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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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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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361 Easy
361 of 998
Application +4 | -1
If a1,a2,a3,….,an+1a_1, a_2, a_3, \ldots ., a_{n+1} are in A.P., then 1a1a2+1a2a3+…⋅1anan+1\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\ldots \cdot \frac{1}{a_n a_{n+1}}, is
View answer and explanation
Correct option: D
na1an+1\frac{n}{a_1 a_{n+1}}
Option D (na1an+1\frac{n}{a_1 a_{n+1}}) is correct because it simplifies the given expression using the properties of arithmetic progressions.
362 Easy
362 of 998
Application +4 | -1
The maximum sum of the series 20+1913+1823+18+…20+19 \frac{1}{3}+18 \frac{2}{3}+18+\ldots is
View answer and explanation
Correct option: A
310
Option A (310) is correct because it is the maximum sum of the series, obtained by summing only the non-negative terms of the AP.
363 Easy
363 of 998
Application +4 | -1
The value of 2.357‾2.\overline {357} is
View answer and explanation
Correct option: A
2355999\frac{{2355}}{{999}}
Option A (2355999\frac{{2355}}{{999}}) is correct because converting the repeating decimal to a fraction and adding it to the integer part yields 2355999\frac{{2355}}{{999}}.
364 Easy
364 of 998
Application +4 | -1
The equation log⁡ex+log⁡e(1+x)=0{\log _e}x + {\log _e}\left( {1 + x} \right) = 0 can be written as
View answer and explanation
Correct option: A
x2+x−1=0{x^2} + x - 1 = 0
The correct answer is x2+x−1=0{x^2} + x - 1 = 0 because by using logarithm properties, the equation can be simplified to this quadratic form.
365 Easy
365 of 998
Application +4 | -1
If the arithmetic mean of two positive numbers aa and b(a>b)b(a>b) is twice their geometric mean, then a:ba: b is
View answer and explanation
Correct option: A
2+3:2−32+\sqrt{3}: 2-\sqrt{3}
Option A (2+3:2−32+\sqrt{3}: 2-\sqrt{3}) is correct because it is derived from the relationship between arithmetic and geometric means given the condition A = 2G.
366 Easy
366 of 998
Application +4 | -1
Let S⊂(0,π)S \subset(0, \pi) denote the set of values of xx satisfying the equation
81+∣cos⁡x∣+cos⁡2x+∣cos⁡3x∣+…to∞=43{8^{1 + |\cos x| + {{\cos }^2}x + \left| {{{\cos }^3}x} \right| + \ldots \mathrm{ to }\infty }} = {4^3}
Then S=S=
View answer and explanation
Correct option: D
{π/3,2π/3}\{\pi / 3,2 \pi / 3\}
Option D ({π/3,2π/3}\{\pi / 3,2 \pi / 3\}) is correct because these are the only values of x in the given interval that satisfy the equation after simplifying the infinite series.
367 Easy
367 of 998
Application +4 | -1
If the sum of first 11 terms of an A.P. a1,a2,a3,…a_1, a_2, a_3, \ldots is 0 , and the sum of the A.P. a1,a3,a5,…,a23a_1, a_3, a_5, \ldots, a_{23} is ka1,a1≠0k a_1, a_1 \neq 0. Then kk is equal to
View answer and explanation
Correct option: C
−725-\frac{72}{5}
Option C (-725\frac{72}{5}) is correct because it is derived from setting up equations based on the given sums and solving for kk.
368 Easy
368 of 998
Application +4 | -1
Let Tr{T_r}, be rrth term of an AP whose first term is aa and common difference is dd. If for some positive integers m,n,m≠n,  Tm=1nm,n,m \ne n,\;{T_m} = \frac{1}{n} and Tn=1m{T_n} = \frac{1}{m}, then a−da - d equals
View answer and explanation
Correct option: A
0
Option A (0) is correct because solving the given conditions for an A.P. leads to a - d = 0.
369 Easy
369 of 998
Application +4 | -1
If a,  b,  ca,\;b,\;c are digits, then the rational number represented by 0. cababab…cababab \ldots is
View answer and explanation
Correct option: D
99c+10a+b990\frac{{99c + 10a + b}}{{990}}
Option D is correct because it accurately converts the repeating decimal into a rational number by setting up equations and solving for the repeating part.
370 Easy
370 of 998
Analyzing +4 | -1
If a1/x=b1/y=c1/z{a^{1/x}} = {b^{1/y}} = {c^{1/z}} and a,  b,ca,\;b,c are in GP, then x,  y,  zx,\;y,\;z will be in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the given conditions lead to the conclusion that x,y,zx, y, z are in Arithmetic Progression.
