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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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211 Easy
211 of 998
Application +4 | -1
If the sum
312+512+22+712+22+32+……+\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+\ldots \ldots+ up to 20 terms is equal
to k21\frac{\mathrm{k}}{21}, then k\mathrm{k} is equal to:
View answer and explanation
Correct option: A
120
Option A (120) is correct because the sum of the series is calculated correctly, and k is found to be 120.
212 Easy
212 of 998
Application +4 | -1
If a(b−c)x2+b(c−a)xy+c(a−b)y2a(b-c) x^2+b(c-a) x y+c(a-b) y^2 is a perfect square, the quantities a,b,ca, b, c are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the given condition for the expression to be a perfect square leads to the relationship that a,b,ca, b, c are in Harmonic Progression.
213 Medium
213 of 998
Application +4 | -1
If a+b+c=3a+b+c=3 and a>0,b>0,c>0a > 0, b > 0, c > 0, then the greatest value of a2b3c2a^2 b^3 c^2 is
View answer and explanation
Correct option: A
310⋅2477\frac{3^{10} \cdot 2^4}{7^7}
Option A (310⋅2477\frac{3^{10} \cdot 2^4}{7^7}) is correct because it represents the greatest value of a2b3c2a^2 b^3 c^2 when a+b+c=3a+b+c=3 and a,b,c>0a, b, c > 0, found using AM-GM inequality.
214 Medium
214 of 998
Analyzing +4 | -1
If first three terms of sequence 116,a,b,16\frac{1}{16}, a, b, \frac{1}{6} are in geometric series and the last three terms are in harmonic series, then the values of aa and bb will be
View answer and explanation
Correct option: C
Both (a) and (b) are true
Option C (Both (a) and (b) are true) is correct because there are two possible sets of values for a and b that satisfy the given conditions for geometric and harmonic series.
215 Easy
215 of 998
Application +4 | -1
For x∈Rx \in \mathbf{R}, let [x][x] denote the greatest integer ≤x\leq x. Largest natural number nn for which

E=[π2]+[1100+π2]+[2100+π2]…+[n100+π2]<43,E=\left[\frac{\pi}{2}\right]+\left[\frac{1}{100}+\frac{\pi}{2}\right]+\left[\frac{2}{100}+\frac{\pi}{2}\right] \ldots+\left[\frac{n}{100}+\frac{\pi}{2}\right]<43, is
View answer and explanation
Correct option: A
41
Option A (41) is correct because it is the largest integer 'n' such that the sum of the greatest integer functions is less than 43.
216 Easy
216 of 998
Application +4 | -1
If mm is the AM of two distinct real numbers ll and nn (l,n>1)(l, n>1) and G1,G2G_1, G_2 and G3G_3 are three geometric means between ll and nn, then G14+2G24+G34G_1^4+2 G_2^4+G_3^4 equals
View answer and explanation
Correct option: C
4 lm2n4 \mathrm{~lm}^2 n
Option C (4 lm2n4 \mathrm{~lm}^2 n) is correct because it is the value of the expression G14+2G24+G34G_1^4+2 G_2^4+G_3^4, derived from the given conditions of AM and geometric means.
217 Medium
217 of 998
Understanding +4 | -1
1+log⁡ex1!+(log⁡ex)22!+(log⁡ex)33!+⋯∞=1+\frac{\log _{e} x}{1 !}+\frac{\left(\log _{e} x\right)^{2}}{2 !}+\frac{\left(\log _{e} x\right)^{3}}{3 !}+\cdots \infty=
View answer and explanation
Correct option: B
xx
Option B (xx) is correct because the given series is the Taylor series expansion of eye^y where y=log⁡exy = \log_e x, so the sum is elog⁡ex=xe^{\log_e x} = x.
218 Medium
218 of 998
Application +4 | -1
If twice the 11th term of an AP is equal to 7 times its 21st term, then its 25th term is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because using the given condition to find a relationship between the first term and common difference leads to the 25th term being zero.
219 Easy
219 of 998
Application +4 | -1
If one A.M. ' aa ' and two G.M.'s pp and qq be inserted between any two numbers, then the value of p3+q3p^3+q^3 is
View answer and explanation
Correct option: B
$2 apq$
Option B ($2 apq$) is correct because it correctly relates the arithmetic mean 'a' and the two geometric means 'p' and 'q' inserted between two numbers.
220 Medium
220 of 998
Application +4 | -1
If the sum of pp terms of an AP is qq and that of qq terms is pp, then the sum of p+qp+q terms will be
View answer and explanation
Correct option: D
−(p+q)-(p+q)
Option D (-(p+q)) is correct because it is the sum of (p+q) terms of the AP, given that the sum of p terms is q and the sum of q terms is p.
