Sequences and Series
309 Questions
Start JEE Mains Test
2021
Q201
JEE Mains
Numerical
14 Mar 2026
Sn(x) = loga1/2x + loga1/3x + loga1/6x + loga1/11x + loga1/18x + loga1/27x + ...... up to n-terms, where a > 1. If S24(x) = 1093 and S12(2x) = 265, then value of a is equal to ____________.
Correct Answer: 16
Explanation:
${S_n}(x) = {\log _a}{x^2} + {\log _a}{x^3} + {\log _a}{x^6} + {\log _a}{x^{11}}$
${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$
${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$
${S_r} = 2 + 3 + 6 + 11$
$ \therefore $ Tn = 2 + (1 + 3 + 5 +......+ (n - 1))
= 2 + ${{n - 1} \over 2}\left[ {2.1 + \left( {n - 2} \right)2} \right]$
= 2 + $\left( {n - 1} \right)\left[ {1 + \left( {n - 2} \right)} \right]$
= n2 - 2n + 3
General term ${T_r} = {r^2} - 2r + 3$
${S_n}(x) = \sum\limits_{r = 1}^n {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = \sum\limits_{r = 1}^{24} {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = {\log _a}x\sum\limits_{r = 1}^{24} {({r^2} - 2r + 3)} $
$1093 = 4372{\log _a}x$
${\log _a}x = {1 \over 4}$
$x = {a^{1/4}}$ .....(i)
${S_{12}}(2x) = {\log _a}(2x)\sum\limits_{r = 1}^{12} {({r^2} - 2r + 3)} $
$265 = 530{\log _a}(2x)$
${\log _a}(2x) = {1 \over 2}$
$2x = {a^{1/2}}$ ....(ii)
From (i) and (ii), we get
$2{a^{{1 \over 4}}} = {a^{{1 \over 2}}}$
$ \Rightarrow $ ${\left( {2{a^{{1 \over 4}}}} \right)^4} = {\left( {{a^{{1 \over 2}}}} \right)^4}$
$ \Rightarrow $ 16$a$ = $a$2
$ \Rightarrow $ $a = 16$
${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$
${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$
${S_r} = 2 + 3 + 6 + 11$
$ \therefore $ Tn = 2 + (1 + 3 + 5 +......+ (n - 1))
= 2 + ${{n - 1} \over 2}\left[ {2.1 + \left( {n - 2} \right)2} \right]$
= 2 + $\left( {n - 1} \right)\left[ {1 + \left( {n - 2} \right)} \right]$
= n2 - 2n + 3
General term ${T_r} = {r^2} - 2r + 3$
${S_n}(x) = \sum\limits_{r = 1}^n {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = \sum\limits_{r = 1}^{24} {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = {\log _a}x\sum\limits_{r = 1}^{24} {({r^2} - 2r + 3)} $
$1093 = 4372{\log _a}x$
${\log _a}x = {1 \over 4}$
$x = {a^{1/4}}$ .....(i)
${S_{12}}(2x) = {\log _a}(2x)\sum\limits_{r = 1}^{12} {({r^2} - 2r + 3)} $
$265 = 530{\log _a}(2x)$
${\log _a}(2x) = {1 \over 2}$
$2x = {a^{1/2}}$ ....(ii)
From (i) and (ii), we get
$2{a^{{1 \over 4}}} = {a^{{1 \over 2}}}$
$ \Rightarrow $ ${\left( {2{a^{{1 \over 4}}}} \right)^4} = {\left( {{a^{{1 \over 2}}}} \right)^4}$
$ \Rightarrow $ 16$a$ = $a$2
$ \Rightarrow $ $a = 16$
2021
Q202
JEE Mains
Numerical
14 Mar 2026
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 ___________.
Correct Answer: 14
Explanation:
${a^2} = {b \over {16}}$ and ${2 \over b} = {1 \over a} + 6$
Solving, we get $a = {1 \over {12}}$ or $a = - {1 \over 4}$ [rejected]
if $a = {1 \over {12}} \Rightarrow b = {1 \over 9}$
$ \therefore $ $72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$
Solving, we get $a = {1 \over {12}}$ or $a = - {1 \over 4}$ [rejected]
if $a = {1 \over {12}} \Rightarrow b = {1 \over 9}$
$ \therefore $ $72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$
2021
Q203
JEE Mains
Numerical
14 Mar 2026
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 ___________.
Correct Answer: 3
Explanation:
A.P. from the set will be 11, 16, 21, 26 .....
