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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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541 Easy
541 of 998
Application +4 | -1
If SnS_n denotes the sum of first nn term of an A.P., then S3n−Sn−1S2n−S2n−1\frac{S_{3 n}-S_{n-1}}{S_{2 n}-S_{2 n-1}} is equal to
View answer and explanation
Correct option: B
2n+12 n+1
Option B (2n+12 n+1) is correct because by substituting the formula for the sum of an A.P., the expression simplifies to 2n+12n+1.
542 Easy
542 of 998
Application +4 | -1
If the sum of all terms of an A.P. obtained by placing nn arithmetic means between 5 and 21 is 117 , then n=n=
View answer and explanation
Correct option: C
7
Option C (7) is correct because it is the number of arithmetic means that satisfy the given sum condition.
543 Easy
543 of 998
Application +4 | -1
In an A.P., the first term is 1 and sum of the first pp terms is 0, then sum of the first (p+q)(p+q) terms is
View answer and explanation
Correct option: A
q(p+q)1−p\frac{q(p+q)}{1-p}
Option A (q(p+q)1−p\frac{q(p+q)}{1-p}) is correct because it is the result of calculating the sum of the first (p+q) terms using the given information about the A.P.
544 Easy
544 of 998
Application +4 | -1
Suppose a1,a2,…an,…a_1, a_2, \ldots a_n, \ldots be an arithmetic progression of natural numbers. If the ration of the sum of first five terms to the sum of first nine terms of the progression is 5:175: 17 and 110 < a15 < 120110\, < \,{a_{15}}\, < \,120, then the sum of the first ten terms of the progression is equal to
View answer and explanation
Correct option: B
380
Option B is correct because by using the given ratio of sums and the condition on a_15, we can find the values of a_1 and d, and then calculate the sum of the first ten terms.
545 Easy
545 of 998
Analyzing +4 | -1
Let a1,a2,a3,…,a10a_1, a_2, a_3, \ldots, a_{10} be in G.P. with ai>0a_i>0 for i=1,2,3,…,10i=1,2,3, \ldots, 10 and SS be the set of pairs (r,k),r,k∈N(r, k), r, k \in N for which
D=∣log⁡ea1ra2klog⁡ea2ra3klog⁡ea3ra4klog⁡ea4ra5klog⁡ea5ra6klog⁡ea6ra7klog⁡ea7ra8klog⁡ea8ra9klog⁡ea9ra10k∣=0D=\left|\begin{array}{lll}\log _e a_1^r a_2^k & \log _e a_2^r a_3^k & \log _e a_3^r a_4^k \\ \log _e a_4^r a_5^k & \log _e a_5^r a_6^k & \log _e a_6^r a_7^k \\ \log _e a_7^r a_8^k & \log _e a_8^r a_9^k & \log _e a_9^r a_{10}^k\end{array}\right|=0
Then the number of elements in SS, is
View answer and explanation
Correct option: A
Infinity many
Option A (Infinity many) is correct because the determinant D is always zero for all values of r and k, making the set S infinitely large.
546 Easy
546 of 998
Application +4 | -1
In 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}\left( {\sqrt 5 - 1} \right)
Option D (12(5−1)\frac{1}{2}\left( {\sqrt 5 - 1} \right)) is correct because it is the positive root of the quadratic equation derived from the given condition of the geometric progression.
547 Easy
547 of 998
Application +4 | -1
The product of x1/2⋅x1/4⋅x1/8x^{1 / 2} \cdot x^{1 / 4} \cdot x^{1 / 8}.. Equals
View answer and explanation
Correct option: C
xx
Option C (xx) is correct because the product of the terms can be simplified using exponent rules and the sum of an infinite geometric series.
548 Easy
548 of 998
Analyzing +4 | -1
If three positive real numbers a,  b,  ca,\;b,\;c are in AP and abc=4abc = 4, then the minimum possible value of bb is
View answer and explanation
Correct option: B
22/3{2^{2/3}}
Option B (22/3{2^{2/3}}) is correct because the minimum value of b is found using the AM-GM inequality, given that a, b, c are in AP and abc = 4.
549 Easy
549 of 998
Application +4 | -1
Consider an infinite geometric series with first term aa and common ratio rr. If its sum is 4 and the second term is 3/43 / 4, then
View answer and explanation
Correct option: D
a=3,r=14a=3, r=\frac{1}{4}
Option D (a=3,r=14a=3, r=\frac{1}{4}) is correct because it satisfies both the sum to infinity formula and the given second term condition for the geometric series.
550 Easy
550 of 998
Analyzing +4 | -1
If 5log⁡abc(a3+b3+c3)=3λ(1+log⁡3(abc)log⁡3(abc))5 \log _{a b c}\left(a^3+b^3+c^3\right)=3 \lambda\left(\frac{1+\log _3(a b c)}{\log _3(a b c)}\right) and (abc)a+b+c=1(a b c)^{a+b+c}=1 and λ=mn\lambda=\frac{m}{n}, where mm and nn are relative primes, the value of ∣m+n∣+∣m−n∣|m+n|+|m-n| is
View answer and explanation
Correct option: B
10
Option B (10) is correct because after simplifying the given equation and using the condition (abc)(a+b+c) = 1, we find m=5 and n=3, leading to |m+n|+|m-n| = 10.