221 Medium
221 of 998
Application +4 | -1
The value of 0.423‾0.4 \overline{23} is
View answer and explanation
Correct option: B
419990\frac{419}{990}
Option B (419990\frac{419}{990}) is correct because the repeating decimal can be expressed as a fraction by summing the non-repeating and repeating parts.
222 Easy
222 of 998
Application +4 | -1
The nth and 10th10 t h terms of aGPa G P 5,25,125,…5,25,125, \ldots, are respectively
View answer and explanation
Correct option: C
5n,5105^n, 5^{10}
Option C (5n,5105^n, 5^{10}) is correct because the nth term of the GP is 5n5^n and the 10th term is 5105^{10}.
223 Medium
223 of 998
Application +4 | -1
If α,β\alpha, \beta and γ\gamma are the roots of the equation x3−6x2+11x−6=0x^3-6 x^2+11 x-6=0. Then, HM of the roots is
View answer and explanation
Correct option: D
1811\frac{18}{11}
Option D (1811\frac{18}{11}) is correct because it is found by comparing the given polynomial with the general form relating roots and AM, GM, HM.
224 Easy
224 of 998
Application +4 | -1
If sum of the series ∑n=0∞rn=S\sum \limits_{n = 0}^\infty {r^n} = S for ∣r∣<1\left| r \right| < 1, then sum of the series ∑n=0∞r2n\sum \limits_{n = 0}^\infty {r^{2n}} , is
View answer and explanation
Correct option: D
S22S−1\frac{{{S^2}}}{{2S - 1}}
Option D (S22S−1\frac{{{S^2}}}{{2S - 1}}) is correct because by substituting the expression for r in terms of S, we find the sum of the series to be S22S−1\frac{{{S^2}}}{{2S - 1}}.
225 Easy
225 of 998
Application +4 | -1
Let {an}n=0∞\left\{a_n\right\}_{n=0}^{\infty} be a sequence such that a0=a1=0a_0=a_1=0 and an+2=2an+1−an+1a_{n+2}=2 a_{n+1}-a_n+1 for all n≥0n \geq 0.
Then ∑n=2∞an7n\sum\limits_{n=2}^{\infty} \frac{a_n}{7^n} is equal to :
View answer and explanation
Correct option: B
7216\frac{7}{216}
Option B (7216\frac{7}{216}) is correct because it is the result of solving the recurrence relation and summing the infinite series.
226 Medium
226 of 998
Application +4 | -1
The sum of nn terms of the series 1+(1+3)+(1+3+5)+……1+(1+3)+(1+3+5)+\ldots \ldots, is
View answer and explanation
Correct option: C
n(n+1)(2n+1)6\frac{n(n+1)(2 n+1)}{6}
Option C (n(n+1)(2n+1)6\frac{n(n+1)(2 n+1)}{6}) is correct because the sum of the n terms of the series is found by recognizing the nth term as n squared and then summing the squares of the first n natural numbers.
227 Easy
227 of 998
Application +4 | -1
The sum of nn terms of an A.P. is 3n2+5.3{n^2} + 5. The number of term which equals 159 is
View answer and explanation
Correct option: C
27
Option C (27) is correct because the nth term of the A.P. is derived from the sum of n terms, and solving for n when the term equals 159 yields 27.
228 Easy
228 of 998
Understanding +4 | -1
If a1,a2,a3,…,an{a_1},{a_2},{a_3}, \ldots ,{a_n} are the nn arithmetic means between aa and bb, then 2∑i=1nai2\sum \limits_{i = 1}^n {a_i} equals
View answer and explanation
Correct option: B
n(a+b)n\left( {a + b} \right)
Option B (n(a+b)n\left( {a + b} \right)) is correct because the sum of n arithmetic means between a and b is n times the average of a and b.
229 Easy
229 of 998
Application +4 | -1
The value of log⁡2log⁡2log⁡4256+2log⁡22,{\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 }}2, is
View answer and explanation
Correct option: C
5
Option C (5) is correct because by simplifying the logarithmic expression step-by-step, the final result is 5.
230 Easy
230 of 998
Application +4 | -1
For x∈R, let[x]x \in R,\,\mathrm{let}\left[ x \right] denote the greatest integer ≤x\le x, then the sum of the series
[−13]+[−13−1100]+[−13−2100]+...+[−13−99100]\left[ { - \frac{1}{3}} \right] + \left[ { - \frac{1}{3} - \frac{1}{{100}}} \right] + \left[ { - \frac{1}{3} - \frac{2}{{100}}} \right] + ... + \left[ { - \frac{1}{3} - \frac{{99}}{{100}}} \right] is
View answer and explanation
Correct option: B
−133- 133
Option B (-133) is correct because it is the sum of the series after evaluating the greatest integer function for each term.
