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<p>Suppose $$a_{1}, a_{2}, 2, a_{3}, a_{4}$$ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $$\frac{49}{2}$$, then $$a_{4}$$ is equal to __________.</p>
Options:
[] | 16
Explanation:
Since, common ratio of A.G.P. is 2 therefore A.G.P. can be taken as
<br/><br/>$$
\begin{aligned}
& \frac{(a-2 d)}{4}, \frac{(c-d)}{2}, a, 2(a+d), 4(a+2 d) \\\\
& \text { or } a_1, a_2, 2, a_3, a_4 \text { (Given) } \\\\
& \Rightarrow a=2
\end{aligned}
$$
<br/><br/>also sum of thes A.G.P. is $\frac{49}{... |
<p>Let $$0 < z < y < x$$ be three real numbers such that $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in an arithmetic progression and $$x, \sqrt{2} y, z$$ are in a geometric progression. If $$x y+y z+z x=\frac{3}{\sqrt{2}} x y z$$ , then $$3(x+y+z)^{2}$$ is equal to ____________.</p>
Options:
[] | 150
Explanation:
$\because \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.
<br/><br/>$$
\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y}
$$ ........... (i)
<br/><br/>and $x, \sqrt{2} y, z$ are in G.P.
<br/><br/>$$
\Rightarrow 2 y^2=x z
$$ .......... (ii)
<br/><br/>from (i), $\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 ... |
If $8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2 p)+\frac{1}{4^3}(3+3 p)+\cdots \cdots \infty$, then the value of $p$ is ____________.
Options:
[] | 9
Explanation:
<p>$$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}$$</p>
<p>$$\text { (sum of infinite terms of A.G.P }=\frac{a}{1-r}+\frac{d r}{(1-r)^2} \text { ) }$$</p>
<p>$$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$$</p> |
<p>If $$1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\ldots$$ upto $$\infty=2+\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right)$$, where a and b are integers with $$\operatorname{gcd}(a, b)=1$$, then $$\mathrm{11 a+1... | 76
Explanation:
<p>$$\begin{aligned}
& S=1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\ldots \infty \\\\
& =1+\frac{(1-\sqrt{2} / \sqrt{3})}{2}+\frac{(1-\sqrt{2} / \sqrt{3})^2}{6}+\frac{(1-\sqrt{2} / \sqrt{3})^3}{12}+\ldots \infty
\end{aligned}$$</p>
<p>$$\t... |
<p>Let the first term of a series be $$T_1=6$$ and its $$r^{\text {th }}$$ term $$T_r=3 T_{r-1}+6^r, r=2,3$$,
............ $$n$$. If the sum of the first $$n$$ terms of this series is $$\frac{1}{5}\left(n^2-12 n+39\right)\left(4 \cdot 6^n-5 \cdot 3^n+1\right)$$, then $$n$$ is equal to ___________.</p>
Options:
[] | 6
Explanation:
<p>$$\begin{aligned}
& T_r=3 T_{r-1}+6^r \\
& \Rightarrow \text { solving homogenous part } \\
& T_r=3 T_{r-1} \\
& \Rightarrow x=3 \text { is the root }
\end{aligned}$$</p>
<p>$$\therefore T_r=a .3^r$$</p>
<p>Solving for particular part</p>
<p>$$\begin{aligned}
& T_r=b .6^r \\
& b .6^r=3 b 6^{r-1}+6^r ... |
Fifth term of a GP is 2, then the product of its 9 terms is
Options:
[{"identifier": "A", "content": "256"}, {"identifier": "B", "content": "512 "}, {"identifier": "C", "content": "1024"}, {"identifier": "D", "content": "none of these"}] | ["B"]
Explanation:
$$a{r^4} = 2$$
<br><br>$$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$$
<br><br>$$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$$ |
l, m, n are the $${p^{th}}$$, $${q^{th}}$$ and $${r^{th}}$$ term of a G.P all positive, $$then\,\left| {\matrix{
{\log \,l} & p & 1 \cr
{\log \,m} & q & 1 \cr
{\log \,n} & r & 1 \cr
} } \right|\,equals$$
Options:
[{"identifier": "A", "content": "- 1"}, {"identifier": "B", "cont... | ["D"]
Explanation:
$$l = A{R^{p - 1}}$$
<br><br>$$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$$
<br><br>$$m = A{R^{q - 1}}$$
<br><br>$$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$$
<br><br>$$n = A{R^{r - 1}}$$
<br><br>$$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$$
<br><... |
Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is
Options:
[{"identifier": "A", "content": "5 "}, {"identifier": "B", "content": "3/5 "}, {"identifier": "C", "content": "8/5 "}, {"identifier": "D", "content": "1/5 "}] | ["B"]
Explanation:
Let $$a=$$ first team of $$G.P.$$ and $$r=$$ common ratio of $$G.P.;$$
<br><br>Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$
<br><br>Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$
<br><br>$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$
<br><br>Also $${a^2} + {a^2}{r^2} ... |
In a geometric progression consisting of positive terms, each term equals the sum of the next two terns. Then the common ratio of its progression is equals
Options:
[{"identifier": "A", "content": "$${\\sqrt 5 }$$ "}, {"identifier": "B", "content": "$$\\,{1 \\over 2}\\left( {\\sqrt 5 - 1} \\right)$$ "}, {"identifier"... | ["B"]
Explanation:
Let the series $$a,ar,$$ $$a{r^2},........$$ are in geometric progression.
<br><br>given, $$a = ar + a{r^2}$$
<br><br>$$ \Rightarrow 1 = r + {r^2}$$
<br><br>$$ \Rightarrow {r^2} + r - 1 = 0$$
<br><br>$$ \Rightarrow r = {{ - 1 \mp \sqrt {1 - 4 \times - 1} } \over 2}$$
<br><br>$$ \Rightarrow r = {{ ... |
The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is
Options:
[{"identifier": "A", "content": "- 4"}, {"identifier": "B", "content": "- 12"}, {"identifier": ... | ["B"]
Explanation:
As per question,
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} =... |
Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :
Options:
[{"identifier": "A", "content": "$$2 - \\sqrt 3 $$ "}, {"identifier": "B", "content": "$$2 + \\sqrt 3 $$ "}, {"identifier": "C", "content": "$$\\sqrt 2 ... | ["B"]
Explanation:
Let $$a,ar,a{r^2}$$ are in $$G.P.$$
<br><br>According to the question
<br><br>$$a,2ar,a{r^2}$$ are in $$A.P.$$
<br><br>$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$
<br><br>$$ \Rightarrow 4r = 1 + {r^2}$$
<br><br>$$ \Rightarrow {r^2} - 4r + 1 = 0$$
<br><br>$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \... |
If the $${2^{nd}},{5^{th}}\,and\,{9^{th}}$$ terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is :
Options:
[{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "$${7 \\over 4}$$ "}, {"identifier": "C", "content": "$${8 \\over 5}$$ "}, {"identifier": "D", "content": "$${4 \\... | ["D"]
Explanation:
<p>The terms of an Arithmetic Progression (A.P.) are given by $a$, $a + d$, $a + 2d$, ..., where $a$ is the first term and $d$ is the common difference.</p>
<p>Given that the 2nd, 5th and 9th terms of an A.P. are in Geometric Progression (G.P.), we can denote them as follows :</p>
<p>2nd term = $a +... |
If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval :
Options:
[{"identifier": "A", "content": "($$-$$ $$\\infty $$, $$-$$10]"}, {"identifier": "B", "content": "($$-$$10, 0)"}, {"identifier": "C", "content": "(0, 10)"}, {"identifier": "D", "content": "[10, $$\\infty $$)"}] | ["C"]
Explanation:
Sum of infinite G.P,<br><br>
S = $$b \over {1-r}$$ where $$\left| r \right| < 1$$<br><br>
$$ \Rightarrow $$ 5 = $$b \over {1-r}$$<br><br>
$$ \Rightarrow $$ 1 - r = $$b \over 5$$<br><br>
$$ \Rightarrow $$ b = 5(1 - r)<br><br>
as $$\left| r \right| < 1$$<br><br>
$$ \therefore $$ -1 < r... |
If a, b, c are in A.P. and a<sup>2</sup>, b<sup>2</sup>, c<sup>2</sup> are in G.P. such that
<br/>a < b < c and a + b + c = $${3 \over 4},$$ then the value of a is :
Options:
[{"identifier": "A", "content": "$${1 \\over 4} - {1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over 4} ... | ["C"]
Explanation:
$$ \because $$$$\,\,\,$$a, b, c are in A.P. then
<br><br>a + c = 2b
<br><br>also it is given that,
<br><br>a + b + c = $${{3 \over 4}}$$ . . . .(1)
<... |
If three distinct numbers a, b, c are in G.P. and the
equations ax<sup>2</sup>
+ 2bx + c = 0 and
dx<sup>2</sup>
+ 2ex + ƒ = 0 have a common root, then
which one of the following statements is
correct?
