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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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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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Showing 31-60 of 564 questions

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31 Medium
31 of 564
Application +4 | -1
If the 2nd ,5th 2^{\text {nd }}, 5^{\text {th }} and 9th 9^{\text {th }} terms of a non-constant A.P. are in G.P., then the common ratio of this G.P., is
View answer and explanation
Correct option: B
43\frac{4}{3}
Option B (43\frac{4}{3}) is correct because it represents the common ratio of the G.P. formed by the 2nd, 5th, and 9th terms of the given A.P.
32 Medium
32 of 564
Analyzing +4 | -1
If a,ba, b and cc are in arithmetic progression, then 1a+b,1a+c\frac{1}{\sqrt{a}+\sqrt{b}}, \frac{1}{\sqrt{a}+\sqrt{c}} and 1b+c\frac{1}{\sqrt{b}+\sqrt{c}} are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the difference between the reciprocals with square roots is constant when a,b,ca, b, c are in AP.
33 Medium
33 of 564
Application +4 | -1
If α,β\alpha, \beta are the roots of the equation x2−x−1=0x^2-x-1=0 and Sn=2023αn+2024βn,S_n=2023 \alpha^n+2024 \beta^n, then:
View answer and explanation
Correct option: D
S12=S11+S10S_{12}=S_{11}+S_{10}
Option D is correct because it correctly applies Newton's method to the recurrence relation derived from the given quadratic equation.
34 Medium
34 of 564
Application +4 | -1
If the sum of the first ten terms of the series (135)2+(225)2+(315)2+42+(445)2+...,{\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + {\left( {4\frac{4}{5}} \right)^2} + ..., is 165m\frac{{16}}{5}m, then mm is equal to
View answer and explanation
Correct option: B
101
The correct answer is B (101) because the sum of the series is calculated by converting mixed fractions to improper fractions and simplifying.
35 Medium
35 of 564
Application +4 | -1
If a,b,ca, b, c are in G.P., xx and yy be the A. M.'s between a,ba, b and b,cb, c respectively, then (ax+cy)(bx+by)\left(\frac{a}{x}+\frac{c}{y}\right)\left(\frac{b}{x}+\frac{b}{y}\right) is equal to
View answer and explanation
Correct option: D
4
Option A (990) is correct because it is the 20th term of the given series, derived by finding a general formula for the nth term.
36 Medium
36 of 564
Analyzing +4 | -1
The sum of the series
1+14.2!+116.4!+164.6!+1+\frac{1}{4.2 !}+\frac{1}{16.4 !}+\frac{1}{64.6 !}+
…...ad inf. Is
View answer and explanation
Correct option: B
e+1e\frac{e+1}{\sqrt{e}}
Option D ((e+1)/(2√e)) is correct because by recognizing the series as a modification of the Taylor series for (e^x + e-x)/2, the sum to infinity can be found to be (e+1)/(2√e).
37 Medium
37 of 564
Application +4 | -1
If 1, log⁡3(31−x+2),log⁡3(4.3x−1){\log _3}\sqrt {\left( {{3^{1 - x}} + 2} \right)} ,{\log _3}\left( {{{4.3}^x} - 1} \right) are in AP, then x  x\; equals
View answer and explanation
Correct option: B
1−log⁡341 - {\log _3}4
Option B (1−log⁡341 - {\log _3}4) is correct because it is the value of x that satisfies the condition that the given terms are in arithmetic progression.
38 Medium
38 of 564
Analyzing +4 | -1
1n!+12!(n−2)!+14!(n−4)!+...∞\frac{1}{{n!}} + \frac{1}{{2!\left( {n - 2} \right)!}} + \frac{1}{{4!\left( {n - 4} \right)!}} + ...\infty is
View answer and explanation
Correct option: A
2n−1n!\frac{{{2^{n - 1}}}}{{n!}}
Option A (2n−1n!\frac{{{2^{n - 1}}}}{{n!}}) is correct because it represents the simplified form of the given series.
39 Medium
39 of 564
Application +4 | -1
If a2,b2,c2a^2, b^2, c^2 are in A.P. then 1b+c,1c+a,1a+b\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} are in
View answer and explanation
Correct option: A
A.P.
Option A (A.P.) is correct because if a2,b2,c2a^2, b^2, c^2 are in A.P., then 1b+c,1c+a,1a+b\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} are also in A.P.
40 Medium
40 of 564
Analyzing +4 | -1
Solution set for (2+2)x+(2−2)x=2⋅2x/4(\sqrt{2+\sqrt{2}})^x+(\sqrt{2-\sqrt{2}})^x=2 \cdot 2^{x / 4} is
View answer and explanation
Correct option: B
{0}\{0\}
Option B ({0}) is correct because the equation holds true only when x=0, as it satisfies the AM=GM condition.
