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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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721 Easy
721 of 998
Application +4 | -1
If a1,a2,a3,a4,a5a_1, a_2, a_3, a_4, a_5 are in HP\mathrm{HP}, then a1a2+a2a3+a3a4+a4a5a_1 a_2+a_2 a_3+a_3 a_4+a_4 a_5 is equal to
View answer and explanation
Correct option: C
4a1a54 a_1 a_5
Option A (n2n^2) is correct because it represents the sum to infinity of the given series, derived using the formula for the sum of an infinite geometric progression.
722 Easy
722 of 998
Analyzing +4 | -1
Let Sn=1.(n−1)+2.(n−2)+3.(n−3)+.....+(n−1).1,n≥4.{S_n} = 1.(n - 1) + 2.(n - 2) + 3.(n - 3) + ..... + (n - 1).1,n \ge 4.
The sum ∑n=4∞(2Snn!−1(n−2)!)\sum\limits_{n = 4}^\infty {\left( {\frac{{2{S_n}}}{{n!}} - \frac{1}{{(n - 2)!}}} \right)} is equal to
View answer and explanation
Correct option: A
e−13\frac{{e - 1}}{3}
Option A (e−13\frac{{e - 1}}{3}) is correct because it is the result of evaluating the given infinite sum by simplifying the expression and using the series expansion of e.
723 Easy
723 of 998
Analyzing +4 | -1
If GG is the GM of the product of rr set of observation with geometric means G1,  G2,…Gr{G_1},\;{G_2}, \ldots {G_r} respectively, then GG is equal to
View answer and explanation
Correct option: B
G1,G2,….,Gr{G_1},{G_2}, \ldots .,{G_r}
Option D (G1,G2,….,Gr{G_1},{G_2}, \ldots .,{G_r}) is incorrect. Option A (log⁡G1+log⁡G2+...+log⁡Gn\log {G_1} + \log {G_2} + ... + \log {G_n}) is incorrect. Option C (log⁡G1,log⁡G2,….,log⁡Gn\log {G_1},\log {G_2}, \ldots .,\log {G_n}) is incorrect. Option B (G1,G2,….,Gr{G_1},{G_2}, \ldots .,{G_r}) is correct because the geometric mean of the product of rr sets of observations is the product of the individual geometric means.
724 Easy
724 of 998
Application +4 | -1
If a,b,ca,b,c are in H.P., then 1b−a+1b−c=\frac{1}{{b - a}} + \frac{1}{{b - c}} =
View answer and explanation
Correct option: B
1a+1c\frac{1}{a} + \frac{1}{c}
Option B, 1a+1c\frac{1}{a} + \frac{1}{c}, is correct because it represents the simplified form of the expression given that a, b, and c are in harmonic progression.
725 Easy
725 of 998
Application +4 | -1
The sum of first 26 terms of an A.P. a1,a2,a3,…a_1, a_2, a_3, \ldots if a2+a6+a9+a18+a21+a25=165a_2+a_6+a_9+a_{18}+a_{21}+a_{25}=165, is
View answer and explanation
Correct option: B
715
Option B (715) is correct because the sum of the first 26 terms of an A.P. can be found using the given relationship between the terms.
726 Easy
726 of 998
Application +4 | -1
If sum of four numbers in A.P. is 28 and product of two middle terms is 45, then product of the first and last terms is
View answer and explanation
Correct option: B
13
Option B (13) is correct because by setting up equations based on the given sum and product, we can solve for the product of the first and last terms.
727 Easy
727 of 998
Application +4 | -1
If log⁡4,log⁡(2x+2),log⁡(2x+2+1)\log 4, \log \left(2^x+2\right), \log \left(2^{x+2}+1\right) are 3 consecutive terms of an arithmetic progression then xx lies between 2 consecutive natural numbers pp and q(p<q)q(p < q) find the value of (p+q)(p+q).
View answer and explanation
Correct option: A
7
The correct answer is A (7) because solving the arithmetic progression condition leads to 2x=122^x = 12, which implies xx lies between 3 and 4, hence p+q=7p+q = 7.
728 Easy
728 of 998
Analyzing +4 | -1
If log⁡yx=log⁡zy=log⁡xz,{\log _y}x = {\log _z}y = {\log _x}z, then
View answer and explanation
Correct option: D
x=y=zx = y = z
Option D (x=y=zx = y = z) is correct because the given logarithmic condition implies that x, y, and z are equal.
