2022
Q151
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum 1 + 2 . 3 + 3 . 32 + ......... + 10 . 39 is equal to :
A.
${{2\,.\,{3^{12}} + 10} \over 4}$
B.
${{19\,.\,{3^{10}} + 1} \over 4}$
D.
${{9\,.\,{3^{10}} + 1} \over 2}$
Show Answer
Correct Answer: B
Explanation:
Let $S = 1\,.\,{3^0} + 2\,.\,{3^1} + 3\,.\,{3^2} + \,\,......\,\, + \,\,10\,.\,{3^9}$
$3S = 1\,.\,{3^1} + 2\,.\,{3^2} + \,\,..........\,\, + \,\,10\,.\,{3^{10}}$
___________________________________________________________
$ - 2S = (1\,.\,{3^0} + 1\,.\,{3^1} + 1\,.\,{3^2} + \,\,........\,\, + \,\,1\,.\,{3^9}) - 10\,.\,{3^{10}}$
$ \Rightarrow S = {1 \over 2}\left[ {10\,.\,{3^{10}} - {{{3^{10}} - 1} \over { - 3 - 1}}} \right]$
$ \Rightarrow S = {{19\,.\,{3^{10}} + 1} \over 4}$
2022
Q152
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let x, y > 0. If x3 y2 = 215 , then the least value of 3x + 2y is
Show Answer
Correct Answer: D
Explanation:
x, y > 0 and x3 y2 = 215
Now, 3x + 2y = (x + x + x) + (y + y)
So, by A.M $\ge$ G.M inequality
${{3x + 2y} \over 5} \ge \root 5 \of {{x^3}\,.\,{y^2}} $
$\therefore$ $3x + 2y \ge 5\root 5 \of {{2^{15}}} \ge 40$
$\therefore$ Least value of $3x + 4y = 40$
2022
Q153
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 :
Show Answer
Correct Answer: B
Explanation:
$\sum\limits_{i = 1}^n {{a_i} = 192} $
$\Rightarrow$ a1 + a2 + a3 + ...... + an = 192
$ \Rightarrow {n \over 2}[{a_1} + {a_n}] = 192$
$ \Rightarrow {a_1} + {a_n} = {{384} \over n}$ ..... (1)
Now, $\sum\limits_{i = 1}^{{n \over 2}} {{a_{2i}} = 120} $
$\Rightarrow$ a2 + a4 + a6 + ...... + an = 120
Here total ${n \over 2}$ terms present.
$\therefore$ ${{{n \over 2}} \over 2}[{a_2} + {a_n}] = 120$
$ \Rightarrow {n \over 4}[{a_1} + 1 + {a_n}] = 120$
$ \Rightarrow {a_1} + {a_n} + 1 = {{480} \over n}$ ..... (2)
Subtracting (1) from (2), we get
$1 = {{480} \over n} - {{384} \over n}$
$ \Rightarrow 1 = {{96} \over n}$
$\Rightarrow$ n = 96
2022
Q154
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 _________.
Show Answer
Correct Answer: 16
Explanation:
Given
$S = {{{a_1}} \over 2} + {{{a_2}} \over {{2^2}}} + {{{a_3}} \over {{2^3}}} + {{{a_4}} \over {{2^4}}}\, + \,.....\,\infty $
${{{1 \over 2}S = {{{a_1}} \over {{2^2}}} + {{{a_2}} \over {{2^3}}}\, + \,.........\,\infty } \over {{S \over 2} = {{{a_1}} \over 2} + {{({a_2} + {a_1})} \over {{2^2}}} + {{({a_3} + {a_2})} \over {{2^3}}}\, + \,......\,\infty }}$
$ \Rightarrow {S \over 2} = {{{a_1}} \over 2} + {d \over 2}$
$ \Rightarrow {a_1} + d = {a_2} = 4 \Rightarrow 4{a_2} = 16$
2022
Q155
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If $\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102}=\frac{\mathrm{k}}{101}$, then 34 k is equal to _________.