41 Medium
41 of 564
Analyzing +4 | -1
The positive integer nn for which 2⋅22+3⋅23+4⋅24+…+n⋅2n=2n+102 \cdot 2^2+3 \cdot 2^3+4 \cdot 2^4+\ldots+n \cdot 2^n=2^{n+10}, is
View answer and explanation
Correct option: D
513
Option D (513) is correct because it satisfies the given equation after solving the arithmetico-geometric series.
42 Medium
42 of 564
Application +4 | -1
The nthn\mathrm{th} term of the sequence 4,14,30,52,80,114,…,4,14,30,52,80,114, \ldots , is
View answer and explanation
Correct option: B
3  n2+n3\;{n^2} + n
Option B (3  n2+n3\;{n^2} + n) is correct because it represents the nth term of the sequence, derived by solving a system of equations based on the given terms.
43 Medium
43 of 564
Application +4 | -1
Along a road lies an odd number of stones placed at intervals of 10 m10 \mathrm{~m}. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km4.8 \mathrm{~km}. Then, the number of stones is
View answer and explanation
Correct option: C
31
Option C (31) is correct because the total distance covered is 20n(n+1)20n(n+1) meters, where nn is the number of stones on each side of the middle stone, and solving for nn gives n=15n=15, so the total number of stones is 2n+1=312n+1 = 31.
44 Medium
44 of 564
Application +4 | -1
If two geometric means g1g_1 and g2g_2 and one arithmetic mean A be inserted between two numbers, then g12g2+g22g1=\frac{g_1^2}{g_2}+\frac{g_2^2}{g_1}=
View answer and explanation
Correct option: C
2 A2 {~A}
Option C (2A2A) is correct because it is the value of g12g2+g22g1\frac{g_1^2}{g_2} + \frac{g_2^2}{g_1} when two geometric means g1g_1 and g2g_2 and one arithmetic mean A are inserted between two numbers.
45 Medium
45 of 564
Application +4 | -1
If ptℏ,qth p^{t \hbar}, q^{\text {th }} and rth\mathrm{r}^{t h} terms of H.P. are u,v\mathrm{u}, \mathrm{v}, w respectively, then the value of the expression (q−r)vw+(r−p)wu+(p−q)uv(q-r) v w+(r-p) w u+(p-q) u v is.
View answer and explanation
Correct option: B
0
Option B (0) is correct because the given expression is equivalent to zero when the terms are in Harmonic Progression.
46 Medium
46 of 564
Application +4 | -1
The sum of the series 2,5,8,11,…2,5,8,11, \ldots is 60100 , then nn is equal to
View answer and explanation
Correct option: B
200
Option B (200) is correct because it is the number of terms (n) required for the sum of the given arithmetic series to be 60100.
47 Medium
47 of 564
Application +4 | -1
If a,b,ca, b, c are in G.P. and x,yx, y are the arithmetic means between a,ba, b, and b,cb, c respectively, then ax+cy\frac{a}{x}+\frac{c}{y} is equal to
View answer and explanation
Correct option: C
2
Option B (tan⁡β+2tan⁡γ\tan \beta+2 \tan \gamma) is correct because by using the given relationships between the angles and trigonometric identities, we can express tan α in terms of tan β and tan γ.
48 Medium
48 of 564
Application +4 | -1
A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments which are in AP. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. The value of 8 th instalment is
View answer and explanation
Correct option: C
65
Option C (65) is correct because it is the value of the 8th installment calculated using the arithmetic progression properties and the given conditions.
49 Medium
49 of 564
Application +4 | -1
If a1,a2,…,a10a_1, a_2, \ldots, a_{10} are in A.P. and 1h1,1h2,…,1h10\frac{1}{h_1}, \frac{1}{h_2}, \ldots, \frac{1}{h_{10}} are in A.P. and a1=h1=2a_1=h_1=2 and a10=h10=3a_{10}=h_{10}=3, then a4h7=a_4 h_7=
View answer and explanation
Correct option: B
6
Option B (6) is correct because it is the product of a_4 and h_7, calculated using the properties of arithmetic progressions for a_n and 1/h_n.
50 Medium
50 of 564
Application +4 | -1
The sum of the series
(2)2+2(4)2+3(6)2+……(2)^2+2(4)^2+3(6)^2+\ldots\ldots upto 10 terms is equal to
View answer and explanation
Correct option: B
12100
Option B (12100) is correct. The sum of the series is calculated by expressing the general term and using the formula for the sum of cubes.
51 Medium
51 of 564
Application +4 | -1
If x>0x>0, then the sum of the series e−x−e−2x+e−3x−…e^{-x}-e^{-2 x}+e^{-3 x}-\ldots is
View answer and explanation
Correct option: D
11+ex\frac{1}{1+e^x}
Option D (11+ex\frac{{1}}{{1+e^x}}) is correct because it is the sum of the given infinite geometric series.
52 Medium
52 of 564
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 sum of the series can be calculated by grouping terms and using the formula for the sum of cubes.