Options:
[{"identifier": "A", "content": "$$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in G.P."}, {"identifier... | ["D"]
Explanation:
Given, a, b, c are in G.P.
<br><br>$$ \therefore $$ b<sup>2</sup> = ac
<br><br>In this equation ax<sup>2</sup>
+ 2bx + c = 0,
<br><br>Discrimant, D = 4b<sup>2</sup> - 4ac
<br><br>= 4ac - 4ac
<br><br>= 0
<br><br>Discrimant = 0 meand roots of the equation are equal.
<br><br>Let both the roots of the ... |
Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$ \ne $$ 0 and 0 < r $$ \le $$ $${1 \over 2}$$
. If 3$$a$$, 7b and 15c are the first three
terms of an A.P., then the 4<sup>th</sup> term of this A.P. is :
Options:
[{"identifier": "A", "content": "$$a$$"}, {"identifier": "B", "content": "$${7 \\over ... | ["A"]
Explanation:
a = a, b = ar and c = ar<sup>2</sup><br><br>
3a, 7b, 15c $$ \to $$ A.P.<br><br>
14b = 3a + 15c<br><br>
14(ar) = 3a + 15(ar<sup>2</sup>)<br><br>
15r<sup>2</sup> – 14r + 3 = 0<br><br>
$$ \Rightarrow r = {1 \over 3},{3 \over 5}(rejected)$$<br><br>
Common difference = 7b – 3a<br><br>
= 7ar – 3a<br><br>
... |
The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is :
Options:
[{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "28"}, {"identifier... | ["B"]
Explanation:
Let terms are $${a \over r},a,ar \to G.P$$
<br><br>$$ \therefore $$ $${a^3}$$ = 512 $$ \Rightarrow $$ a = 8
<br><br>$${8 \over r} + 4,12,8r \to A.P.$$
<br><br>24 = $${8 \over r} + 4 + 8r$$
<br><br>r = 2, r = $${1 \over 2}$$
<br><br>r = 2(4, 8, 16)
<br><br>r = $${1 \over 2}$$ (1... |
Let a<sub>1</sub>, a<sub>2</sub>, . . . . . ., a<sub>10</sub> be a G.P. If $${{{a_3}} \over {{a_1}}} = 25,$$ then $${{{a_9}} \over {{a_5}}}$$ equals
Options:
[{"identifier": "A", "content": "5<sup>3</sup>"}, {"identifier": "B", "content": "2(5<sup>2</sup>)"}, {"identifier": "C", "content": "4(5<sup>2</sup>)"}, ... | ["D"]
Explanation:
a<sub>1</sub>, a<sub>2</sub>, . . . . ., a<sub>10</sub> are in G.P.,
<br><br>Let the common ratio be r
<br><br>$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$
<br><br>$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}... |
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $${{27} \over {19}}$$.Then the common ratio of this series is :
Options:
[{"identifier": "A", "content": "$${4 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${... | ["C"]
Explanation:
$${a \over {1 - r}} = 3\,\,\,\,\,\,\,.....(1)$$
<br><br>$${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$$
<br><br>$$ \Rightarrow 6{r^2} - 13r + 6 = 0$$
<br><br>$$ \Rightarrow r = {2 \over 3}\,\,$$
<br><br>as ... |
Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... a<sub>10</sub> be in G.P. with a<sub>i</sub> > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k $$ \in $$ N (the set of natural numbers) for which
<br/><br/>$$\left| {\matrix{
{{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} &am... | ["D"]
Explanation:
Apply
<br><br>C<sub>3</sub> $$ \to $$ C<sub>3</sub><sub></sub> $$-$$ C<sub>2</sub>
<br><br>C<sub>2</sub><sub></sub> $$ \to $$ C<sub>2</sub> $$-$$ C<sub>1</sub>
<br><br>We get D = 0 |
If a, b, c be three distinct real numbers in G.P. and a + b + c = xb , then x <b>cannot</b> be
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "-3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "-2"}] | ["A"]
Explanation:
a, b, c are in G.P.
<br><br>So, b = ar
<br><br>and c = ar<sup>2</sup>
<br><br>given a + b + c = xb
<br><br>$$ \Rightarrow $$ a + br + ar<sup>2</sup> = x(ar)
<br><br>$$ \Rightarrow $$ 1 + r + r<sup>2</sup> = xr
<br><br>$$ \Rightarrow $$ x = 1 + r + $${1 \o... |
Let $${a_1}$$
, $${a_2}$$
, $${a_3}$$
,....... be a G.P. such that <br/>$${a_1}$$
< 0, $${a_1}$$
+ $${a_2}$$
= 4 and $${a_3}$$
+ $${a_4}$$
= 16. <br/>If $$\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $$, then $$\lambda $$ is
equal to:
Options:
[{"identifier": "A", "content": "171"}, {"identifier": "B", "content":... | ["B"]
Explanation:
$${a_1}$$
+ $${a_2}$$
= 4
<br><br>$$ \Rightarrow $$ $${a_1}$$
+ $${a_1}$$r
= 4 ...(1)
<br><br>$${a_3}$$
+ $${a_4}$$
= 16
<br><br>$$ \Rightarrow $$ $${a_1}$$r<sup>2</sup>
+ $${a_1}$$r<sup>3</sup>
= 16 ...(2)
<br><br>Doing (1) $$ \div $$ (2), we get
<br><br>r = $$ \pm $$ 2
<br><br>If r = 2, th... |
Let a<sub>n</sub> be the n<sup>th</sup> term of a G.P. of positive terms.<br/><br/>
$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$ and $$\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $$,
<br/><br/>
then $$\sum\limits_{n = 1}^{200} {{a_n}} $$ is equal to :
Options:
[{"identifier": "A", "content": "150"}, {"identifie... | ["A"]
Explanation:
$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$
<br><br>$$ \Rightarrow $$ a<sub>3</sub> + a<sub>5</sub> + a<sub>7</sub> + .... + a<sub>201</sub> = 200
<br><br>$$ \Rightarrow $$ $$a{r^2}{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}$$ = 200 ....(1)
<br><br>$$\sum\limits_{n ... |
The sum of the first three terms of a G.P. is S and
their product is 27. Then all such S lie in :
Options:
[{"identifier": "A", "content": "[-3, $$\\infty $$)"}, {"identifier": "B", "content": "(-$$ \\propto $$, 9]"}, {"identifier": "C", "content": "(-$$ \\propto $$, -9] $$ \\cup $$ [-3, $$\\infty $$)"}, {"identifier"... | ["D"]
Explanation:
Let three terms of G.P. are $${a \over r}$$, a, ar
<br><br>$$ \therefore $$ $$a\left( {{1 \over r} + 1 + r} \right)$$ = S ...(1)
<br><br>and a<sup>3</sup>
= 27
<br><br>$$ \Rightarrow $$ a = 3
<br><br>$$ \therefore $$ $$3\left( {{1 \over r} + 1 + r} \right)$$ = S
<br><br>$$ \Rightarrow $$ $${{1 \ove... |
The value of $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$ is equal to ______.