G.P. from the set will be 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192 .....
So common terms are 16, 256, 4096.
G.P. from the set will be 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192 .....
So common terms are 16, 256, 4096.
2021
Q204
JEE Mains
Numerical
14 Mar 2026
The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _________.
Correct Answer: 1000
Explanation:
Let N be the four digit number
gcd(N, 18) = 3
Hence N is an odd integer which is divisible by 3 but not by 9.
4 digit odd multiples of 3
1005, 1011, ..........., 9999 $ \to $ 1500
4 digit odd multiples of 9
1017, 1035, ..........., 9999 $ \to $ 500
Hence number of such N = 1000
gcd(N, 18) = 3
Hence N is an odd integer which is divisible by 3 but not by 9.
4 digit odd multiples of 3
1005, 1011, ..........., 9999 $ \to $ 1500
4 digit odd multiples of 9
1017, 1035, ..........., 9999 $ \to $ 500
Hence number of such N = 1000
2021
Q205
JEE Mains
Numerical
14 Mar 2026
If the arithmetic mean and geometric mean of the pth and qth terms of the
sequence $-$16, 8, $-$4, 2, ...... satisfy the equation
4x2 $-$ 9x + 5 = 0, then p + q is equal to __________.
sequence $-$16, 8, $-$4, 2, ...... satisfy the equation
4x2 $-$ 9x + 5 = 0, then p + q is equal to __________.
Correct Answer: 10
Explanation:
Given, $4{x^2} - 9x + 5 = 0$
$ \Rightarrow (x - 1)(4x - 5) = 0$
$ \Rightarrow $ A. M. $ = {5 \over 4}$, G. M. = 1 (As A. M. $ \ge $ G. M)
Again, for the series
$-$16, 8, $-$4, 2 ..........
${p^{th}}$ term ${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$
${q^{th}}$ term ${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$
Now, A. M. = ${{{t_p} + {t_q}} \over 2} = {5 \over 4}$ & G. M. = $\sqrt {{t_p}{t_q}} = 1$
$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$
$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$
$ \Rightarrow p + q = 10$
$ \Rightarrow (x - 1)(4x - 5) = 0$
$ \Rightarrow $ A. M. $ = {5 \over 4}$, G. M. = 1 (As A. M. $ \ge $ G. M)
Again, for the series
$-$16, 8, $-$4, 2 ..........
${p^{th}}$ term ${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$
${q^{th}}$ term ${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$
Now, A. M. = ${{{t_p} + {t_q}} \over 2} = {5 \over 4}$ & G. M. = $\sqrt {{t_p}{t_q}} = 1$
$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$
$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$
$ \Rightarrow p + q = 10$
2021
Q206
JEE Mains
Numerical
14 Mar 2026
Let A1, A2, A3, ....... be squares such that for each n $ \ge $ 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one, is __________.
Correct Answer: 9
Explanation:
$ \therefore $ Side lengths are in G.P.
${T_n} = {{12} \over {{{\left( {\sqrt 2 } \right)}^{n - 1}}}}$
$ \therefore $ Area $ = {{144} \over {{2^{n - 1}} }}$ < 1
$ \Rightarrow {2^{n - 1}} > 144$
Smallest n = 9
2021
Q207
JEE Mains
Numerical
14 Mar 2026
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 _________.
Correct Answer: 3
Explanation:
Let the terms are $a,ar,a{r^2},a{r^3}$
$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$ ..........(1)
${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$
${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$ ...............(2)
Doing ${{(1)} \over {(2)}},$
${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$
Also given, ${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$
${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$
$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$
$2\alpha = 3$
$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$ ..........(1)
${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$
${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$ ...............(2)
Doing ${{(1)} \over {(2)}},$
${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$
Also given, ${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$
${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$
$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$
$2\alpha = 3$
2020
Q208
JEE Mains
MCQ
14 Mar 2026
The common difference of the A.P.
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an. If
a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an. If
a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
A.
127
B.
81
C.
–127
D.
-81
2020
Q209
JEE Mains
MCQ
14 Mar 2026
Let a , b, c , d and p be any non zero distinct real numbers such that
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
A.
a, c, p are in G.P.
B.
a, b, c, d are in G.P.
C.
a, b, c, d are in A.P.
D.
a, c, p are in A.P.
2020
Q210
JEE Mains
MCQ
14 Mar 2026
If the sum of the first 20 terms of the series
${\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...$ is 460,
then x is equal to :
${\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...$ is 460,
then x is equal to :
A.
e2
B.