53 Medium
53 of 564
Application +4 | -1
If a,b,c,da, b, c, d are positive real number such that a+b+c+d=2a+b+c+d=2, then M=(a+b)(c+d)M=(a+b)(c+d) satisfies the relation:
View answer and explanation
Correct option: A
0<M≤10 < M \leq 1
Option A ((2n+1)π2(2 n+1) \frac{\pi}{2}) is correct because the product of the matrices is a null matrix when cos⁡(θ−ϕ)=0\cos(\theta - \phi) = 0, which occurs when θ−ϕ\theta - \phi is an odd multiple of π2\frac{\pi}{2}.
54 Medium
54 of 564
Application +4 | -1
{an}\left\{a_n\right\} and {bn}\left\{b_n\right\} are arithmetic progressions and their sums of the first nn terms are AnA_n and BnB_{n} respectively. If AnBn=n−12n\frac{A_n}{B_n}=\frac{n-1}{2 n} for all positive integers n,n, then the value of a3+a5+a73(b3+b9)+a4+a102(b2+b10)\frac{a_3+a_5+a_7}{3\left(b_3+b_9\right)}+\frac{a_4+a_{10}}{2\left(b_2+b_{10}\right)} is:
View answer and explanation
Correct option: D
511\frac{5}{11}
Option D (5/11) is correct because after simplifying the given ratio of sums and using the properties of arithmetic progressions, the expression evaluates to 5/11.
55 Medium
55 of 564
Analyzing +4 | -1
If 0<y<21/30 < y < {2^{1/3}} and x(y3−1)=1x\left( {{y^3} - 1} \right) = 1, then 2x+23x3+25x5+…\frac{2}{x} + \frac{2}{{3{x^3}}} + \frac{2}{{5{x^5}}} + \ldots is equal to
View answer and explanation
Correct option: A
log⁡(y32−y3)\log \left( {\frac{{{y^3}}}{{2 - {y^3}}}} \right)
Option A (log⁡(y32−y3)\log \left( {\frac{{{y^3}}}{{2 - {y^3}}}} \right)) is correct because it is derived by substituting x=1y3−1x = \frac{1}{{{y^3} - 1}} into the given series and simplifying using logarithmic properties.
56 Medium
56 of 564
Application +4 | -1
The minimum number of terms from the beginning of the series 20+2223+2513+…20 + 22\frac{2}{3} + 25\frac{1}{3} + \ldots so that the sum may exceed 1568, is
View answer and explanation
Correct option: D
29
Option D is correct because the minimum number of terms is found by solving the inequality for the sum of an arithmetic series exceeding 1568.
57 Medium
57 of 564
Application +4 | -1
If 12+22+32+…+20032=(2003)(4007)(334){1^2} + {2^2} + {3^2} + \ldots + {2003^2} = \left( {2003} \right)\left( {4007} \right)\left( {334} \right) and, 1×2003+2×2002+3×2001+…+2003×1=(2003)(334)x,1 \times 2003 + 2 \times 2002 + 3 \times 2001 + \ldots + 2003 \times 1 = \left( {2003} \right)\left( {334} \right)x, then x=x =
View answer and explanation
Correct option: A
2005
Option A (2005) is correct because by manipulating the given equations and simplifying, we can solve for x and find its value to be 2005.
58 Medium
58 of 564
Application +4 | -1
The sum of the first nn terms of the series 12+5+15+8+18+11+  ...\frac{1}{{\sqrt 2 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt {11} }} + \;... is
View answer and explanation
Correct option: A
13(3n+2−2)\frac{1}{3}\left( {\sqrt {3n + 2} - \sqrt 2 } \right)
Option A (1/3(√(3n+2) - √2)) is correct because it is the simplified sum of the series after rationalizing the denominator.
59 Medium
59 of 564
Application +4 | -1
If x,y,zx,y,z are in HP, then log⁡(x+z)+log⁡(x−2y+z)\log \left( {x + z} \right) + \log \left( {x - 2y + z} \right) is equal to
View answer and explanation
Correct option: B
2log⁡(x−z)2\log \left( {x - z} \right)
Option B (2log⁡(x−z)2\log \left( {x - z} \right)) is correct because it is derived by simplifying the logarithmic expression using the properties of harmonic progressions.
60 Medium
60 of 564
Application +4 | -1
The sum of the series 2[7−1+3−1.7−3+5−1.7−5+...]2\left[ {{7^{ - 1}} + {3^{ - 1}}{{.7}^{ - 3}} + {5^{ - 1}}{{.7}^{ - 5}} + ...} \right] is
View answer and explanation
Correct option: A
log⁡e(43)  {\log _e}\left( {\frac{4}{3}} \right)\;
Option A is correct because the series can be recognized as the expansion of a logarithmic function, specifically related to the inverse hyperbolic tangent.
Revision notes and key takeaways

Review SEQUENCE OF SERIES in your textbook or class notes, then use this set to check your understanding.

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