Options:
[] | 4
Explanation:
Given, $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$
<br><br>As sum of GP upto infinity = $${a \over {1 - r}}$$
<br><br>$$ \therefore $$ $${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $$ = $${{{1 \over 3}} \over {1 - {... |
If the sum of the second, third and fourth terms
of a positive term G.P. is 3 and the sum of its
sixth, seventh and eighth terms is 243, then the
sum of the first 50 terms of this G.P. is :
Options:
[{"identifier": "A", "content": "$${2 \\over {13}}\\left( {{3^{50}} - 1} \\right)$$"}, {"identifier": "B", "content": "$... | ["D"]
Explanation:
Let first term = a > 0
<br><br>Common ratio = r > 0
<br><br>ar + ar<sup>2</sup>
+ ar<sup>3</sup>
= 3 ....(i)
<br><br>ar<sup>5</sup>
+ ar<sup>6</sup>
+ ar<sup>7</sup>
= 243 ....(ii)
<br><br>$$ \Rightarrow $$ r<sup>4</sup>(ar + ar<sup>2</sup> + ar<sup>3</sup>) = 243
<br><br>$$ \Rightarrow $... |
Let a , b, c , d and p be any non zero distinct real numbers such that
<br/>(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>)p<sup>2</sup> – 2(ab + bc + cd)p + (b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) = 0. Then :
Options:
[{"identifier": "A", "content": "a, c, p are in G.P."}, {"identifier": "B", "content": "a, b... | ["B"]
Explanation:
(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>)p<sup>2</sup> – 2(ab + bc + cd)p + (b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) = 0
<br><br>$$ \Rightarrow $$ (a<sup>2</sup>p<sup>2</sup>
+ 2abp + b<sup>2</sup>
) + (b<sup>2</sup>p<sup>2</sup>
+ 2bcp + c<sup>2</sup>
) + (c<sup>2</sup>
p<sup>2</sup>
... |
The sum of first four terms of a geometric progression (G. P.) is $${{65} \over {12}}$$ and the sum of their respective reciprocals is $${{65} \over {18}}$$. If the product of first three terms of the G.P. is 1, and the third term is $$\alpha$$, then 2$$\alpha$$ is _________.
Options:
[] | 3
Explanation:
Let the terms are $$a,ar,a{r^2},a{r^3}$$<br><br>$$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$$ ..........(1)<br><br>$${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$$<br><br>$${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over ... |
Let A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, ....... be squares such that for each n $$ \ge $$ 1, the length of the side of A<sub>n</sub> equals the length of diagonal of A<sub>n+1</sub>. If the length of A<sub>1</sub> is 12 cm, then the smallest value of n for which area of A<sub>n</sub> is less than one, is _____... | 9
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266517/exam_images/rfwj0y3zp4ddisjyhq5w.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Mathematics - Sequences and Series Question 146 Englis... |
In an increasing geometric series, the sum of the second and the sixth term is $${{25} \over 2}$$ and the product of the third and fifth term is 25. Then, the sum of 4<sup>th</sup>, 6<sup>th</sup> and 8<sup>th</sup> terms is equal to :
Options:
[{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "32"... | ["D"]
Explanation:
a, ar, ar<sup>2</sup>, .....<br><br>$${T_2} + {T_6} = {{25} \over 2} \Rightarrow ar(1 + {r^4}) = {{25} \over 2}$$<br><br>$${a^2}{r^2}{(1 + {r^4})^2} = {{625} \over 4}$$ .... (1)<br><br>$${T_3}.{T_5} = 25 \Rightarrow (a{r^2})(a{r^4}) = 25$$<br><br>$${a^2}{r^6} = 25$$ .....(2)<br><br>On dividing (1) b... |
Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ___________.
Options:
[] | 3
Explanation:
A.P. from the set will be 11, 16, 21, 26 .....
<br><br>G.P. from the set will be 4, 8, 16, 32, 64, 128, 256,
512, 1024, 2048, 4096, 8192 .....
<br><br>So common terms are 16, 256, 4096. |
Let $${1 \over {16}}$$, a and b be in G.P. and $${1 \over a}$$, $${1 \over b}$$, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ___________.
Options:
[] | 14
Explanation:
$${a^2} = {b \over {16}}$$ and $${2 \over b} = {1 \over a} + 6$$<br><br>Solving, we get $$a = {1 \over {12}}$$ or $$a = - {1 \over 4}$$ [rejected]<br><br>if $$a = {1 \over {12}} \Rightarrow b = {1 \over 9}$$<br><br>$$ \therefore $$ $$72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$$ |
If the sum of an infinite GP a, ar, ar<sup>2</sup>, ar<sup>3</sup>, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar<sup>2</sup>, ar<sup>4</sup>, ar<sup>6</sup>, ....... is :
Options:
[{"identifier": "A", "content": "$${5 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 2}... | ["B"]
Explanation:
Sum of infinite terms :<br><br>$${a \over {1 - r}} = 15$$ ..... (i)<br><br>Series formed by square of terms :<br><br>a<sup>2</sup>, a<sup>2</sup>r<sup>2</sup>, a<sup>2</sup>r<sup>4</sup>, a<sup>2</sup>r<sup>6</sup> .......<br><br>Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$<br><br>$$ \Rightarrow {a \... |
Let a<sub>1</sub>, a<sub>2</sub>, ......., a<sub>10</sub> be an AP with common difference $$-$$ 3 and b<sub>1</sub>, b<sub>2</sub>, ........., b<sub>10</sub> be a GP with common ratio 2. Let c<sub>k</sub> = a<sub>k</sub> + b<sub>k</sub>, k = 1, 2, ......, 10. If c<sub>2</sub> = 12 and c<sub>3</sub> = 13, then $$\sum\li... | 2021
Explanation:
$$a_{1}, a_{2}, a_{3}, \ldots, a_{10}$$ are in AP common difference $$=-3$$<br/><br/> $$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$$ are in GP common ratio $$=2$$<br/><br/> Since, $$c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$$<br/><br/>
$$\therefore c_{2} =a_{2}+b_{2}=12$$<br/><br/>
$$ c_{3} =a_{3}+b_{3}=... |
Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r<sup>2</sup>, then r<sup>2</sup> $$-$$ d is equal to :
Options:
[{"identifier": "A", "content": "... | ["B"]
Explanation:
Let numbers be $${a \over r}$$, a, ar $$\to$$ G.P.<br><br>$${a \over r}$$, 2a, ar $$\to$$ A.P. $$\Rightarrow$$ 4a = $${a \over r}$$ + ar $$\Rightarrow$$ r + $${1 \over r}$$ = 4<br><br>r = 2 $$\pm$$ $$\sqrt 3 $$<br><br>4<sup>th</sup> form of G.P. = 3r<sup>2</sup> $$\Rightarrow$$ ar<sup>2</sup> = 3r<s... |
<p>Let for n = 1, 2, ......, 50, S<sub>n</sub> be the sum of the infinite geometric progression whose first term is n<sup>2</sup> and whose common ratio is $${1 \over {{{(n + 1)}^2}}}$$. Then the value of <br/><br/>$${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$ is eq... | 41651
Explanation:
<p>$${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$$</p>
<p>Now $${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$</p>
<p>$$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\l... |
<p>Let A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, ....... be an increasing geometric progression of positive real numbers. If A<sub>1</sub>A<sub>3</sub>A<sub>5</sub>A<sub>7</sub> = $${1 \over {1296}}$$ and A<sub>2</sub> + A<sub>4</sub> = $${7 \over {36}}$$, then the value of A<sub>6</sub> + A<sub>8</sub> + A<sub>10</... | ["C"]