551 Easy
551 of 998
Application +4 | -1
If S is the sum to infinity of a G.P., whose first term is a,a, then the sum of the first nn terms is
View answer and explanation
Correct option: B
S{1−(1−as)n}S\left\{1-\left(1-\frac{a}{s}\right)^n\right\}
Option B (S{1−(1−as)n}S\{1-(1-\frac{a}{s})^n\}) is correct because it correctly expresses the sum of the first n terms of the G.P. in terms of S, a, and n.
552 Easy
552 of 998
Application +4 | -1
For a sequence <an>,a1=2< {a_n} > ,{a_1} = 2 and an+1an=13.\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{1}{3}. Then, ∑r=120ar\sum \limits_{r = 1}^{20} {a_r} is
View answer and explanation
Correct option: B
3(1−1320)3\left( {1 - \frac{1}{{{3^{20}}}}} \right)
Option B (3(1−1320)3\left( {1 - \frac{1}{{{3^{20}}}}} \right)) is correct because it is the sum of the first 20 terms of the geometric progression.
553 Easy
553 of 998
Application +4 | -1
Every term of a G.P. is positive and also every term is the sum of two preceding terms. Then, the common ratio of the G.P. is
View answer and explanation
Correct option: B
5+12\frac{\sqrt{5}+1}{2}
Option B (5+12\frac{{\sqrt{{5}}+1}}{{2}}) is correct because it is the positive root of the quadratic equation derived from the given condition.
554 Easy
554 of 998
Analyzing +4 | -1
If a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n are in H.P. then a1a2+a3+…+an,a2a1+a3+…+an,a3a1+a2+a4+…+an\frac{a_1}{a_2+a_3+\ldots+a_n}, \frac{a_2}{a_1+a_3+\ldots+a_n}, \frac{a_3}{a_1+a_2+a_4+\ldots+a_n}, …,ana1+a2+…+an−1\ldots, \frac{a_n}{a_1+a_2+\ldots+a_{n-1}} are in
View answer and explanation
Correct option: C
H.P.
Option A (A.P.) is incorrect. Option B (G.P.) is incorrect. Option C (H.P.) is correct because if a1, a2, ..., an are in HP, then the given expression is in HP.
555 Easy
555 of 998
Application +4 | -1
The GM of roots of the equation x2−18x+9=0{x^2} - 18x + 9 = 0 is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the geometric mean (GM) of the roots of the quadratic equation is the square root of the product of the roots, which equals 3.
556 Easy
556 of 998
Application +4 | -1
Let x∈(1,∞)x \in \left( {1,\infty } \right) and nn be a positive integer greater than 1. If fn(x)=n1log⁡2x+1log⁡3x+…+1log⁡nx{f_n}\left( x \right) = \frac{n}{{\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_3}x}} + \ldots + \frac{1}{{{{\log }_n}x}}}}, then (n!)fn(x){\left( {n!} \right)^{{f_n}\left( x \right)}} equals to
View answer and explanation
Correct option: B
xn{x^n}
Option B (xn{x^n}) is correct because it is the simplified form of the expression after applying logarithm properties.
557 Easy
557 of 998
Analyzing +4 | -1
If a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n are in A.P., where ai>0a_i>0 for all ii, then the 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}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}, is
View answer and explanation
Correct option: D
n−1a1−an\frac{n-1}{\sqrt{a_1}-\sqrt{a_n}}
Option D (n−1a1−an\frac{n-1}{\sqrt{a_1}-\sqrt{a_n}}) is correct because it simplifies the sum of the given series by rationalizing the denominators and using the properties of an arithmetic progression.
558 Easy
558 of 998
Application +4 | -1
The sum of nn terms of the following series 1+(1+x)+(1+x+x2)+…1 + \left( {1 + x} \right) + \left( {1 + x + {x^2}} \right) + \ldots is
View answer and explanation
Correct option: C
n(1−x)−x(1−xn)(1−x)2\frac{{n\left( {1 - x} \right) - x\left( {1 - {x^n}} \right)}}{{{{\left( {1 - x} \right)}^2}}}
Option C (n(1−x)−x(1−xn)(1−x)2\frac{{n\left( {1 - x} \right) - x\left( {1 - {x^n}} \right)}}{{{{\left( {1 - x} \right)}^2}}}) is correct because it is derived by summing the series and using the formula for the sum of a geometric progression.
559 Easy
559 of 998
Analyzing +4 | -1
The sum of the series log⁡42−log⁡82+log⁡162−…,{\log _4}2 - {\log _8}2 + {\log _{16}}2 - \ldots , is
View answer and explanation
Correct option: D
1−log⁡e21 - {\log _e}2
Option D (1−log⁡e21 - {\log _e}2) is correct because the series simplifies to 1−log⁡e21 - \log_e 2 using logarithm properties and the Maclaurin series for log⁡(1+x)\log(1+x).