371 Easy
371 of 998
Analyzing +4 | -1
If a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n are non-zero real numbers such that
(a12+a22+…+an−12)(a22+a32+…+an2)≤(a1a2+a2a3+…+an−1an)2\begin{aligned} \left(a_1^2+a_2^2+\ldots+a_{n-1}^2\right)\left(a_2^2\right.&\left.+a_3^2+\ldots+a_n^2\right) \\ & \leq\left(a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n\right)^2 \end{aligned}
then, a1,a2,…,ana_1, a_2, \ldots, a_n are in
View answer and explanation
Correct option: B
G.P.
The correct answer is G.P. because the given inequality, when simplified using Cauchy-Schwarz inequality, leads to the condition where the terms must be in geometric progression.
372 Easy
372 of 998
Analyzing +4 | -1
2log⁡x−log⁡(x+1)−log⁡(x−1)2\log x - \log \left( {x + 1} \right) - \log \left( {x - 1} \right) is equal to
View answer and explanation
Correct option: B
1x2+12  x4+13  x6+…\frac{1}{{{x^2}}} + \frac{1}{{2\;{x^4}}} + \frac{1}{{3\;{x^6}}} + \ldots
Option B (1x2+12  x4+13  x6+…\frac{1}{{{x^2}}} + \frac{1}{{2\;{x^4}}} + \frac{1}{{3\;{x^6}}} + \ldots) is the correct expansion after simplifying the logarithmic expression.
373 Easy
373 of 998
Analyzing +4 | -1
If ∣x∣<1\left| x \right| < 1 and y=x−x22+x33−x44+...y = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ..., then xx is equal to
View answer and explanation
Correct option: C
y+y22!+y33!+...y + \frac{{{y^2}}}{{2!}} + \frac{{{y^3}}}{{3!}} + ...
Option C (y+y22!+y33!+...y + \frac{{{y^2}}}{{2!}} + \frac{{{y^3}}}{{3!}} + ...) is correct because it represents the exponential function ey−1e^y - 1, which is equal to x when y=log⁡(1+x)y = \log(1+x).
374 Easy
374 of 998
Application +4 | -1
The sum to nn terms of the series 113+1+213+23+1+2+313+23+33+…\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+\ldots is
View answer and explanation
Correct option: C
2nn+1\frac{2 n}{n+1}
Option C (2nn+1\frac{{2n}}{{n+1}}) is correct because simplifying the general term of the series and summing it results in the given expression.
375 Easy
375 of 998
Application +4 | -1
log⁡2a3=log⁡2b4=log⁡2c5λ\frac{{{{\log }_2}a}}{3} = \frac{{{{\log }_2}b}}{4} = \frac{{{{\log }_2}c}}{{5\mathrm{\lambda }}} andand a−3b−4c=1{a^{ - 3}}{b^{ - 4}}c = 1 then λ=\mathrm{\lambda } =
View answer and explanation
Correct option: C
5
Option C (5) is correct because it is the value of lambda that satisfies the given logarithmic equation and the condition a−3b−4c=1{a^{ - 3}}{b^{ - 4}}c = 1.
376 Easy
376 of 998
Application +4 | -1
If the AM and GM of roots of a quadratic equations are 8 and 5 respectively, then the quadratic equation will be
View answer and explanation
Correct option: C
x2−16x+25=0{x^2} - 16x + 25 = 0
Option C (x² - 16x + 25 = 0) is correct because it is the quadratic equation formed using the given arithmetic mean and geometric mean of the roots.
377 Easy
377 of 998
Analyzing +4 | -1
Consider an A.P. with first term ' aa ' and common difference ' dd '. Let SkS_k denote the sum of the first kk terms. If SkxSx\frac{S_{k x}}{S_x} is independent of xx, then
View answer and explanation
Correct option: C
2a=d2 a=d
Option C (2a=d2 a=d) is correct because it makes the expression SkxSx\frac{S_{k x}}{S_x} independent of xx, as shown in the solution.
378 Easy
378 of 998
Application +4 | -1
If 21−x+21+x,f(x),3x+3−x2^{1-x}+2^{1+x}, f(x), 3^x+3^{-x} are in A.P., then the minimum value of f(x)f(x) is
View answer and explanation
Correct option: C
3
Option C is correct because the minimum value of f(x) is found by using AM ≥ GM inequality on the given arithmetic progression, resulting in a minimum value of 3.
379 Easy
379 of 998
Analyzing +4 | -1
Let A1,G1,H1A_1, G_1, H_1 denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For n>2n>2, let An−1A_{n-1} and Hn−1H_{n-1} have arithmetic, geometric and harmonic means as An,Gn,HnA_n, G_n, H_n respectively. Then, which one of the following statements is correct?
View answer and explanation
Correct option: C
G1=G2=G3=…G_1=G_2=G_3=\ldots
Option C (G1=G2=G3=...) is correct because the geometric mean remains constant in this recursive process.