Explanation:
<p>$${{{A_4}} \over {{r^3}}}.\,{{{A_4}} \over r}.\,{A_4}r\,.\,{A_4}{r^3} = {1 \over {1296}}$$</p>
<p>$${A_4} = {1 \over 6}$$</p>
<p>$${A_2} = {7 \over {36}} - {1 \over 6} = {1 \over {36}}$$</p>
<p>So $${A_6} + {A_8} + {A_{10}} = 1 + 6 + 36 = 43$$</p> |
<p>If a<sub>1</sub> (> 0), a<sub>2</sub>, a<sub>3</sub>, a<sub>4</sub>, a<sub>5</sub> are in a G.P., a<sub>2</sub> + a<sub>4</sub> = 2a<sub>3</sub> + 1 and 3a<sub>2</sub> + a<sub>3</sub> = 2a<sub>4</sub>, then a<sub>2</sub> + a<sub>4</sub> + 2a<sub>5</sub> is equal to ___________.</p>
Options:
[] | 40
Explanation:
<p>Let G.P. be a<sub>1</sub> = a, a<sub>2</sub> = ar, a<sub>3</sub> = ar<sup>2</sup>, .........</p>
<p>$$\because$$ 3a<sub>2</sub> + a<sub>3</sub> = 2a<sub>4</sub></p>
<p>$$\Rightarrow$$ 3ar + ar<sup>2</sup> = 2ar<sup>3</sup></p>
<p>$$\Rightarrow$$ 2ar<sup>2</sup> $$-$$ r $$-$$ 3 = 0</p>
<p>$$\therefor... |
<p>Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be $$\frac{98}{25}$$. Then the sum of the first 21 terms of an AP, whose first term is $$10\mathrm{a r}, \mathrm{n}^{\text {th }}$$ term is $$\mathrm{a}_{\mathrm{n}}$$ and the common differenc... | ["A"]
Explanation:
<p>Let first term of G.P. be a and common ratio is r</p>
<p>Then, $${a \over {1 - r}} = 5$$ ...... (i)</p>
<p>$$a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}$$</p>
<p>$$\therefore$$ $${r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3... |
<p>If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is</p>
Options:
[{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\fra... | ["A"]
Explanation:
$\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}, \mathrm{ar}^{3}(\mathrm{a}, \mathrm{r}>0)$
<br/><br/>$a^{4} r^{6}=1296$
<br/><br/>$a^{2} r^{3}=36$
<br/><br/>$a=\frac{6}{r^{3 / 2}}$
<br/><br/>$a+a r+a r^{2}+a r^{3}=126$
<br/><br/>$\frac{1}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}}{\mathrm{r}^{3 / 2}}+\fr... |
Let $a, b, c>1, a^3, b^3$ and $c^3$ be in A.P., and $\log _a b, \log _c a$ and $\log _b c$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\frac{a+4 b+c}{3}$ and the common difference is $\frac{a-8 b+c}{10}$ is $-444$, then $a b c$ is equal to :
Options:
[{"identifier": "A", "content": "343... | ["B"]
Explanation:
<p>$$2{b^3} = {a^3} + {c^3}$$</p>
<p>$${\left( {{{\log a} \over {\log c}}} \right)^2} = \left( {{{\log b} \over {\log a}}} \right)\left( {{{\log c} \over {\log b}}} \right)$$</p>
<p>$$ \Rightarrow {(\log a)^3} = {(\log c)^3}$$</p>
<p>$$ \Rightarrow \log a = \log c$$</p>
<p>$$ \Rightarrow a = c$$</p>... |
<p>Let $$\{ {a_k}\} $$ and $$\{ {b_k}\} ,k \in N$$, be two G.P.s with common ratios $${r_1}$$ and $${r_2}$$ respectively such that $${a_1} = {b_1} = 4$$ and $${r_1} < {r_2}$$. Let $${c_k} = {a_k} + {b_k},k \in N$$. If $${c_2} = 5$$ and $${c_3} = {{13} \over 4}$$ then $$\sum\limits_{k = 1}^\infty {{c_k} - (12{a_6} +... | 9
Explanation:
<p>$$\{ {a_k}\} $$ be a G.P. with $${a_1} = 4,r = {r_1}$$</p>
<p>And</p>
<p>$$\{ {b_k}\} $$ be G.P. with $${b_1} = 4,r = {r_2}$$ $$({r_1} < {r_2})$$</p>
<p>Now</p>
<p>$${C_k} = {a_k} + {b_k}$$</p>
<p>$${c_1} = 4 + 4 = 8$$ and $${c_2} = 5$$<p>
<p>$${a_2} + {b_2} = 5$$</p>
<p>$$\therefore$$ $${r_1} + {r_2... |
<p>Let $$a_1,a_2,a_3,...$$ be a $$GP$$ of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $$a_1a_9+a_2a_4a_9+a_5+a_7$$ is equal to __________.</p>
Options:
[] | 60
Explanation:
Let $r$ be the common ratio of the G.P
<br/><br/>
$\therefore a_{1} r^{3} \times a_{1} r^{5}=9$
<br/><br/>
$a_{1}^{2} r^{8}=9 \Rightarrow a_{1} r^{4}=3$
<br/><br/>
And
<br/><br/>
$$
\begin{aligned}
& a_{1}\left(r^{4}+r^{6}\right)=24 \\\\
\Rightarrow & 3\left(1+r^{2}\right)=24 \\\\
\therefore & r^{2}=7 ... |
<p>For the two positive numbers $$a,b,$$ if $$a,b$$ and $$\frac{1}{18}$$ are in a geometric progression, while $$\frac{1}{a},10$$ and $$\frac{1}{b}$$ are in an arithmetic progression, then $$16a+12b$$ is equal to _________.</p>
Options:
[] | 3
Explanation:
$$
\begin{aligned}
& \mathrm{a}, \mathrm{b}, \frac{1}{18} \rightarrow \mathrm{GP} \\\\
& \frac{\mathrm{a}}{18}=\mathrm{b}^2\quad...(i) \\\\
& \frac{1}{\mathrm{a}}, 10, \frac{1}{\mathrm{~b}} \rightarrow \mathrm{AP} \\\\
& \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=20 \\\\
& \Rightarrow \mathrm{a}+\mathrm... |
<p>The 4$$^\mathrm{th}$$ term of GP is 500 and its common ratio is $$\frac{1}{m},m\in\mathbb{N}$$. Let $$\mathrm{S_n}$$ denote the sum of the first n terms of this GP. If $$\mathrm{S_6 > S_5 + 1}$$ and $$\mathrm{S_7 < S_6 + \frac{1}{2}}$$, then the number of possible values of m is ___________</p>
Options:
[] | 12
Explanation:
$T_{4}=500$
<br/><br/>
$$
a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}}
$$
<br/><br/>
Now,
<br/><br/>
$$
\begin{aligned}
& S_{6} > S_{5}+1 \\\\
& \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\
& a r^{5} > 1 \\\\
& \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}... |
<p>Let a$$_1$$, a$$_2$$, a$$_3$$, .... be a G.P. of increasing positive numbers. Let the sum of its 6<sup>th</sup> and 8<sup>th</sup> terms be 2 and the product of its 3<sup>rd</sup> and 5<sup>th</sup> terms be $$\frac{1}{9}$$. Then $$6(a_2+a_4)(a_4+a_6)$$ is equal to</p>
Options:
[{"identifier": "A", "content": "2$$\... | ["D"]
Explanation:
<p>Given the conditions :</p>
<ol>
<li>$a_6 + a_8 = 2 \Rightarrow a r^5 + a r^7 = 2$</li>
<li>$a_3 \cdot a_5 = \frac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4 = \frac{1}{9} \Rightarrow a r^3 = \frac{1}{3}$</li>
</ol>
<p>From this, we can form the equation $\frac{r^2}{3} + \frac{r^4}{3} = 2$, which s... |
<p>Let the first term $$\alpha$$ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to</p>
Options:
[{"identifier": "A", "content": "241"}, {"identifier": "B", "content": "231"}, {"identifier": "C",... | ["B"]
Explanation:
Given that the first term $a$ and common ratio $r$ of a geometric progression be positive integer. So, their 1st three terms are $a, a r, a r^2$
<br/><br/>According to the question, $a^2+a^2 r^2+a^2 r^4=33033$
<br/><br/>$$
\begin{aligned}
\Rightarrow a^2\left(1+r^2+r^4\right) & =3 \times 7 \times 11... |
<p>If
<br/><br/>$$(20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}=k(20)^{19}$$,
<br/><br/>then $$k$$ is equal to ___________.</p>
Options:
[] | 400
Explanation:
$\begin{aligned} &(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17} \\ & \quad+\ldots \ldots+20(21)^{19}=k(20)^{19} \\\\ & \Rightarrow(20)^{19}\left[1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right]=k(20)^{19} \\\\ & \Rightarrow k=1+2\left(\frac{21}{20... |
If three successive terms of a G.P. with common ratio $\mathrm{r}(\mathrm{r}>1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to _____________.