231 Easy
231 of 998
Application +4 | -1
If log⁡42+log⁡44+log⁡416+log⁡4x=6,{\log _4}2 + {\log _4}4 + {\log _4}16 + {\log _4}x = 6, then x=x =
View answer and explanation
Correct option: C
32
Option C (32) is correct because by using logarithm properties and solving the resulting equation, x is found to be 32.
232 Easy
232 of 998
Application +4 | -1
Suppose for each n∈N,14+24+34+…+n4=an5+bn4+cn3+dn2+en+fn \in \mathbf{N}, 1^4+2^4+3^4+\ldots+n^4=a n^5+b n^4+c n^3+d n^2+e n+f, then value of bb is
View answer and explanation
Correct option: B
1/21 / 2
Option B (1/2) is correct because by comparing coefficients after subtracting the sum of powers formula for n and n+1, we can determine the value of b to be 1/2.
233 Easy
233 of 998
Application +4 | -1
The number of solutions of log⁡2(x−1)=2log⁡2(x−3){\log _2}\left( {x - 1} \right) = 2{\log _2}\left( {x - 3} \right) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because solving the logarithmic equation yields two possible solutions, but only one satisfies the domain restrictions of the logarithms.
234 Medium
234 of 998
Application +4 | -1
If a1,a2,a3,…,  an{a_1},{a_2},{a_3}, \ldots ,\;{a_n} are in AP, where ai>0{a_i} > 0 for all i, then value of 1a1+a2+1a2+a3+...+1an−1+an\frac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \frac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + ... + \frac{1}{{{a_{n - 1}} + \sqrt {{a_n}} }} is equal to
View answer and explanation
Correct option: A
n−1a1+an\frac{{n - 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}
Option A (n−1a1+an\frac{{n - 1}}{{\sqrt {{{a_1}}} + \sqrt {{{a_n}}} }}) is correct because it is the simplified form of the given series after rationalization and telescoping.
235 Easy
235 of 998
Application +4 | -1
The number of terms in an A.P. is even; the sum of the odd terms in it is 24 and that the even terms is 30 . If the last term exceeds the first term by 101210 \frac{1}{2}, then the number of terms in the A.P. is:
View answer and explanation
Correct option: B
8
Option B (8) is correct because it is the number of terms that satisfies the given conditions for the arithmetic progression.
236 Medium
236 of 998
Application +4 | -1
13−23+33−43+…+93={1^3} - {2^3} + {3^3} - {4^3} + \ldots + {9^3} =
View answer and explanation
Correct option: A
425
Option A (425) is correct because the alternating sum of cubes from 1 to 9 can be simplified and calculated to be 425.
237 Easy
237 of 998
Analyzing +4 | -1
If x,y,zx,y,z are in AP, then 1x+y,1z+x,1y+z\frac{1}{{\sqrt x + \sqrt y }},\frac{1}{{\sqrt z + \sqrt x }},\frac{1}{{\sqrt y + \sqrt z }} are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because if x, y, z are in AP, then the reciprocals with the given square roots are also in AP.
238 Easy
238 of 998
Understanding +4 | -1
If a,  b,  c,  d,  e,  f  a,\;b,\;c,\;d,\;e,\;f\; are in AP, then the value of e−ce - c will be
View answer and explanation
Correct option: B
2 (d−c)\left( {d - c} \right)
Option D (d−cd - c) is incorrect. The correct answer is 2(d−cd - c). Since a,b,c,d,e,f are in A.P. e-c = 2(d-c)
239 Medium
239 of 998
Application +4 | -1
Jairam purchased a house in Rs 15000 and paid Rs 5000 at once. Rest money he promised to pay in annual installment of Rs 1000 with 10% per annum interest. How much money is to be paid by Jairam?
View answer and explanation
Correct option: C
Rs 20500
Option C (Rs 20500) is correct because it is the total amount Jairam has to pay, including the initial payment and the sum of all installments with interest.
240 Easy
240 of 998
Analyzing +4 | -1
If the sum of an infinitely decreasing G.P. is 3, and the sum of the squares of its terms is 9/2,9/2, the sum of the cubes of the terms is
View answer and explanation
Correct option: B
10813\frac{{108}}{{13}}
Option B (108/13) is correct because it is derived by finding the first term and common ratio of the G.P. and then calculating the sum of the cubes of the terms.
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