380 Easy
380 of 998
Analyzing +4 | -1
The sum and product of five consecutive terms of an A.P. are 25 and 2520 respectively. If one of the terms is 12\frac{1}{2}, then the value of the greatest term is
View answer and explanation
Correct option: B
16
Option B (16) is correct because solving the equations for the sum and product of the arithmetic progression, and considering the term 1/2, leads to the greatest term being 16.
381 Easy
381 of 998
Application +4 | -1
If a geometric progression consisting of positive terms, each term equals the sum of the next two terms, then the common ratio of this progression equals
View answer and explanation
Correct option: D
12(5−1)\frac{1}{2}(\sqrt{5}-1)
Option D (12(5−1)\frac{1}{2}(\sqrt{5}-1)) is correct because it is the positive root of the quadratic equation derived from the given condition of the geometric progression.
382 Easy
382 of 998
Application +4 | -1
The sum of all natural numbers ′n′'n' such that 100 < n<200100\, < \,n < 200 and HCF (91, n) >1(91,\,n)\, > 1 is
View answer and explanation
Correct option: D
3121
Option D (3121) is correct because it is the sum of all natural numbers between 100 and 200 that have a HCF greater than 1 with 91.
383 Easy
383 of 998
Application +4 | -1
If 32sin⁡2θ−1,14,34−2sin⁡2θ3^{2 \sin 2 \theta-1}, 14,3^{4-2 \sin 2 \theta} are first three terms of an A.P., then its fifth term is given by
View answer and explanation
Correct option: C
53
Option B (n2(n+1)/2n^2(n+1) / 2) is correct. When n is odd, the sum of the series can be found by subtracting the last term from the sum when n is even.
384 Easy
384 of 998
Analyzing +4 | -1
Let a1,a2,a3,….,ana_{1}, a_{2}, a_{3}, \ldots ., a_{n} be nn positive consecutive terms of an arithmetic progression. If d>0d>0 is its common difference, then

lim⁡n→∞dn(1a1+a2+1a2+a3+…+1an−1+an)\lim \limits_{{n \rightarrow \infty}} \sqrt{\frac{d}{n}}\left(\begin{array}{c} \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+ \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \end{array}\right) is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the limit simplifies to 1 after rationalizing the denominator and applying limit properties.
385 Easy
385 of 998
Analyzing +4 | -1
If ab+c,bc+a,ca+b\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}} are in AP, then
View answer and explanation
Correct option: C
a2,b2,c2{a^2},{b^2},{c^2} are in AP
Option C (a2,b2,c2{a^2},{b^2},{c^2} are in AP) is correct because by manipulating the given condition that ab+c,bc+a,ca+b\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}} are in AP, we can derive the relationship 2b2=a2+c22b^2 = a^2 + c^2, which means a2,b2,c2{a^2},{b^2},{c^2} are in AP.
386 Easy
386 of 998
Analyzing +4 | -1
Let Sn=∑k=14n(−1)k(k+1)2k2S_n=\sum\limits_{k=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} k^2. Then, SnS_n can take the value (s)(s)
View answer and explanation
Correct option: A
1056 and 1332
Option A (1056 and 1332) is correct because these values can be obtained by substituting integer values of n into the derived formula for SnS_n, which is 4n(4n+1)4n(4n+1).
387 Easy
387 of 998
Application +4 | -1
If the product of nn positive numbers is unity, then their sum is
View answer and explanation
Correct option: D
never less than nn
Option D (never less than n) is correct because, by the AM-GM inequality, the sum of n positive numbers whose product is unity is always greater than or equal to n.
388 Easy
388 of 998
Application +4 | -1
Let aa and bb be two given numbers. If AA denotes the single AM and SS denotes the sum of nn AM's between aa and bb, then S/AS / A depends on
View answer and explanation
Correct option: D
nn
Option D (nn) is correct because the ratio S/A simplifies to n, showing it only depends on the number of arithmetic means.
389 Easy
389 of 998
Analyzing +4 | -1
If four distinct integers form an increasing A.P. such that one of these numbers is equal to the sum of the squares of the remaining three numbers, then the numbers are
View answer and explanation
Correct option: C
−1,0,1,2-1,0,1,2
Option C (-1, 0, 1, 2) is correct because these numbers form an increasing arithmetic progression and satisfy the given condition.
390 Easy
390 of 998
Application +4 | -1
If a1,a2,a3,a4,a5{a_1},{a_2},{a_3},{a_4},{a_5} and a6{a_6} are six arithmetic means between 3 and 31, then a6−a5{a_6} - {a_5} and a1+a6{a_1} + {a_6} are respectively equals to
View answer and explanation
Correct option: C
4 and 34
Option C (4 and 34) is correct because the common difference is 4, and a6 - a5 = d = 4, and a1 + a6 = 7 + 27 = 34.
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