Options:
[] | 1
Explanation:
<p>To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $ar^2$, where $a$ is the first term and $r > 1$. These three terms represent the lengths of the sides of a triangle.</p>
<p>According to the triangle i... |
<p>Let $$2^{\text {nd }}, 8^{\text {th }}$$ and $$44^{\text {th }}$$ terms of a non-constant A. P. be respectively the $$1^{\text {st }}, 2^{\text {nd }}$$ and $$3^{\text {rd }}$$ terms of a G. P. If the first term of the A. P. is 1, then the sum of its first 20 terms is equal to -</p>
Options:
[{"identifier": "A", "c... | ["D"]
Explanation:
<p>$$\begin{aligned}
& 1+d, \quad 1+7 d, 1+43 d \text { are in GP } \\
& (1+7 d)^2=(1+d)(1+43 d) \\
& 1+49 d^2+14 d=1+44 d+43 d^2 \\
& 6 d^2-30 d=0 \\
& d=5 \\
& S_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] \\
& \quad=10[2+95] \\
& \quad=970
\end{aligned}$$</p> |
<p>If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to</p>
Options:
[{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["B"]
Explanation:
<p>$$\begin{aligned}
& a+a r+a r^2+a r^3+\ldots .+a r^{63} \\
& =7\left(a+a r^2+a r^4 \ldots .+a r^{62}\right) \\
& \Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2} \\
& r=6
\end{aligned}$$</p> |
<p>If each term of a geometric progression $$a_1, a_2, a_3, \ldots$$ with $$a_1=\frac{1}{8}$$ and $$a_2 \neq a_1$$, is the arithmetic mean of the next two terms and $$S_n=a_1+a_2+\ldots . .+a_n$$, then $$S_{20}-S_{18}$$ is equal to</p>
Options:
[{"identifier": "A", "content": "$$-2^{15}$$\n"}, {"identifier": "B", "con... | ["A"]
Explanation:
<p>Let $$r^{\prime}$$th term of the GP be $$a^{n-1}$$. Given,</p>
<p>$$\begin{aligned}
& 2 a_r=a_{r+1}+a_{r+2} \\
& 2 a r^{n-1}=a r^n+a r^{n+1} \\
& \frac{2}{r}=1+r \\
& r^2+r-2=0
\end{aligned}$$</p>
<p>Hence, we get, $$r=-2$$ (as $$r \neq 1$$)</p>
<p>So, $$\mathrm{S}_{20}-\mathrm{S}_{18}=$$ (Sum up... |
<p>Let $$a$$ and $$b$$ be be two distinct positive real numbers. Let $$11^{\text {th }}$$ term of a GP, whose first term is $$a$$ and third term is $$b$$, is equal to $$p^{\text {th }}$$ term of another GP, whose first term is $$a$$ and fifth term is $$b$$. Then $$p$$ is equal to</p>
Options:
[{"identifier": "A", "con... | ["C"]
Explanation:
<p>The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified ... |
<p>Let $$a, a r, a r^2$$, ............ be an infinite G.P. If $$\sum_\limits{n=0}^{\infty} a r^n=57$$ and $$\sum_\limits{n=0}^{\infty} a^3 r^{3 n}=9747$$, then $$a+18 r$$ is equal to</p>
Options:
[{"identifier": "A", "content": "27"}, {"identifier": "B", "content": "38"}, {"identifier": "C", "content": "31"}, {"identi... | ["C"]
Explanation:
<p>$$\begin{array}{ll}
\sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\
\sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)}
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=... |
<p>Let the first three terms 2, p and q, with $$q \neq 2$$, of a G.P. be respectively the $$7^{\text {th }}, 8^{\text {th }}$$ and $$13^{\text {th }}$$ terms of an A.P. If the $$5^{\text {th }}$$ term of the G.P. is the $$n^{\text {th }}$$ term of the A.P., then $n$ is equal to:</p>
Options:
[{"identifier": "A", "cont... | ["C"]
Explanation:
<p>$$\begin{aligned}
& \text { Let } p=2 r, q=2 r^2 \\
& T_7=2, T_8=2 r, T_{13}=2 r^2 \\
& d=2 r-2=2(r-1) \\
& 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\
& \Rightarrow r^2-6 r+5=0 \\
& \Rightarrow(r-1)(r-5)=0 \\
& \therefore r=1,5 \\
& r=1 \text { (rejected) as } q \neq 2 \\
& \therefore r=5
\end{aligned}... |
<p>In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $$\frac{70}{3}$$ and the product of the third and fifth terms is 49. Then the sum of the $$4^{\text {th }}, 6^{\text {th }}$$ and $$8^{\text {th }}$$ terms is equal to:</p>
Options:
[{"identifier": "A", "content": "78... | ["C"]
Explanation:
<p>Let's denote the first term of the geometric progression by $$a$$ and the common ratio by $$r$$. The terms of the geometric progression can be written as follows:</p>
<p>First term: $$a$$</p>
<p>Second term: $$ar$$</p>
<p>Third term: $$(ar^2)$$</p>
<p>Fourth term: $$(ar^3)$$</p>
<p>Fifth ter... |
<p>Let $$A B C$$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $$A B C$$ and the same process is repeated infinitely many times. If $$\mathrm{P}$$ is the sum of perimeters and $$Q$$ is be the sum of areas of all the triangles formed in this process, then ... | ["B"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwac04ho/c0187de2-d2dd-45ec-8dc9-5238b7084f24/51988fc0-141b-11ef-aad1-15919f5484e3/file-1lwac04hp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwac04ho/c0187de2-d2dd-45ec-8dc9-5238b7084f24/51988fc0-141b-11... |
If $$x = \sum\limits_{n = 0}^\infty {{a^n},\,\,y = \sum\limits_{n = 0}^\infty {{b^n},\,\,z = \sum\limits_{n = 0}^\infty {{c^n},} } } \,\,$$ where a, b, c are in A.P and $$\,\left| a \right| < 1,\,\left| b \right| < 1,\,\left| c \right| < 1$$ then x, y, z are in
Options:
[{"identifier": "A", "content": "G.... | ["D"]
Explanation:
$$x = \sum\limits_{n = 0}^\infty {{a^n}} = {1 \over {1 - a}}\,\,\,\,\,\,\,\,\,\,a = 1 - {1 \over x}$$
<br><br>$$y = \sum\limits_{n = 0}^\infty {{b^n}} = {1 \over {1 - b}}\,\,\,\,\,\,\,\,\,\,b = 1 - {1 \over y}$$
<br><br>$$z = \sum\limits_{n = 0}^\infty {{c^n}} = {1 \over {1 - c}}\,\,\,\,\,\,\,... |
If $${{a_1},{a_2},....{a_n}}$$ are in H.P., then the expression $${{a_1}\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$$ is equal to
Options:
[{"identifier": "A", "content": "$$n({a_1}\\, - {a_n})$$ "}, {"identifier": "B", "content": "$$(n - 1)({a_1}\\, - {a_n})$$ "}, {"identifier": "C", "content": "$$n{a_1}... | ["D"]
Explanation:
$${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$$ (say)
<br><br>Then $${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$$
<br><br>$... |
<p>$$x = \sum\limits_{n = 0}^\infty {{a^n},y = \sum\limits_{n = 0}^\infty {{b^n},z = \sum\limits_{n = 0}^\infty {{c^n}} } } $$, where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc $$\ne$$ 0, then :</p>
Options:
[{"identifier": "A", "content": "x, y, z are in A.P."}, {"identifier": "B", "content": ... | ["C"]
Explanation:
<p>$$x = \sum\limits_{n = 0}^\infty {{a^n} = {1 \over {1 - a}};\,y = \sum\limits_{n = 0}^\infty {{b^n} = {1 \over {1 - b}};\,z = \sum\limits_{n = 0}^\infty {{c^n} = {1 \over {1 - c}}} } } $$</p>
<p>Now,</p>
<p>a, b, c $$\to$$ AP</p>
<p>1 $$-$$ a, 1 $$-$$ b, 1 $$-$$ c $$\to$$ AP</p>
<p>$${1 \over ... |
$${1^3} - \,\,{2^3} + {3^3} - {4^3} + ... + {9^3} = $$
Options:
[{"identifier": "A", "content": "425"}, {"identifier": "B", "content": "- 425"}, {"identifier": "C", "content": "475"}, {"identifier": "D", "content": "- 475"}] | ["A"]
Explanation:
$${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$$
<br><br>$$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$$
<br><br>$$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right... |
The value of $$\,{2^{1/4}}.\,\,{4^{1/8}}.\,{8^{1/16}}...\infty $$ is
Options:
[{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "3/2 "}, {"identifier": "D", "content": "4"}] | ["B"]
Explanation:
The product is $$p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$$
<br><br>$$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$$
<br><br>Now let
<br><br>$$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$$
<br><br>$${1 \over 2}S = {1 \over 8} + {2 \over {16}} + ..... |
The sum of the serier $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..............$$ up to $$\infty $$ is equal to
Options:
[{"identifier": "A", "content": "$$\\log {\\,_e}\\left( {{4 \\over e}} \\right)\\,\\,$$ "}, {"identifier": "B", "content": "$$2\\,\\log {\\,_e}2$$ "}, {"identifier": "C", "content": "$$\\... | ["A"]
Explanation:
$${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $$
<br><br>$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$
<br><br>$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$
<br><br>$$ = \left( {{1 \over 1}... |