71/2
C.
72
D.
746/21
2020
Q211
JEE Mains
MCQ
14 Mar 2026
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 :
A.
${2 \over {13}}\left( {{3^{50}} - 1} \right)$
B.
${1 \over {13}}\left( {{3^{50}} - 1} \right)$
C.
${1 \over {26}}\left( {{3^{49}} - 1} \right)$
D.
${1 \over {26}}\left( {{3^{50}} - 1} \right)$
2020
Q212
JEE Mains
MCQ
14 Mar 2026
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:
A.
66
B.
81
C.
65
D.
78
2020
Q213
JEE Mains
MCQ
14 Mar 2026
If 210 + 29.31 + 28
.32 +.....+ 2.39 + 310 = S - 211, then S is equal to :
A.
${{{3^{11}}} \over 2} + {2^{10}}$
B.
311 — 212
C.
2.311
D.
311
2020
Q214
JEE Mains
MCQ
14 Mar 2026
Let a1, a2, ..., an be a given A.P. whose
common difference is an integer and
Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $ \le $ n $ \le $ 50, then
the ordered pair (Sn-4, an–4) is equal to:
common difference is an integer and
Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $ \le $ n $ \le $ 50, then
the ordered pair (Sn-4, an–4) is equal to:
A.
(2480, 249)
B.
(2480, 248)
C.
(2490, 248)
D.
(2490, 249)
2020
Q215
JEE Mains
MCQ
14 Mar 2026
The minimum value of 2sinx + 2cosx is :
A.
${2^{-1 + \sqrt 2 }}$
B.
${2^{1 - {1 \over {\sqrt 2 }}}}$
C.
${2^{1 - \sqrt 2 }}$
D.
${2^{-1 + {1 \over {\sqrt 2 }}}}$
2020
Q216
JEE Mains
MCQ
14 Mar 2026
If 1+(1–22.1)+(1–42.3)+(1-62.5)+......+(1-202.19)= $\alpha $ - 220$\beta $,
then an ordered pair $\left( {\alpha ,\beta } \right)$ is equal to:
then an ordered pair $\left( {\alpha ,\beta } \right)$ is equal to:
A.
(11, 103)
B.
(10, 103)
C.
(10, 97)
D.
(11, 97)
2020
Q217
JEE Mains
MCQ
14 Mar 2026
If the sum of the series
20 + 19${3 \over 5}$ + 19${1 \over 5}$ + 18${4 \over 5}$ + ...
upto nth term is 488 and the nth term is negative, then :
20 + 19${3 \over 5}$ + 19${1 \over 5}$ + 18${4 \over 5}$ + ...
upto nth term is 488 and the nth term is negative, then :
A.
n = 41
B.
n = 60
C.
nth term is –4
D.
nth term is -4${2 \over 5}$
2020
Q218
JEE Mains
MCQ
14 Mar 2026
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 :
A.
${1 \over 4}$
B.
${1 \over 5}$
C.
${1 \over 7}$
D.
${1 \over 6}$
2020
Q219
JEE Mains
MCQ
14 Mar 2026
Let S be the sum of the first 9 terms of the
series :
{x + k$a$} + {x2 + (k + 2)$a$} + {x3 + (k + 4)$a$}
+ {x4 + (k + 6)$a$} + .... where a $ \ne $ 0 and x $ \ne $ 1.
If S = ${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$, then k is equal to :
{x + k$a$} + {x2 + (k + 2)$a$} + {x3 + (k + 4)$a$}
+ {x4 + (k + 6)$a$} + .... where a $ \ne $ 0 and x $ \ne $ 1.
If S = ${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$, then k is equal to :
A.
-3
B.
1
C.
-5
D.
3
2020
Q220
JEE Mains
MCQ
14 Mar 2026
If the sum of first 11 terms of an A.P.,
a1, a2, a3, .... is 0 (a $ \ne $ 0), then the sum of the A.P.,
a1 , a3 , a5 ,....., a23 is ka1 , where k is equal to :
a1, a2, a3, .... is 0 (a $ \ne $ 0), then the sum of the A.P.,
a1 , a3 , a5 ,....., a23 is ka1 , where k is equal to :
A.
${{121} \over {10}}$
B.
-${{121} \over {10}}$
C.
${{72} \over 5}$
D.
-${{72} \over 5}$
2020
Q221
JEE Mains
MCQ
14 Mar 2026
The sum of the first three terms of a G.P. is S and
their product is 27. Then all such S lie in :
A.
[-3, $\infty $)
B.