Show Answer
Correct Answer: 286
Explanation:
$S = {1 \over {2 \times 3 \times 4}} + {1 \over {3 \times 4 \times 5}} + {1 \over {4 \times 5 \times 6}}\, + \,....\, + \,{1 \over {100 \times 101 \times 102}}$
$ = {1 \over {(3 - 1)\,.\,1}}\left[ {{1 \over {2 \times 3}} - {1 \over {101 \times 102}}} \right]$
$ = {1 \over 2}\left( {{1 \over 6} - {1 \over {101 \times 102}}} \right)$
$ = {{143} \over {102 \times 101}} = {k \over {101}}$
$\therefore$ $34k = 286$
2022
Q156
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
${6 \over {{3^{12}}}} + {{10} \over {{3^{11}}}} + {{20} \over {{3^{10}}}} + {{40} \over {{3^9}}} + \,\,...\,\, + \,\,{{10240} \over 3} = {2^n}\,.\,m$, where m is odd, then m . n is equal to ____________.
Show Answer
Correct Answer: 12
Explanation:
${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + {5 \over 5}\left( {{1 \over {{3^{12}}}}\,.\,{2^{12}}\,.\,{3^{12}} - {1 \over {{3^{12}}}}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + {2^{12}} - {1 \over {{3^{12}}}} = {2^n}\,.\,m$
$ \Rightarrow {2^n}\,.\,m = {2^{12}}$
$ \Rightarrow m = 1$ and $n = 12$
$m\,.\,n = 12$
2022
Q157
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
$
\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\cdots+
\frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$ is equal to _____________.
Show Answer
Correct Answer: 120
Explanation:
${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$
$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$
$ = {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$
$ = {{2n(2{n^2} + 3n + 1) - 3{n^2} - 3n + n} \over {n(4n + 3)}}$
$ = {{{n^2}(4n + 3)} \over {n(4n + 3)}} = n$
$\therefore$ ${T_n} = n$
${S_n} = \sum\limits_{n = 1}^{15} {{T_n} = {{15 \times 16} \over 2} = 120} $
2022
Q158
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If $\sum\limits_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$, where m and n are co-prime, then $m+n$ is equal to _____________.
Show Answer
Correct Answer: 166
Explanation:
$\sum\limits_{k = 1}^{10} {{k \over {{k^4} + {k^2} + 1}}} $
$ = {1 \over 2}\left[ {\sum\limits_{k = 1}^{10} {\left( {{1 \over {{k^2} - k + 1}} - {1 \over {{k^2} + k + 1}}} \right)} } \right.$
$ = {1 \over 2}\left[ {1 - {1 \over 3} + {1 \over 3} - {1 \over 7} + {1 \over 7} - {1 \over {13}}\, + \,...\, + \,{1 \over {91}} - {1 \over {111}}} \right]$
$ = {1 \over 2}\left[ {1 - {1 \over {111}}} \right] = {{110} \over {2\,.\,111}} = {{55} \over {111}} = {m \over n}$
$\therefore$ $m + n = 55 + 111 = 166$
2022
Q159
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 ___________.
Show Answer
Correct Answer: 53
Explanation:
${d_1} = {{199 - 100} \over 2} \notin I$
${d_2} = {{199 - 100} \over 3} = 33$
${d_3} = {{199 - 100} \over 4} \notin I$
${d_n} = {{199 - 100} \over {i + 1}} \in I$
${d_i} = 33 + 11,\,9$
Sum of CD's $ = 33 + 11 + 9$
$ = 53$
2022
Q160
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The series of positive multiples of 3 is divided into sets : $\{3\},\{6,9,12\},\{15,18,21,24,27\}, \ldots$ Then the sum of the elements in the $11^{\text {th }}$ set is equal to ____________.
Show Answer
Correct Answer: 6993
Explanation:
Given series
$\therefore$ 11th set will have $1 + (10)2 = 21$ term
Also upto 10th set total $3 \times k$ type terms will be $1 + 3 + 5\, + \,......\, + \,19 = 100 - $ term
$\therefore$ Set $11 = \{ 3 \times 101,\,3 \times 102,\,......\,3 \times 121\} $
$\therefore$ Sum of elements $ = 3 \times (101 + 102\, + \,...\, + \,121)$
$ = {{3 \times 222 \times 21} \over 2} = 6993$
2022
Q161
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let $a, b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0$ and $\mathrm{q}$ and s are the roots of the equation $x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0$, such that $\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}}$ are in A.P., then $\mathrm{a}^{-1}-\mathrm{b}^{-1}$ is equal to _____________.
Show Answer
Correct Answer: 38
Explanation:
$\because$ 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 $\frac{1}{s}$.
2022
Q162
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let $a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$ and $b_{n}=a_{n}+b_{n-1}$ for every
Show Answer
Correct Answer: 27560
Explanation:
Given,
${a_n} = {a_{n - 1}} + 2$
$ \Rightarrow {a_n} - {a_{n - 1}} = 2$
$\therefore$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.