560 Easy
560 of 998
Analyzing +4 | -1
If a+b1−ab,b,b+c1−bc\frac{{a + b}}{{1 - ab}},b,\frac{{b + c}}{{1 - bc}} are in A.P., then a,1b,ca,\frac{1}{b},c are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the given condition leads to the relationship 2b=1a+1c2b = \frac{1}{a} + \frac{1}{c}, which defines a harmonic progression for a,1b,ca, \frac{1}{b}, c.
561 Easy
561 of 998
Application +4 | -1
If 3,A\sqrt{3}, A and 2\sqrt{2} are in APA P, then 3+22\frac{\sqrt{3}+\sqrt{2}}{2} is greater than or equal to
View answer and explanation
Correct option: B
6\sqrt{6}
Option B (6\sqrt{6}) is correct because, since 3,A,2\sqrt{3}, A, \sqrt{2} are in AP, A is the AM of 3\sqrt{3} and 2\sqrt{2}, and AM >= GM.
562 Easy
562 of 998
Application +4 | -1
In an increasing A.P., the sum of three consecutive terms is 51 and the product of first and third terms is 273. Its third term is
View answer and explanation
Correct option: D
21
Option D (21) is correct because by setting up equations based on the properties of an arithmetic progression and solving for the terms, the third term is found to be 21.
563 Easy
563 of 998
Analyzing +4 | -1
If f(n)=∑r=1nr4f(n)=\sum\limits_{{r=1}}^n r^4, then the value of ∑r=1nr(n−r)3\sum\limits_{{r=1}}^n r(n-r)^3 is equal to
View answer and explanation
Correct option: B
14{n3(n+1)2−4f(n)}\frac{1}{4}\left\{n^3(n+1)^2-4 f(n)\right\}
Option B is correct because by using the given function and properties of summation, the expression can be simplified to 14{n3(n+1)2−4f(n)}\frac{{1}}{{4}}\{n^3(n+1)^2-4 f(n)\}.
564 Easy
564 of 998
Analyzing +4 | -1
The sum of the series 1+13.14+15.142+17.143+…∞1 + \frac{1}{3}.\frac{1}{4} + \frac{1}{5}.\frac{1}{{{4^2}}} + \frac{1}{7}.\frac{1}{{{4^3}}} + \ldots \infty is
View answer and explanation
Correct option: C
log⁡e3{\log _e}3
Option C (log⁡e3{\log _e}3) is correct because the given series can be related to the logarithmic series expansion after a suitable substitution.
565 Easy
565 of 998
Application +4 | -1
The third term of an AP is 18 and the seventh term is 30 , then the sum of the first 17 terms is
View answer and explanation
Correct option: C
612
Option C (612) is correct because it is the sum of the first 17 terms of the arithmetic progression, calculated by first finding the first term and common difference.
566 Easy
566 of 998
Application +4 | -1
If a,b,ca,b,c are in H.P., then ab+c,bc+a,ca+b\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}} will be in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because if a,b,ca, b, c are in H.P., then ab+c,bc+a,ca+b\frac{a}{{b + c}}, \frac{b}{{c + a}}, \frac{c}{{a + b}} are also in H.P.
567 Easy
567 of 998
Analyzing +4 | -1
If α, β\alpha ,\,\beta are natural numbers, such that 100α−199β=(100)(100)+(99)(101)+(98)(102)+....+(1)(199){100^\alpha } - 199\beta = (100)(100) + (99)(101) + (98)(102) + .... + (1)(199), then the slope of the line passing through (α, β)(\alpha ,\,\beta ) and origin is
View answer and explanation
Correct option: B
550
Option B (550) is correct. The slope is calculated by simplifying the given equation and then finding the values of α\alpha and β\beta.
568 Easy
568 of 998
Application +4 | -1
If the sum to 2n2n terms of the AP 2, 5, 8, 11,… is equal to the sum to n terms of the AP 57, 59, 61, 63,…, then nn is equal to
View answer and explanation
Correct option: B
11
Option B (11) is correct because equating the sum of 2n terms of the first AP to the sum of n terms of the second AP and solving for n yields n=11.
569 Easy
569 of 998
Application +4 | -1
If a,b,ca, b, c are in H.P., then ab+c−a,bc+a−b,ca+b−c\frac{a}{b+c-a}, \frac{b}{c+a-b}, \frac{c}{a+b-c} are in
View answer and explanation
Correct option: C
H.P.
Option C is correct because if a, b, c are in H.P., then the given expression is also in H.P.
570 Easy
570 of 998
Application +4 | -1
If x=1+a+a2+...∞x = 1 + a + {a^2} + ...\infty and y=1+b+b2+...∞y = 1 + b + {b^2} + ...\infty where aa and bb are proper fractions, then 1+ab+a2b2+...∞1 + ab + {a^2}{b^2} + ...\infty equals
View answer and explanation
Correct option: A
xyy+x−1\frac{{xy}}{{y + x - 1}}
Option A (xyy+x−1\frac{{xy}}{{y + x - 1}}) is correct because it is derived by expressing aa and bb in terms of xx and yy, and then substituting them into the infinite geometric series formula.
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