(-$ \propto $, 9]
C.
(-$ \propto $, -9] $ \cup $ [-3, $\infty $)
D.
(-$ \propto $, -3] $ \cup $ [9, $\infty $)
2020
Q222
JEE Mains
MCQ
14 Mar 2026
If |x| < 1, |y| < 1 and x $ \ne $ y, then the sum to infinity
of the following series
(x + y) + (x2+xy+y2) + (x3+x2y + xy2+y3) + ....
(x + y) + (x2+xy+y2) + (x3+x2y + xy2+y3) + ....
A.
${{x + y - xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}$
B.
${{x + y - xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$
C.
${{x + y + xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}$
D.
${{x + y + xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$
2020
Q223
JEE Mains
MCQ
14 Mar 2026
Let an be the nth term of a G.P. of positive terms.
$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $ and $\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $,
then $\sum\limits_{n = 1}^{200} {{a_n}} $ is equal to :
$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $ and $\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $,
then $\sum\limits_{n = 1}^{200} {{a_n}} $ is equal to :
A.
150
B.
175
C.
225
D.
300
2020
Q224
JEE Mains
MCQ
14 Mar 2026
The product ${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$ ... to $\infty $ is equal
to :
A.
${2^{{1 \over 4}}}$
B.
${2^{{1 \over 2}}}$
C.
1
D.
2
2020
Q225
JEE Mains
MCQ
14 Mar 2026
If the 10th term of an A.P. is ${1 \over {20}}$ and its 20th term
is ${1 \over {10}}$, then the sum of its first 200 terms is
A.
100
B.
$100{1 \over 2}$
C.
$50{1 \over 4}$
D.
50
2020
Q226
JEE Mains
MCQ
14 Mar 2026
Let ƒ : R $ \to $ R be such that for all
x $ \in $ R
(21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P.,
then the minimum value of ƒ(x) is
(21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P.,
then the minimum value of ƒ(x) is
A.
2
B.
0
C.
3
D.
4
2020
Q227
JEE Mains
MCQ
14 Mar 2026
If the sum of the first 40 terms of the series,
3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to :
3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to :
A.
20
B.
5
C.
10
D.
25
2020
Q228
JEE Mains
MCQ
14 Mar 2026
Let ${a_1}$
, ${a_2}$
, ${a_3}$
,....... be a G.P. such that
${a_1}$ < 0, ${a_1}$ + ${a_2}$ = 4 and ${a_3}$ + ${a_4}$ = 16.
If $\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $, then $\lambda $ is equal to:
${a_1}$ < 0, ${a_1}$ + ${a_2}$ = 4 and ${a_3}$ + ${a_4}$ = 16.
If $\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $, then $\lambda $ is equal to:
A.
171
B.
-171
C.
-513
D.
${{511} \over 3}$
2020
Q229
JEE Mains
MCQ
14 Mar 2026
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:
A.
${{21} \over 2}$
B.
27
C.
7
D.
16
2020
Q230
JEE Mains
Numerical
14 Mar 2026
If m arithmetic means (A.Ms) and three
geometric means (G.Ms) are inserted between
3 and 243 such that 4th A.M. is equal to 2nd
G.M., then m is equal to _________ .
Correct Answer: 39
Explanation:
Given m arithmetic means (A.Ms) present between 3 and 243
$ \therefore $ Common difference, $d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$
$ \therefore $ 4th A.M. = a + 4d
= 3 + 4 $ \times $ ${{240} \over {m + 1}}$
Also there are 3 G.M between 3 and 243
$ \therefore $ Common ratio (r) = ${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$
where n = number of G.M inserted.
$ \therefore $ r = ${\left( {{{243} \over 3}} \right)^{{1 \over {3 + 1}}}} = 3$
Given,
4th A.M = 2nd G.M
$ \Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$
$ \Rightarrow {{960} \over {m + 1}} = 24$
$ \Rightarrow m = 39$
$ \therefore $ Common difference, $d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$
$ \therefore $ 4th A.M. = a + 4d
= 3 + 4 $ \times $ ${{240} \over {m + 1}}$
Also there are 3 G.M between 3 and 243
$ \therefore $ Common ratio (r) = ${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$
where n = number of G.M inserted.
$ \therefore $ r = ${\left( {{{243} \over 3}} \right)^{{1 \over {3 + 1}}}} = 3$
Given,
4th A.M = 2nd G.M
$ \Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$
$ \Rightarrow {{960} \over {m + 1}} = 24$
$ \Rightarrow m = 39$
2020
Q231
JEE Mains
Numerical
14 Mar 2026
The value of ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$ is equal to ______.