The sum of the first n terms of the series $${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}$$ when n is even. When n is odd the sum is
Options:
[{"identifier": "A", "content": "$${\\left[ {{{n(n + 1)} \\over 2}} \\right]^2}$$ "}, {"identifier": "B", "content": "$${{{n^2}(n ... | ["B"]
Explanation:
If $$n$$ is odd, the required sum is
<br><br>$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$
<br><br>$$ = {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$$
<br><br>[ As $$\left( {n - 1} \... |
The sum of series $${1 \over {2\,!}} + {1 \over {4\,!}} + {1 \over {6\,!}} + ........$$ is
Options:
[{"identifier": "A", "content": "$${{\\left( {{e^2} - 2} \\right)} \\over e}\\,$$ "}, {"identifier": "B", "content": "$${{{{\\left( {e - 1} \\right)}^2}} \\over {2e}}$$ "}, {"identifier": "C", "content": "$${{\\left( {... | ["B"]
Explanation:
We know that
<br><br>$$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$$
<br><br>and
<br><br>$${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$$
<br><br>$$\therefore$$ $$e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \rig... |
The sum of the series $$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$ ad inf. is
Options:
[{"identifier": "A", "content": "$${{e - 1} \\over {\\sqrt e }}\\,$$ "}, {"identifier": "B", "content": "$${{e + 1} \\over {\\sqrt e }}$$ "}, {"identifier": "C", "content": "$${{e - 1} \\over {2\\sqrt e... | ["D"]
Explanation:
$${{{e^x} + {e^{ - x}}} \over 2}$$
<br><br>$$ = 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........$$
<br><br>Putting $$x = {1 \over 2}$$ we get
<br><br>$$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$
<br><br>$$\infty = {{{e^{{1 \over 2}}} + {e... |
The sum of series $${1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - .......$$ upto infinity is
Options:
[{"identifier": "A", "content": "$${e^{ - {1 \\over 2}}}$$ "}, {"identifier": "B", "content": "$${e^{ + {1 \\over 2}}}$$"}, {"identifier": "C", "content": "$${e^{ - 2}}$$ "}, {"identifier": "D", "content": "$${e... | ["D"]
Explanation:
We know that $${e^x} = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ........\infty $$
<br><br>Put $$x=-1$$
<br><br>$$\therefore$$ $${e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty $$
<br><br>$$\therefore$$ $${e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} ... |
The sum to infinite term of the series $$1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + .....$$ is
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["A"]
Explanation:
We have
<br><br>$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........\infty \,\,\,\,\,...\left( 1 \right)$$
<br><br>Multiplying both sides by $${1 \over 3}$$ we get
<br><br>$${1 \over 3}S = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + {{10} \over... |
<p> <b> Statement-1: </b> The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +.....+ (361 + 380 + 400) is 8000.
</p><p> <b> Statement-2: </b> $$\sum\limits_{k = 1}^n {\left( {{k^3} - {{(k - 1)}^3}} \right)} = {n^3}$$, for any natural number n.</p>
Options:
[{"identifier": "A", "content": "Statement... | ["B"]
Explanation:
$$n$$<sup>th </sup> term of the given series
<br><br>$$ = {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$$
<br><br>$$ = {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$$
<br><br>$$ = {n^3} - {\left( {n - 1} \right)^3}$$
<br><br>$$ \R... |
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,........,is
Options:
[{"identifier": "A", "content": "$${7 \\over {81}}\\left( {179 - {{10}^{ - 20}}} \\right)$$ "}, {"identifier": "B", "content": "$$\\,{7 \\over 9}\\left( {99 - {{10}^{ - 20}}} \\right)$$ "}, {"identifier": "C", "content": "$${7 \\over {81}... | ["C"]
Explanation:
Given sequence can be written as
<br><br>$${7 \over {10}} + {{77} \over {100}} + {{777} \over {{{10}^3}}} + ..... + $$ up to $$20$$ terms
<br><br>$$ = 7\left[ {{1 \over {10}} + {{11} \over {100}} + {{111} \over {{{10}^3}}} + ...... + } \right.\,\,$$ up to $$20$$ terms ]
<br><br>Multiply and divide b... |
If $${(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + ......... + 10{(11)^9} = k{(10)^9},$$, then k is equal to :
Options:
[{"identifier": "A", "content": "100 "}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "$${{121} \\over {10}}$$ "}, {"identifier": "D", "content": "$${{441} \\over {... | ["A"]
Explanation:
Let $${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$
<br><br>Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\lef... |
The sum of first 9 terms of the series.
<br/><br/>$${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......$$
Options:
[{"identifier": "A", "content": "142"}, {"identifier": "B", "content": "192"}, {"identifier": "C", "content": "71"}, {"identifier": "D", "content": ... | ["D"]
Explanation:
$${n^{th}}$$ term of series
<br><br>$$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$
<br><br>Sum of $$n$$ term $$ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$
<br><br>$$ = {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \rig... |
If the sum of the first ten terms of the series $${\left( {1{3 \over 5}} \right)^2} + {\left( {2{2 \over 5}} \right)^2} + {\left( {3{1 \over 5}} \right)^2} + {4^2} + {\left( {4{4 \over 5}} \right)^2} + .......is\,{{16} \over 5}m,$$ then m is equal to :
Options:
[{"identifier": "A", "content": "100"}, {"identifier": "B... | ["D"]
Explanation:
$${\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}$$
<br><br>$$S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4... |
Let z = 1 + ai be a complex number, a > 0, such that z<sup>3</sup> is a real number.
<br/><br/>Then the sum 1 + z + z<sup>2</sup> + . . . . .+ z<sup>11</sup> is equal to :
Options:
[{"identifier": "A", "content": "$$ - 1250\\,\\sqrt 3 \\,i$$ "}, {"identifier": "B", "content": "$$ 1250\\,\\sqrt 3 \\,i$$ "}, {"id... | ["D"]
Explanation:
z = 1 + ai
<br><br>z<sup>2</sup> = 1 $$-$$ a<sup>2</sup> + 2ai
<br><br>z<sup>2</sup> . z = {(1 $$-$$ a<sup>2</sup>) + 2ai} {1 + ai}
<br><br>= (1 $$-$$ a<sup>2</sup>) + 2ai + (1 $$-$$ a<sup>2</sup>) ai $$-$$ 2a<sup>2</sup>
<br><br>$$ \because $$ z<sup>3</sup... |
For any three positive real numbers a, b and c,
<br/><br/>9(25$${a^2}$$ + b<sup>2</sup>) + 25(c<sup>2</sup> - 3$$a$$c) = 15b(3$$a$$ + c).
<br/>Then
Options:
[{"identifier": "A", "content": "b, c and $$a$$ are in G.P."}, {"identifier": "B", "content": "b, c and $$a$$ are in A.P."}, {"identifier": "C", "content": "$$a$$... | ["B"]
Explanation:
9(25$${a^2}$$ + b<sup>2</sup>) + 25(c<sup>2</sup> - 3$$a$$c) = 15b(3$$a$$ + c)
<br><br> $$ \Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc$$
<br><br>$$ \Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac = 45ab + 15bc$$
<br><br>$$ \Rightarrow ... |
Let
<br/><br/>S<sub>n</sub> = $${1 \over {{1^3}}}$$$$ + {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.$$
<br/><br/>If 100 S<sub>n</sub> = n, then n is equal to :
Options:
[{"identifier": "A", "content": "199... | ["A"]
Explanation:
n<sup>th</sup> term, T<sub>n</sub> = $${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$$
<br><br>T<sub>n</sub> = $${{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ T<sub>n</sub> = $${2 \over {n\left( {n +... |
If the sum of the first n terms of the series $$\,\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$$ is $$435\sqrt 3 ,$$ then n equals :
Options:
[{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "29"}] | ["B"]
Explanation:
Given,
<br><br>$$\sqrt 3 $$ + $$\sqrt {75} $$ + $$\sqrt {243} $$ + $$\sqrt {507} $$ + . . . . . .+ n terms
<br><br>= $$\sqrt 3 $$ + $$\sqrt {25 \times 3} $$ + $$\sqrt {81 \times 3} $$ + $$\sqrt {169 \times 3} $$ + . . . . . .+ n terms
<br><br>= $$\sqrt 3 $$ + 5$$\sqrt 3 $$ + 9$$\sqrt 3 $$ + 13$$\s... |
Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series
<br/>1<sup>2</sup> + 2.2<sup>2</sup> + 3<sup>2</sup> + 2.4<sup>2</sup> + 5<sup>2</sup> + 2.6<sup>2</sup> ...........