Also given ${a_1} = 1$
$\therefore$ Series is = 1, 3, 5, 7 ......
$\therefore$ ${a_n} = 1 + (n - 1)2 = 2n - 1$
Also ${b_n} = {a_n} + {b_{n - 1}}$
When $n = 2$ then
${b_2} - {b_1} = {a_2} = 3$
$ \Rightarrow {b_2} - 1 = 3$ [Given ${b_1} = 1$]
$ \Rightarrow {b_2} = 4$
When $n = 3$ then
${b_3} - {b_2} = {a_3}$
$ \Rightarrow {b_3} - 4 = 5$
$ \Rightarrow {b_3} = 9$
$\therefore$ Series is = 1, 4, 9 ......
= 12 , 22 , 32 ....... n2
$\therefore$ ${b_n} = {n^2}$
Now, $\sum\limits_{n = 1}^{15} {\left( {{a_n}\,.\,{b_n}} \right)} $
$ = \sum\limits_{n = 1}^{15} {\left[ {(2n - 1){n^2}} \right]} $
$ = \sum\limits_{n = 1}^{15} {2{n^3} - \sum\limits_{n = 1}^{15} {{n^2}} } $
$ = 2\left( {{1^3} + {2^3} + \,\,...\,\,{{15}^3}} \right) - \left( {{1^2} + {2^2} + \,\,...\,\,{{15}^2}} \right)$
$ = 2 \times {\left( {{{15 \times 16} \over 2}} \right)^2} - \left( {{{15(16) \times 31} \over 6}} \right)$
$ = 27560$
2022
Q163
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let for $f(x) = {a_0}{x^2} + {a_1}x + {a_2},\,f'(0) = 1$ and $f'(1) = 0$. If a0 , a1 , a2 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 _____________.
Show Answer
Correct Answer: 2
Explanation:
Given,
$f(x) = {a_0}{x^2} + {a_1}x + {a_2}$
$f'(0) = 1$
$f'(1) = 0$
a0 , a1 , a2 are in A. G. P
Common difference of $AP = 1$
Common ratio of $GP = 2$
A.P terms = a, a + 1, a + 2
G.P terms = y, ry, r2 y
$\therefore$ AGP terms = ay, (a+1)ry, (a+2)r2 y
$\therefore$ ${a_0} = ay$
${a_1} = (a + 1)ry = (a + 1)2y$
${a_2} = (a + 2){r^2}y = (a + 2)4y$
Now, $f'(x) = 2x{a_0} + {a_1}$
$\therefore$ $f'(0) = {a_1} = 1$
and $f'(1) = 2{a_0} + {a_1} = 0$
$ \Rightarrow 2{a_0} + 1 = 0$
$ \Rightarrow {a_0} = - {1 \over 2}$
$\therefore$ $ay = - {1 \over 2}$
and $(a + 1)2y = 1$
$ \Rightarrow 2ay + 2y = 1$
$ \Rightarrow 2 \times \left( { - {1 \over 2}} \right) + 2y = 1$
$ \Rightarrow 2y = + \,2$
$ \Rightarrow y = + \,1$
$\therefore$ $a = - {1 \over 2}$
$\therefore$ ${a_2} = (a + 2)4y$
$ = \left( { - {1 \over 2} + 2} \right) \times 4\,.\,1$
$ = 6$
$\therefore$ $f(x) = - {1 \over 2}{x^2} + x + 6$
$\therefore$ $f(4) = - {1 \over 2}{(4)^2} + 4 + 6$
$ = - 8 + 10$
$ = 2$
2022
Q164
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 ________.
Show Answer
Correct Answer: 2223
Explanation:
1st AP :
3, 6, 9, 12, ....... upto 78 terms
t78 = 3 + (78 $-$ 1)3
= 3 + 77 $\times$ 3
= 234
2nd AP :
5, 9, 13, 17, ...... upto 59 terms
t59 = 5 + (59 $-$ 1)4
= 5 + 58 $\times$ 4
= 237
Common term's AP :
First term = 9
Common difference of first AP = 3
And common difference of second AP = 4
$\therefore$ Common difference of common terms
AP = LCM (3, 4) = 12
$\therefore$ New AP = 9, 21, 33, .......