Correct Answer: 4
Explanation:
Given, ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$
As sum of GP upto infinity = ${a \over {1 - r}}$
$ \therefore $ ${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $ = ${{{1 \over 3}} \over {1 - {1 \over 3}}}$ = ${1 \over 2}$
$ \therefore $ ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$
= ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{{16} \over {100}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{4 \over {10}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {{{10} \over 4}} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {2.5} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${{{\left( {2.5} \right)}^{ - 2{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}}$
= ${{{\left( {{1 \over 2}} \right)}^{ - 2}}}$ = 4
As sum of GP upto infinity = ${a \over {1 - r}}$
$ \therefore $ ${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $ = ${{{1 \over 3}} \over {1 - {1 \over 3}}}$ = ${1 \over 2}$
$ \therefore $ ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$
= ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{{16} \over {100}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{4 \over {10}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {{{10} \over 4}} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {2.5} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${{{\left( {2.5} \right)}^{ - 2{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}}$
= ${{{\left( {{1 \over 2}} \right)}^{ - 2}}}$ = 4
2020
Q232
JEE Mains
Numerical
14 Mar 2026
The number of terms common to the two A.P.'s
3, 7, 11, ....., 407 and 2, 9, 16, ....., 709 is ______.
Correct Answer: 14
Explanation:
First A.P. is 3, 7, 11, 15, 19, 23, ..... 407
d1 = 4
Second A.P. is 2, 9, 16, 23, ..... 709
d2 = 7
First common term = 23
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
Last term $ \le $ 407
$ \Rightarrow $ 23 + (n – 1) (28) $ \le $ 407
$ \Rightarrow $ n $ \le $ 14.7
$ \therefore $ n = 14
d1 = 4
Second A.P. is 2, 9, 16, 23, ..... 709
d2 = 7
First common term = 23
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
Last term $ \le $ 407
$ \Rightarrow $ 23 + (n – 1) (28) $ \le $ 407
$ \Rightarrow $ n $ \le $ 14.7
$ \therefore $ n = 14
2020
Q233
JEE Mains
Numerical
14 Mar 2026
The sum, $\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $ is equal to
________.
Correct Answer: 504
Explanation:
$\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $
= ${1 \over 4}\sum\limits_{n = 1}^7 {\left( {2{n^3} + 3{n^2} + n} \right)} $
= ${1 \over 2}\sum\limits_{n = 1}^7 {{n^3}} $ + ${3 \over 4}\sum\limits_{n = 1}^7 {{n^2}} $ + ${1 \over 4}\sum\limits_{n = 1}^7 n $
= ${1 \over 2}{\left( {{{7\left( {7 + 1} \right)} \over 2}} \right)^2}$ + ${3 \over 4}\left( {{{7\left( {7 + 1} \right)\left( {14 + 1} \right)} \over 6}} \right)$ + ${1 \over 4}{{7\left( 8 \right)} \over 2}$
= (49)(8) + (15$ \times $7) + (7)
= 392 + 105 + 7 = 504
= ${1 \over 4}\sum\limits_{n = 1}^7 {\left( {2{n^3} + 3{n^2} + n} \right)} $
= ${1 \over 2}\sum\limits_{n = 1}^7 {{n^3}} $ + ${3 \over 4}\sum\limits_{n = 1}^7 {{n^2}} $ + ${1 \over 4}\sum\limits_{n = 1}^7 n $
= ${1 \over 2}{\left( {{{7\left( {7 + 1} \right)} \over 2}} \right)^2}$ + ${3 \over 4}\left( {{{7\left( {7 + 1} \right)\left( {14 + 1} \right)} \over 6}} \right)$ + ${1 \over 4}{{7\left( 8 \right)} \over 2}$
= (49)(8) + (15$ \times $7) + (7)
= 392 + 105 + 7 = 504
2020
Q234
JEE Mains
Numerical
14 Mar 2026
The sum $\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $ is :
Correct Answer: 1540
Explanation:
$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $
= $\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $
= $\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $
= ${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 \over 2} \times {{20 \times 21} \over 2}$
= 1540
= $\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $
= $\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $
= ${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 \over 2} \times {{20 \times 21} \over 2}$
= 1540
2019
Q235
JEE Mains
MCQ
14 Mar 2026
If a1, a2, a3, ..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :
A.
120
B.
200
C.
150
D.