729 Easy
729 of 998
Analyzing +4 | -1
The sixth term of an AP is equal to 2. The value of the common difference of the AP which makes the product a1a4a5a_1 a_4 a_5 least, is given by
View answer and explanation
Correct option: C
23\frac{2}{3}
Option C (23\frac{2}{3}) is correct because it is the common difference that minimizes the product a1a4a5a_1 a_4 a_5 by satisfying the condition for a minimum in the derivative.
730 Easy
730 of 998
Understanding +4 | -1
If arithmetic mean of two positive numbers is AA, their geometric mean is GG and harmonic mean is HH, then HH is equal to
View answer and explanation
Correct option: A
G2/A{G^2}/A
Option A is correct because the harmonic mean (H) can be expressed as the square of the geometric mean (G) divided by the arithmetic mean (A).
731 Easy
731 of 998
Application +4 | -1
If the first and the nth {n}^{\text {th }} term of aa G.P. are aa and bb respectively and PP is the product of the first nn terms, then P2=P^2=
View answer and explanation
Correct option: B
(ab)n(a b)^n
Option B ((ab)n(ab)^n) is correct because it is the square of the product of the first n terms of the G.P.
732 Easy
732 of 998
Application +4 | -1
If the first, second and last terms of an arithmetic series are a,ba,b and ccrespectively, then the number of terms is
View answer and explanation
Correct option: A
b+c−2ab−a\frac{{b + c - 2a}}{{b - a}}
Option A (b+c−2ab−a\frac{{b + c - 2a}}{{b - a}}) is correct because, using the formula for the last term of an arithmetic series and solving for the number of terms (n), we get this expression.
733 Easy
733 of 998
Application +4 | -1
Let the sum of the first nn terms of a non-constant AP a1, a2, a3....  {a_1},\,{a_2},\,{a_3}....\,\, be 50n+n(n−7)2A, 50n + \frac{{n(n - 7)}}{2}A,\, where AA is a constant. If dd is the common difference of this AP, then the ordered pair (d, a50)(d,\,{a_{50}}) is equal to
View answer and explanation
Correct option: A
(A, 50+46A)(A,\,50 + 46A)
Option A ((A, 50+46A)(A,\,50 + 46A)) is correct because after calculating the common difference and the 50th term, the ordered pair matches this option.
734 Easy
734 of 998
Application +4 | -1
The nnth term of an AP\mathrm{AP} is 4n+14 n+1. The sum of the first 15 terms is
View answer and explanation
Correct option: A
495
Option A (495) is correct because it is the sum of the first 15 terms of the arithmetic progression, where the nth term is given by 4n+1.
735 Easy
735 of 998
Analyzing +4 | -1
The value of 3+log⁡log⁡3432+12log⁡log⁡(494)+13log⁡log⁡(1125)\frac{{3 + \log \log 343}}{{2 + \frac{1}{2}\log \log \left( {\frac{{49}}{4}} \right) + \frac{1}{3}\log \log \left( {\frac{1}{{125}}} \right)}} is
View answer and explanation
Correct option: A
33
Answer A (3) is correct because by simplifying the logarithmic expression using logarithm properties, the value of the given expression is equal to 3.
736 Easy
736 of 998
Analyzing +4 | -1
If sum of xx terms of a series is Sx=1(2x+3)(2x+1)S_x=\frac{1}{(2 x+3)(2 x+1)} whose rth r^{\text {th }} term is TrT_r. Then, ∑r=1n1Tr\sum\limits_{{r=1}}^n \frac{1}{T_r} is equal to
View answer and explanation
Correct option: B
−14∑(2r+1)(2r−1)(2r+3)-\frac{1}{4} \sum(2 r+1)(2 r-1)(2 r+3)
Option B is correct because by finding the rth term of the series and then taking the reciprocal, the summation can be expressed as the negative of 1/4 times the sum of the product of the given terms.
737 Easy
737 of 998
Application +4 | -1
The coefficient of x3{x^3} in the expansion of 3x{3^x} is
View answer and explanation
Correct option: D
(log⁡3)36\frac{{{{\left( {\log 3} \right)}^3}}}{6}
Option D ((log⁡3)36\frac{{{{\left( {\log 3} \right)}^3}}}{6}) is correct because it is the coefficient of x3{x^3} in the Maclaurin series expansion of 3x{3^x}.