<br/>If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to
Options:
[{"identifier": "A", "content": "496"... | ["C"]
Explanation:
<b><u>Note</u> : </b>
<br><br>Sum of square of first n odd terms
<br><br>1<sup>2</sup> + 3<sup>2</sup> + 5<sup>2</sup> + . . . . .+ n<sup>2</sup> = $${{n\left( {2n - 1} \right)\left( {2n + 1} \right)} \over 3}$$
<br><br>Given,
<br><br>1<sup>2</sup> + 2. 2<sup>2</sup> + 3<sup>2</sup> + 2.4<sup>2... |
The sum of the first 20 terms of the series
<br/><br/>$$1 + {3 \over 2} + {7 \over 4} + {{15} \over 8} + {{31} \over {16}} + ...,$$ is :
Options:
[{"identifier": "A", "content": "$$38 + {1 \\over {{2^{19}}}}$$"}, {"identifier": "B", "content": "$$38 + {1 \\over {{2^{20}}}}$$"}, {"identifier": "C", "content": "$$39 +... | ["A"]
Explanation:
1 + $${3 \over 2}$$ + $${7 \over 4}$$ + $${15 \over 8}$$ + $${31 \over 16}$$ + . . . .
<br><br>= (2 $$-$$ 1) + (2 $$-$$ $${1 \over 2}$$ ) + (2 $$-$$ $${1 \over 4}$$) + (2 $$-$$ $${1 \over 8}$$) + . . . . .+ 20 terms
<br><br>= (2 + 2 + . . . . . 20 terms) $$-$$ (1 + $${1 \over 2}$$ +... |
Let A<sub>n</sub> = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3}$$ $$-$$. . . . . + ($$-$$1)<sup>n-1</sup> $${\left( {{3 \over 4}} \right)^n}$$ and B<sub>n</sub> = 1 $$-$$ A<sub>n</sub>.
<br/>Then, the least dd natural numbr p, so that B<sub>n</sub> >... | ["B"]
Explanation:
A<sub>n</sub> = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3} - .... + {\left( { - 1} \right)^{n - 1}}{\left( {{3 \over 4}} \right)^n}$$
<br><br>Which in a G.P. with a = $${{3 \over 4}}$$, r = $${{{ - 3} \over 4}}$$ and number of terms = n
<br>... |
For x $$\varepsilon $$ R, let [x] denote the greatest integer $$ \le $$ x, then the sum of the series
$$\left[ { - {1 \over 3}} \right] + \left[ { - {1 \over 3} - {1 \over {100}}} \right] + \left[ { - {1 \over 3} - {2 \over {100}}} \right] + .... + \left[ { - {1 \over 3} - {{99} \over {100}}} \right]$$ is :
Options:
[... | ["C"]
Explanation:
$$\left[ {{{ - 1} \over 3}} \right] + \left[ {{{ - 1} \over 3} - {1 \over {100}}} \right] + \left[ {{{ - 1} \over 3} - {2 \over {100}}} \right] +$$<br><br>
$$ ....... + \left[ {{{ - 1} \over 3} - {{99} \over {100}}} \right]$$<br><br>
$$ \Rightarrow ( - 1 - 1 - 1 - .....67\,times) + ( - 2 - 2 - 2 - ... |
The sum
<br/>$$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$$ is equal to :
Options:
[{"identifier": "A", "content": "620"}, {"identifier": "B",... | ["A"]
Explanation:
$$Sum = \sum\limits_{n = 1}^{15} {{{{1^3} + {2^3} + .... + {n^3}} \over {1 + 2 + .... + n}}} - {1 \over 2}{{15 \times 16} \over 2}$$<br><br>
$$ = \sum\limits_{n = 1}^{15} {{{n(n + 1)} \over 2}} - 60$$<br><br>
$$ = {1 \over 2}\sum\limits_{n = 1}^{15} {{n^2}} + {1 \over 2}\sum\limits_{n = 1}^{15} {... |
The sum
<br/>$${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$$ upto 10 terms is:
Options:
[{"identifier": "A", "content": "600"}, {"identifier": "B", "content": "660"}, {"identifier": "C"... | ["B"]
Explanation:
$${T_r} = {{(2r + 1)({1^3} + {2^3} + {3^3} + ...... + {r^3})} \over {{1^2} + {2^2} + {3^2} + ...... + {r^2}}}$$<br><br>
$${T_r} = (2r + 1){\left( {{{r(r + 1)} \over 2}} \right)^2} \times {6 \over {r(r + 1)(2r + 1)}}$$<br><br>
$${T_r} = {{3r(r + 1)} \over 2}$$<br><br>
Now, $$S = \sum\limits_{r = 1}^{... |
The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term is :-
Options:
[{"identifier": "A", "content": "945"}, {"identifier": "B", "content": "916"}, {"identifier": "C", "content": "915"}, {"identifier": "D", "content": "946"}] | ["D"]
Explanation:
S = 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term
<br><br>General term T<sub>n</sub> = n (2n - 1)
<br><br>$$ \therefore $$ S<sub>n</sub> = $$\sum {{T_n}} $$
<br><br>$$ \Rightarrow $$ S<sub>n</sub> = $$\sum {\left( {2{n^2} - n} \right)} $$
<br><br>$$ \Rightarrow $$ S<sub>n</sub> = $$2\sum {{n^2} - \... |
The sum
$$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $$ is equal to
Options:
[{"identifier": "A", "content": "$$2 - {11 \\over {{2^{19}}}}$$"}, {"identifier": "B", "content": "$$2 - {3 \\over {{2^{17}}}}$$"}, {"identifier": "C", "content": "$$1 - {11 \\over {{2^{20}}}}$$"}, {"identifier": "D", "content": "$$2 - {21... | ["A"]
Explanation:
Let S = $$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $$
<br><br>$$ \Rightarrow $$ S = $${1 \over 2} + {2 \over {{2^2}}} + {3 \over {{2^3}}} + {4 \over {{2^4}}} + ....... + {{20} \over {{2^{20}}}}$$ .......(1)
<br><br>This is an Arithmetic Geometric Sequence. Here numerator is in A.P and denominat... |
If the sum of the first 15 terms of the series $${\left( {{3 \over 4}} \right)^3} + {\left( {1{1 \over 2}} \right)^3} + {\left( {2{1 \over 4}} \right)^3} + {3^3} + {\left( {3{3 \over 4}} \right)^3} + ....$$ is equal to 225 k, then k
is equal to :
Options:
[{"identifier": "A", "content": "9"}, {"identifier": "B", "cont... | ["C"]
Explanation:
S = $${\left( {{3 \over 4}} \right)^3} + {\left( {{6 \over 4}} \right)^3} + {\left( {{9 \over 4}} \right)^3} + {\left( {{{12} \over 4}} \right)^3} + \,........15$$ term
<br><br>= $${{27} \over {64}}$$ $$\sum\limits_{r = 1}^{15} {{r^3}} $$
<br><br>= $${{27} \over {64.}}{\left[ {{{15\left( {15 + 1} \r... |
Let S<sub>k</sub> = $${{1 + 2 + 3 + .... + k} \over k}.$$ If $$S_1^2 + S_2^2 + .....\, + S_{10}^2 = {5 \over {12}}$$A, then A is equal to :
Options:
[{"identifier": "A", "content": "283"}, {"identifier": "B", "content": "156"}, {"identifier": "C", "content": "301"}, {"identifier": "D", "content": "303"}] | ["D"]
Explanation:
S<sub>k</sub> = $${{K + 1} \over 2}$$
<br><br>$$\sum {S_k^2} = {5 \over {12}}$$ A
<br><br>$$\sum\limits_{K = 1}^{10} {{{\left( {{{K + 1} \over 2}} \right)}^2}} = {{{2^2} + {3^2} + - - + {{11}^2}} \over 4} = {5 \over {12}}$$ A
<br><br>$${{11 \times 12 \times 23} \over 6} - 1 = {5 \over 3}$$ A
<b... |
The sum of the following series
<br/><br/>$$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9}$$
<br/><br/> $$ + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....$$ up to 15 terms, is :
Options:
[{"identifier": "A", "con... | ["D"]
Explanation:
$$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9} + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....\,15$$
<br><br>$$ = {{3\left( {{1^2}} \right)} \over 3} + {{6\left( {{1^2} + {2^2}} \right)} \over 5} + ... |
Some identical balls are arranged in rows to form
an equilateral triangle. The first row consists of one