tn = 9 + (n $-$ 1)12 $\le$ 234
$ \Rightarrow n \le {{237} \over {12}}$
$ \Rightarrow n = 19$
$\therefore$ ${S_{19}} = {{19} \over 2}\left[ {2.9 + (19 - 1)12} \right]$
$ = 19(9 + 108)$
$ = 2223$
2022
Q165
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let for n = 1, 2, ......, 50, Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is ${1 \over {{{(n + 1)}^2}}}$. Then the value of
Show Answer
Correct Answer: 41651
Explanation:
${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$
Now ${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $
$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\left\{ {({n^2} - n) + 2\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} \right\}} $
$ = {1 \over {26}} + {{50 \times 51 \times 101} \over 6} - {{50 \times 51} \over 2} + 2\left( {{1 \over 2} - {1 \over {52}}} \right)$
$ = 1 + 25 \times 17(101 - 3)$
$ = 41651$
2022
Q166
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let A = {1, a1 , a2 ....... a18 , 77} be a set of integers with 1 < a1 < a2 < ....... < a18 < 77. 1 + a2 + ...... + a18 is equal to _____________.
Show Answer
Correct Answer: 702
Explanation:
If we write the elements of $A+A$, we can certainly find 39 distinct elements as $1+1,1+a_{1}, 1+a_{2}, \ldots .1$ $+a_{18}, 1+77, a_{1}+77, a_{2}+77, \ldots \ldots a_{18}+77,77+77$.
2022
Q167
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the sum of the first ten terms of the series
${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$
is ${m \over n}$, where m and n are co-prime numbers, then m + n is equal to ______________.
Show Answer
Correct Answer: 276
Explanation:
${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$
$ = {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$
$ = {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$
${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $
$ = {1 \over 4}\left[ {1 - {1 \over 5} + {1 \over 5} - {1 \over {13}} + \,\,....\,\, + \,\,{1 \over {181}} - {1 \over {221}}} \right]$
$ \Rightarrow {S_{10}} = {1 \over 4}\,.\,{{220} \over {221}} = {{55} \over {221}} = {m \over n}$
$\therefore$ $m + n = 276$
2022
Q168
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If a1 (> 0), a2 , a3 , a4 , a5 are in a G.P., a2 + a4 = 2a3 + 1 and 3a2 + a3 = 2a4 , then a2 + a4 + 2a5 is equal to ___________.
Show Answer
Correct Answer: 40
Explanation:
Let G.P. be a1 = a, a2 = ar, a3 = ar2 , .........
$\because$ 3a2 + a3 = 2a4
$\Rightarrow$ 3ar + ar2 = 2ar3
$\Rightarrow$ 2ar2 $-$ r $-$ 3 = 0
$\therefore$ r = $-$1 or ${3 \over 2}$
$\because$ a1 = a > 0 then r $\ne$ $-$1
Now, a2 + a4 = 2a3 + 1
ar + ar3 = 2ar2 + 1
$a\left( {{3 \over 2} + {{27} \over 8} - {9 \over 2}} \right) = 1$
$\therefore$ a = ${8 \over 3}$
$\therefore$ a2 + a4 + 2a5 = a(r + r3 + 2r4 )
$ = {8 \over 3}\left( {{3 \over 2} + {{27} \over 8} + {{81} \over 8}} \right) = 40$
2022
Q169
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
For a natural number n, let ${\alpha _n} = {19^n} - {12^n}$. Then, the value of ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}}$ is ___________.
Show Answer
Correct Answer: 4
Explanation:
${\alpha _n} = {19^n} - {12^n}$
Let equation of roots 12 & 19 i.e.
${x^2} - 31x + 228 = 0$
$ \Rightarrow (31 - x) = {{228} \over x}$ (where x can be 19 or 12)
$\therefore$ ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}} = {{31({{19}^9} - {{12}^9}) - ({{19}^{10}} - {{12}^{10}})} \over {57({{19}^8} - {{12}^8})}}$
$ = {{{{19}^9}(31 - 19) - {{12}^9}(31 - 12)} \over {57({{19}^8} - {{12}^8})}}$
$ = {{228({{19}^8} - {{12}^8})} \over {57({{19}^8} - {{12}^8})}} = 4$.
2022
Q170
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The greatest integer less than or equal to the sum of first 100 terms of the sequence ${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$ ...... is equal to ___________.
Show Answer
Correct Answer: 98
Explanation:
$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, + $ ....
$ = \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)} $
$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$
$ = 98 + 2{\left( {{2 \over 3}} \right)^{100}}$
$\therefore$ $[S] = 98$
2021
Q171
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let Sn = 1 . (n $-$ 1) + 2 . (n $-$ 2) + 3 . (n $-$ 3) + ..... + (n $-$ 1) . 1, n $\ge$ 4.