280
2019
Q236
JEE Mains
MCQ
14 Mar 2026
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 :
A.
- 153
B.
- 135
C.
- 133
D.
- 131
2019
Q237
JEE Mains
MCQ
14 Mar 2026
Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6= – 48, then S10 is equal to :
A.
- 320
B.
- 380
C.
- 460
D.
- 210
2019
Q238
JEE Mains
MCQ
14 Mar 2026
Let a1, a2, a3,......be an A.P. with a6 = 2. Then the common difference of this A.P., which maximises the
product a1a4a5, is :
A.
${3 \over 2}$
B.
${6 \over 5}$
C.
${8 \over 5}$
D.
${2 \over 3}$
2019
Q239
JEE Mains
MCQ
14 Mar 2026
The sum
$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}}$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$ is equal to :
$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$ is equal to :
A.
620
B.
1240
C.
1860
D.
660
2019
Q240
JEE Mains
MCQ
14 Mar 2026
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 4th term of this A.P. is :
A.
$a$
B.
${7 \over 3}a$
C.
5$a$
D.
${2 \over 3}a$
2019
Q241
JEE Mains
MCQ
14 Mar 2026
If a1, a2, a3, ............... an are in A.P. and a1 + a4 + a7 + ........... + a16 = 114, then a1 + a6 + a11 + a16 is equal to :
A.
38
B.
98
C.
76
D.
64
2019
Q242
JEE Mains
MCQ
14 Mar 2026
The sum
${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$ upto 10 terms is:
${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$ upto 10 terms is:
A.
600
B.
660
C.
680
D.
620
2019
Q243
JEE Mains
MCQ
14 Mar 2026
Some identical balls are arranged in rows to form
an equilateral triangle. The first row consists of one
ball, the second row consists of two balls and so
on. If 99 more identical balls are addded to the total
number of balls used in forming the equilaterial
triangle, then all these balls can be arranged in a
square whose each side contains exactly 2 balls
less than the number of balls each side of the
triangle contains. Then the number of balls used to
form the equilateral triangle is :-
A.
262
B.
190
C.
157
D.
225
2019
Q244
JEE Mains
MCQ
14 Mar 2026
If the sum and product of the first three term in
an A.P. are 33 and 1155, respectively, then a value
of its 11th term is :-
A.
–25
B.
–36
C.
25
D.
–35
2019
Q245
JEE Mains
MCQ
14 Mar 2026
The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term is :-
A.
945
B.
916
C.
915
D.
946
2019
Q246
JEE Mains
MCQ
14 Mar 2026
Let the sum of the first n terms of a non-constant
A.P., a1, a2, a3, ..... 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, a50) is equal to
A.
(A, 50+45A)
B.
(50, 50+45A)
C.
(A, 50+46A)
D.
(50, 50+46A)
2019
Q247
JEE Mains
MCQ
14 Mar 2026
If three distinct numbers a, b, c are in G.P. and the
equations ax2
+ 2bx + c = 0 and
dx2
+ 2ex + ƒ = 0 have a common root, then
which one of the following statements is
correct?
A.
$d \over a$, $e \over b$, $f \over c$ are in G.P.
B.
d, e, ƒ are in A.P
C.
d, e, ƒ are in G.P
D.
$d \over a$, $e \over b$, $f \over c$ are in A.P.
2019
Q248
JEE Mains
MCQ
14 Mar 2026
The sum
$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $ is equal to
A.
$2 - {11 \over {{2^{19}}}}$
B.
$2 - {3 \over {{2^{17}}}}$
C.
$1 - {11 \over {{2^{20}}}}$
D.
$2 - {21 \over {{2^{20}}}}$
2019
Q249
JEE Mains
MCQ
14 Mar 2026
The sum of all natural numbers 'n' such that
100 < n < 200 and H.C.F. (91, n) > 1 is :
A.
3221
B.
3121
C.
3203
D.
3303
2019
Q250
JEE Mains
MCQ
14 Mar 2026
If sin4$\alpha $ + 4 cos4$\beta $ + 2 = 4$\sqrt 2 $ sin $\alpha $ cos $\beta $; $\alpha $, $\beta $ $ \in $ [0, $\pi $],
then cos($\alpha $ + $\beta $) $-$ cos($\alpha $ $-$ $\beta $) is equal to :
then cos($\alpha $ + $\beta $) $-$ cos($\alpha $ $-$ $\beta $) is equal to :
A.
$ - \sqrt 2 $
B.
0
C.
$-$ 1
D.
$\sqrt 2 $