ball, the second row consists of two balls and so
on. If 99 more identical balls are addded to the total
number of balls used in forming the equilaterial
triangle, then all these balls can be arranged in a
square wh... | ["B"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264552/exam_images/ooaanpllsv7wqqhp4zns.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Sequences and Series Question 185 English... |
If 2<sup>10</sup> + 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> = S - 2<sup>11</sup>, then S is equal to :
Options:
[{"identifier": "A", "content": "$${{{3^{11}}} \\over 2} + {2^{10}}$$"}, {"identifier": "B", "content": "3<sup>11</sup> \u2014 2<sup>12</sup>"}, {"... | ["D"]
Explanation:
Let S<sub>1</sub> = 2<sup>10</sup> + 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> ....(1)
<br><br>Also $${{3{S_1}} \over 2}$$ = 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> + $${{{3^{11}}} ... |
If the sum of the first 20 terms of the series
<br/>$${\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...$$ is 460,
<br/>then x is equal to :
Options:
[{"identifier": "A", "content": "e<sup>2</sup>"}, {"identifier": "B", "content": "7<sup>1/2</su... | ["C"]
Explanation:
460 = log<sub>7</sub> <sup>x</sup>·(2 + 3 + 4 + ..... + 20 + 21)
<br><br>$$ \Rightarrow $$ 460 = log<sub>7</sub> <sup>x</sup>. $$\left( {{{21 \times 22} \over 2} - 1} \right)$$
<br><br>$$ \Rightarrow $$ 460 = 230. log<sub>7</sub> <sup>x</sup>
<br><br>$$ \Rightarrow $$ log<sub>7</sub> <sup>x</sup> = ... |
If 1+(1–2<sup>2</sup>.1)+(1–4<sup>2</sup>.3)+(1-6<sup>2</sup>.5)+......+(1-20<sup>2</sup>.19)= $$\alpha $$ - 220$$\beta $$, <br/>then an ordered pair $$\left( {\alpha ,\beta } \right)$$ is equal to:
Options:
[{"identifier": "A", "content": "(11, 103)"}, {"identifier": "B", "content": "(10, 103)"}, {"identifier": "C",... | ["A"]
Explanation:
$$1 + (1 - {2^2}.1) + (1 - {4^2}.3) + (1 - {6^2}.5) + ....(1 - {20^2}.19)$$<br><br>$$S = 1 + \sum\limits_{r = 1}^{10} {\left[ {1 - {{(2r)}^2}(2r - 1)} \right] = 1 + \sum\limits_{r = 1}^{10} {\left( {1 - 8{r^3} + 4{r^2}} \right)} = 1 + 10 - } \sum\limits_{r = 1}^{10} {\left( {8{r^3} - 4{r^2}} \right... |
If the sum of the series
<br/><br/>20 + 19$${3 \over 5}$$ + 19$${1 \over 5}$$ + 18$${4 \over 5}$$ + ...
<br/><br/>upto nth term is 488
and the n<sup>th</sup> term is negative, then :
Options:
[{"identifier": "A", "content": "n = 41"}, {"identifier": "B", "content": "n = 60"}, {"identifier": "C", "content": "n<sup>th</... | ["C"]
Explanation:
$$S = {{100} \over 5} + {{98} \over 5} + {{96} \over 5} + {{94} \over 5} + ...\,n$$<br><br>$$\,{S_n} = {n \over 2}\left( {2 \times {{100} \over 5} + (n - 1)\left( {{{ - 2} \over 5}} \right)} \right) = 188$$<br><br>$$ \Rightarrow $$ $$n(100 - n + 1) = 488 \times 5$$<br><br>$$ \Rightarrow $$ $${n^2} -... |
Let S be the sum of the first 9 terms of the
series :
<br/>{x + k$$a$$} + {x<sup>2</sup> + (k + 2)$$a$$} + {x<sup>3</sup> + (k + 4)$$a$$}
<br/>+ {x<sup>4</sup> + (k + 6)$$a$$} + .... where a $$ \ne $$ 0 and x $$ \ne $$ 1.
<br/><br/>If S = $${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$$, then k is equal... | ["A"]
Explanation:
S = {x + k$$a$$} + {x<sup>2</sup> + (k + 2)$$a$$} + {x<sup>3</sup> + (k + 4)$$a$$}
<br>+ {x<sup>4</sup> + (k + 6)$$a$$} + ....<br><br>
$$S = \left( {x + {x^2} + {x^3} + ....\,9terms} \right) + $$<br> $$a\left( {k + \left( {k + 2} \right) + (k + 4) + (k + 6) + ....9terms} \right)$$<b... |
The product $${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ... to $$\infty $$ is equal
to :
Options:
[{"identifier": "A", "content": "$${2^{{1 \\over 4}}}$$"}, {"identifier": "B", "content": "$${2^{{1 \\over 2}}}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "con... | ["B"]
Explanation:
$${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ...
<br><br>= $${2^{{1 \over 4} + {2 \over {16}} + {3 \over {48}} + ...\infty }}$$
<br><br>= $${2^{{1 \over 4} + {1 \over 8} + {1 \over {16}} + ...\infty }}$$
<br><br>= $${2^{\left( {{{{1 \over 4}} \over {1 - {1 \... |
The sum, $$\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $$ is equal to
________.
Options:
[] | 504
Explanation:
$$\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $$
<br><br>= $${1 \over 4}\sum\limits_{n = 1}^7 {\left( {2{n^3} + 3{n^2} + n} \right)} $$
<br><br>= $${1 \over 2}\sum\limits_{n = 1}^7 {{n^3}} $$ + $${3 \over 4}\sum\limits_{n = 1}^7 {{n^2}} $$ + $${1 \over 4}\sum\lim... |
The sum $$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $$ is :
Options:
[] | 1540
Explanation:
$$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $$
<br><br>= $$\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $$
<br><br>= $$\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $$
<br><br>= $${1 \over 2} \times {{20 \times 21 \times... |
If the sum of the first 40 terms of the series, <br/>3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to :
Options:
[{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "25"}] | ["A"]
Explanation:
3 + 4 + 8 + 9 + 13 + 14 +…….upto 40 terms
<br><br>$$ \Rightarrow $$ 7 + 17 + 27 +…….20 terms
<br><br>S = $${{20} \over 2}$$ [2 × 7 + 19 × 10]
<br><br> = 102 × 20 = 102 m
<br><br>$$ \therefore $$ m = 20 |
If |x| < 1, |y| < 1 and x $$ \ne $$ y, then the sum to infinity
of the following series
<br/><br/>(x + y) + (x<sup>2</sup>+xy+y<sup>2</sup>) + (x<sup>3</sup>+x<sup>2</sup>y + xy<sup>2</sup>+y<sup>3</sup>) + ....
Options:
[{"identifier": "A", "content": "$${{x + y - xy} \\over {\\left( {1 + x} \\right)\\left( {1 ... | ["B"]
Explanation:
(x + y) + (x<sup>2</sup>+xy+y<sup>2</sup>) + (x<sup>3</sup>+x<sup>2</sup>y + xy<sup>2</sup>+y<sup>3</sup>) + ....
<br><br>By multiplying and dividing x – y :
<br><br>$${{\left( {{x^2} - {y^2}} \right) + \left( {{x^3} - {y^3}} \right) + \left( {{x^4} - {y^4}} \right) + ...} \over {x - y}}$$
<br><br>=... |
If $$0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi } $$ and $$z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi } $$ then :
Options:
[{"identifier": "A", "content": "xy $$-$$ z = (x + y)z"}, ... | ["C"]
Explanation:
$$x = 1 + {\cos ^2}\theta + ..........\infty $$<br><br>$$x = {1 \over {1 - {{\cos }^2}\theta }} = {1 \over {{{\sin }^2}\theta }}$$ .......(1)<br><br>$$y = 1 + {\sin ^2}\phi + ........\infty $$<br><br>$$y = {1 \over {1 - {{\sin }^2}\phi }} = {1 \over {{{\cos }^2}\phi }}$$ ....... (2)<br><br>$$z = {... |
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