Show Answer
Correct Answer: A
Explanation:
Let Tr = r(n $-$ r)r = nr $-$ r2
2021
Q172
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let a1 , a2 , ..........., a21 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 a6 a16 is equal to :
Show Answer
Correct Answer: B
Explanation:
$\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} } $1 + 10d = 9 ..... (2)1 = 3 & d = ${3 \over 5}$1 = 15 & d = $ - {3 \over 5}$6 . a16 = (a1 + 5d) (a1 + 15d)6 a16 = 72
2021
Q173
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let a1 , a2 , a3 , ..... 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 :
Show Answer
Correct Answer: C
Explanation:
${{{{10} \over 2}(2{a_1} + 9d)} \over {{p \over 2}(2{a_1} + (p - 1)d)}} = {{100} \over {{p^2}}}$
2021
Q174
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of 10 terms of the series
Show Answer
Correct Answer: B
Explanation:
$S = {{{2^2} - {1^2}} \over {{1^2} \times {2^2}}} + {{{3^2} - {2^2}} \over {{2^2} \times {3^2}}} + {{{4^2} - {3^2}} \over {{3^2} \times {4^2}}} + ...$
2021
Q175
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 r2 , then r2 $-$ d is equal to :
Show Answer
Correct Answer: B
Explanation:
Let numbers be ${a \over r}$, a, ar $\to$ G.P.th form of G.P. = 3r2 $\Rightarrow$ ar2 = 3r2 $\Rightarrow$ a = 32 $-$ d = (2 + $\sqrt 3 $)2 $-$ 3$\sqrt 3 $
2021
Q176
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If 0 < x < 1 and $y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$, then the value of e1 + y at $x = {1 \over 2}$ is :
Show Answer
Correct Answer: A
Explanation:
$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$
2021
Q177
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If 0 < x < 1, then ${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$, is equal to :
A.
$x\left( {{{1 + x} \over {1 - x}}} \right) + {\log _e}(1 - x)$
B.
$x\left( {{{1 - x} \over {1 + x}}} \right) + {\log _e}(1 - x)$
C.
${{1 - x} \over {1 + x}} + {\log _e}(1 - x)$
D.
${{1 + x} \over {1 - x}} + {\log _e}(1 - x)$
Show Answer
Correct Answer: A
Explanation:
Let $t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty $
2021
Q178
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If for x, y $\in$ R, x > 0, y = log10 x + log10 x1/3 + log10 x1/9 + ...... upto $\infty$ terms
Show Answer
Correct Answer: D
Explanation:
${{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}$6 , 9)
2021
Q179
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the series
A.
$1 + {{{2^{101}}} \over {{4^{101}} - 1}}$
B.
$1 + {{{2^{100}}} \over {{4^{101}} - 1}}$
C.
$1 - {{{2^{100}}} \over {{4^{100}} - 1}}$
D.
$1 - {{{2^{101}}} \over {{2^{400}} - 1}}$
Show Answer
Correct Answer: D
Explanation:
$S = {1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$
2021
Q180
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the sum of an infinite GP a, ar, ar2 , ar3 , ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar2 , ar4 , ar6 , ....... is :
Show Answer
Correct Answer: B
Explanation:
Sum of infinite terms :2 , a2 r2 , a2 r4 , a2 r6 .......2 , ar4 , ar6
2021
Q181
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let Sn be the sum of the first n terms of an arithmetic progression. If S3n = 3S2n , then the value of ${{{S_{4n}}} \over {{S_{2n}}}}$ is :
Show Answer
Correct Answer: A
Explanation:
Let a be first term and d be common diff. of this A.P.3n = 3S2n
2021
Q182
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 $-$ S6 is equal to:
Show Answer
Correct Answer: A
Explanation:
Let first term of A.P. be a and common difference is d.
