["D"] | null | Let common difference of series
<br>a<sub>1</sub>
, a<sub>2</sub>
, a<sub>3</sub>
,..., a<sub>n</sub>
be d.
<br><br>$$ \because $$ a<sub>40</sub> = a<sub>1</sub> + 39d == –159 ...(i)
<br><br>and a<sub>100</sub> = a<sub>1</sub> + 99d = –399 ...(ii)
<br><br>From eqn. (ii) and (i)
<br>d = –4 and a<sub>1</sub>
= –3.
<br>... | mcq | jee-main-2020-online-6th-september-evening-slot |
qMKL8WjZUIAD5hRjSpjgy2xukfg6jne3 | maths | sequences-and-series | arithmetic-progression-(a.p) | If $${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 = ... | 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 ... | [{"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>$${S_{n - ... | mcq | jee-main-2020-online-4th-september-evening-slot |
p6030E7IzHd7wg2E0Jjgy2xukf0p91yb | maths | sequences-and-series | arithmetic-progression-(a.p) | If the first term of an A.P. is 3 and the sum of
its first 25 terms is equal to the sum of its next
15 terms, then the common difference of this
A.P. is : | [{"identifier": "A", "content": "$${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 $$ ... | mcq | jee-main-2020-online-3rd-september-morning-slot |
7cFUZoeafMOtnxiXOLjgy2xukez7017n | maths | sequences-and-series | arithmetic-progression-(a.p) | If the sum of first 11 terms of an A.P.,
<br/>a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ....
is 0 (a $$ \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}}$$ .....(1... | mcq | jee-main-2020-online-2nd-september-evening-slot |
m38rP9Tkzz02doHgLE7k9k2k5hj1bjo | maths | sequences-and-series | arithmetic-progression-(a.p) | If the 10<sup>th</sup> term of an A.P. is $${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}}$$
<br><br>$$ \Rightarr... | 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... | 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 su... | 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[... | 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>$$... | mcq | jee-main-2021-online-22th-july-evening-shift |
1krub79cd | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all the elements in the set {n$$\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}$$(3 + 99) $$-$$ 365 $$-$$ 17<br><b... | integer | jee-main-2021-online-22th-july-evening-shift |
1krvs2cgo | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>n</sub> be the sum of the first n terms of an arithmetic progression. If S<sub>3n</sub> = 3S<sub>2n</sub>, then the value of $${{{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... | 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> =... | 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 $$\matrix{
{\underline 2 } & {\underline 3 } & {\underline 1 } \cr
{\underline 3 } & {\underline 4 } & {\underline 1 } \cr
{\underline 4 } & {\underline 5 } & {... | integer | jee-main-2021-online-26th-august-evening-shift |
1ktk9gc86 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... be an A.P. If $${{{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 \ov... | 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>$$\Rightarr... | integer | jee-main-2021-online-31st-august-evening-shift |
1kto9lnec | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, ..........., a<sub>21</sub> be an AP such that $$\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}}}}} \righ... | 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 t... | 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> + ........ | [] | 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... | 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<... | 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, $$\sum\limits_{i = 1}^{{n \over 2}} {{a_{... | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6dx827f | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$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}... | [] | 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-... | 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>$$ \Rightarrow 16{... | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6p3bg7l | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$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}}} + ... | 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 t... | 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 \Righta... | 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}}-... | 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/>
$$
\begin{aligned}
& \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\\\
& y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\\\
& \Rightarrow \frac{7}{6} p q^{... | mcq | jee-main-2023-online-24th-january-morning-shift |
1lgpy2jk4 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$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$$.
<br/><br/>Now, we need to find the sum of the first 12 terms for each A.P. The formula ... | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgutxklh | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$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] =20... | 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/>$$
\begin{aligned}
& \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\\\
& \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\rig... | mcq | jee-main-2023-online-8th-april-morning-shift |
lsamv2vn | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $\mathrm{S}_{15}-\mathrm{S}_5$ is equal to : | [{"identifier": "A", "content": "800"}, {"identifier": "B", "content": "890"}, {"identifier": "C", "content": "790"}, {"identifier": "D", "content": "690"}] | ["C"] | null | <p>To solve this problem, we will start by using the properties of an arithmetic progression (AP).</p>
<p>The sum of the first $n$ terms of an AP can be calculated using the formula:
$$ S_n = \frac{n}{2} (2a + (n-1)d) $$
where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common differ... | mcq | jee-main-2024-online-1st-february-evening-shift |
lsapy2m8 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $3,7,11,15, \ldots, 403$ and $2,5,8,11, \ldots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ___________. | [] | null | 6699 | <p>To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).
<p>The first AP is:</p></p>
<p>$$3, 7, 11, 15, \ldots, 403$$</p... | integer | jee-main-2024-online-1st-february-morning-shift |
lsbl2vr8 | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of common terms in the progressions <br/><br/>$4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and <br/><br/>$3,6,9,12, \ldots \ldots$, up to $37^{\text {th }}$ term is : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p>$$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$$</p>
<p>Common differen... | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscn8k0e | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>$$\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}=... | 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-8001-48... | 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>$$\begin{aligned}
& \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text {... | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsgb2052 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$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... | 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,UklGRjIHAABXRUJQVlA4ICYHAADwYgCdASoAA/MAP4G+2mW2L6wnILMJgsAwCWlu/HyZcetQ2f1w/1/rfMBOYO0eXvb59Dcv6Ip7TtadrTtadrTrnAmZpMwrqGmNsPadrTtadrTtTNrvzQeUko135oPKRhlK3bVgTkiVPHpm7CNG42w9p2tNko135oPKSUa8DEsz2N5RXJ3p1I6IU7Wna07Wna02... | [] | 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-aab7-199c55... | 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 ____... | [] | null | 910 | <p>Let $$a_n=a+(n-1) d \forall n \in N$$</p>
<p>$$\begin{aligned}
A_k & =\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\ldots a_{2 k-1}^2-a_{2 k}^2 \\
& =(-d)\left(a_1+a_2+\ldots+a_{2 k}\right) \\
& A_k=(-d k)(2 a+(2 k-1) d) \\
\Rightarrow & A_3=(-3 d)(2 a+5 d)=-153 \\
\Rightarrow & d(2 a+5 d)=51 \quad \text{... (i... | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s20ci | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>For $$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}$$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.</p>
<p>For three numbers to be in an arithmetic progression, the middle term must b... | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294px | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $$\mathrm{m}$$ is equal to... | [{"identifier": "A", "content": "125"}, {"identifier": "B", "content": "160"}, {"identifier": "C", "content": "150"}, {"identifier": "D", "content": "180"}] | ["C"] | null | <p>To determine the value of $$\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}$$ computers, the assignmen... | mcq | jee-main-2024-online-6th-april-evening-shift |
1l5w0vbmu | maths | sequences-and-series | arithmetico-geometric-progression | <p>Let for $$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>... | integer | jee-main-2022-online-30th-june-morning-shift |