2021
Q183
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If sum of the first 21 terms of the series ${\log _{{9^{1/2}}}}x + {\log _{{9^{1/3}}}}x + {\log _{{9^{1/4}}}}x + .......$, where x > 0 is 504, then x is equal to
Show Answer
Correct Answer: D
Explanation:
$s = 2{\log _9}x + 3{\log _9}x + ....... + 22{\log _9}x$
2021
Q184
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 $-$ S1 ) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :
Show Answer
Correct Answer: C
Explanation:
S1 = ${{2n} \over 2}$[2a + (2n $-$ 1)d]2 = ${{4n} \over 2}$[2a + (4n $-$ 1)d]1 and d is common difference)2 $-$ S1 $ \Rightarrow $ 2n[2a + (4n $-$ 1)d] $-$ n[2a + (2n $-$ 1)d] = 10006 = ${{6n} \over 2}$[2a + (6n $-$ 1)d] = 3(S2 $-$ S1 ) = 3000
2021
Q185
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If $\alpha$, $\beta$ are natural numbers such that $\alpha$ $-$ 199$\beta$ = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through ($\alpha$, $\beta$) and origin is :
Show Answer
Correct Answer: B
Explanation:
RHS = $\sum\limits_{r = 0}^{99} {(100 - r)(100 + r)} $$\alpha$ $-$ (199)$\beta$
2021
Q186
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
${1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}}$ is equal to
Show Answer
Correct Answer: B
Explanation:
$S = \sum\limits_{r = 1}^{100} {{1 \over {{{(2n + 1)}^2} - 1}}} $
2021
Q187
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the series
A.
${{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10$
B.
${{41} \over 8}e - {{19} \over 8}{e^{ - 1}} - 10$
C.
${{41} \over 8}e + {{19} \over 8}{e^{ - 1}} + 10$
D.
$ - {{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10$
Show Answer
Correct Answer: B
Explanation:
$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $
2021
Q188
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the infinite series
Show Answer
Correct Answer: B
Explanation:
$S = 1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + ....$
2021
Q189
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 4th , 6th and 8th terms is equal to :
Show Answer
Correct Answer: D
Explanation:
a, ar, ar2 , .....
2021
Q190
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The minimum value of $f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$, where a, $x \in R$ and a > 0, is equal to :
Show Answer
Correct Answer: D
Explanation:
We know, $AM \ge GM$
2021
Q191
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 :
Show Answer
Correct Answer: C
Explanation:
$x = 1 + {\cos ^2}\theta + ..........\infty $
2021
Q192
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________.
Show Answer
Correct Answer: 5143
Explanation:
A = 4-digit numbers divisible by 3
2021
Q193
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If $S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$, then 160 S is equal to ________.
Show Answer
Correct Answer: 305
Explanation:
$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$
2021
Q194
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 _____________.
Show Answer
Correct Answer: 7744
Explanation:
209, 220, 231, ..........., 495th place $\matrix{
{\underline 3 } & {\underline 1 } & {\underline 9 } \cr
{\underline 4 } & {\underline 1 } & {\underline 8 } \cr
} $
2021
Q195
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let a1 , a2 , ......., a10 be an AP with common difference $-$ 3 and b1 , b2 , ........., b10 be a GP with common ratio 2. Let ck = ak + bk , k = 1, 2, ......, 10. If c2 = 12 and c3 = 13, then $\sum\limits_{k = 1}^{10} {{c_k}} $ is equal to _________.
Show Answer
Correct Answer: 2021
Explanation:
$a_{1}, a_{2}, a_{3}, \ldots, a_{10}$ are in AP common difference $=-3$
2021
Q196
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
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 _____________.
Show Answer
Correct Answer: 3
Explanation:
$2{\log _3}({2^x} - 5) = {\log _2} + {\log _3}\left( {{2^x} - {7 \over 2}} \right)$x = 4 or 2x = 8
2021
Q197
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the value of2 is equal to _______________.
Show Answer
Correct Answer: 3
Explanation:
$l = {\left( {\underbrace {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}}}_S + ....} \right)^{{{\log }_{0.25}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ...} \right)}}$2 = 3
2021
Q198
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of all the elements in the set {n$\in$ {1, 2, ....., 100} | H.C.F. of n and 2040 is 1} is equal to _____________.
Show Answer
Correct Answer: 1251
Explanation:
2040 = 23 $\times$ 3 $\times$ 5 $\times$ 17
2021
Q199
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
For k $\in$ N, let ${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $, where $\alpha > 0$. Then the value of $100{\left( {{{{A_{14}} + {A_{15}}} \over {{A_{13}}}}} \right)^2}$ is equal to _____________.
Show Answer
Correct Answer: 9
Explanation:
${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $
2021
Q200
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let $\left\{ {{a_n}} \right\}_{n = 1}^\infty $ be a sequence such that a1 = 1, a2 = 1 and ${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$ for all n $\ge$ 1. Then the value of $47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}} $ is equal to ______________.
Show Answer
Correct Answer: 7
Explanation:
${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$, let $\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} = P$n we get