738 Easy
738 of 998
Application +4 | -1
In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of AP be equal to the middle term of the GP, then the middle term of the entire sequence is
View answer and explanation
Correct option: A
−1031- \frac{{10}}{{31}}
Option A (-10/31) is correct because it is the middle term calculated using the properties of AP and GP and the given conditions.
739 Easy
739 of 998
Application +4 | -1
If every even term of a series is a times the term before it and every odd term is cc times the term before it, the first term being unity, then the sum to 2n2 n terms is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the sum of the given series to 2n terms, calculated based on the provided conditions, does not match any of the given options.
740 Easy
740 of 998
Application +4 | -1
Let ana_n be the nth term an A.P. If ∑r=1100a2r=α\sum\limits_{r=1}^{100} a_{2 r}=\alpha and ∑r=1100a2r−1=β\sum\limits_{r=1}^{100} a_{2 r-1}=\beta, then the common difference of the A.P., is
View answer and explanation
Correct option: A
α−β100\frac{\alpha-\beta}{100}
Option A (α−β100\frac{\alpha-\beta}{100}) is correct because it determines the common difference of the A.P. based on the given sums of even and odd terms.
741 Easy
741 of 998
Analyzing +4 | -1
If 1,5,25,125,625,…1,5,25,125,625, \ldots forms aa GPG P, then the sequence consisting odd term of the given GPG P, is
View answer and explanation
Correct option: B
GPG P
Option B (GP) is correct because the sequence of odd terms (1, 25, 625, ...) maintains a constant ratio between consecutive terms, which is the defining characteristic of a geometric progression.
742 Easy
742 of 998
Application +4 | -1
The nnth term of the sequence 212,1713,119,2023…2 \frac{1}{2}, 1 \frac{7}{13}, 1 \frac{1}{9}, \frac{20}{23} \ldots is
View answer and explanation
Correct option: C
205n+3\frac{20}{5 n+3}
Option C is correct because the nth term of the sequence can be found by taking the reciprocal of the terms, recognizing an A.P., and then finding the reciprocal of the nth term of that A.P.
743 Easy
743 of 998
Application +4 | -1
If the sum of the first 2 n terms of the A.P. 2,5,8, is equal to the sum of the first nn terms of the A.P. 57, 59, 61,...., then nn equals
View answer and explanation
Correct option: C
11
Option C (11) is correct because it is the value of n that equates the sum of the first 2n terms of the first A.P. to the sum of the first n terms of the second A.P.
744 Easy
744 of 998
Analyzing +4 | -1
The value of [(0.16)log⁡0.25(13+132+133+...∞)]1/2{\left[ {{{\left( {0.16} \right)}^{{{\log }_{0.25}}\left( {\frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ...\infty } \right)}}} \right]^{1/2}} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the simplified value is 210\frac{2}{{\sqrt {10} }}, which is not among the given options.
745 Easy
745 of 998
Application +4 | -1
Let a1,a2,a3,…,an,…….a_1, a_2, a_3, \ldots, a_n, \ldots \ldots .. be a G.P such that a4a6=14\frac{a_4}{a_6}=\frac{1}{4} and a2+a5=216a_2+a_5=216. Then, a1=a_1=
View answer and explanation
Correct option: A
12 or, 1087\frac{108}{7}
Option A (12 or, 108/7) is correct because solving the equations for the given G.P. yields two possible values for the first term.
746 Easy
746 of 998
Application +4 | -1
The value of 1−log⁡2+(log⁡2)22!−(log⁡2)33!+…1 - \log 2 + \frac{{{{\left( {\log 2} \right)}^2}}}{{2!}} - \frac{{{{\left( {\log 2} \right)}^3}}}{{3!}} + \ldots is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the given series is the expansion of e−log2e^{-log2} which simplifies to 12\frac{1}{2}.
747 Easy
747 of 998
Application +4 | -1
Three distinct numbers a,b,ca, b, c form a GP in that order and the numbers a+b,b+c,c+aa+b, b+c, c+a form an AP in that order. Then the common ratio of the GP is
View answer and explanation
Correct option: C
-2
Option C (-2) is correct because solving the equations for the GP and AP yields a common ratio of -2.
748 Easy
748 of 998
Application +4 | -1
The geometric mean of the numbers 3,32,33,…,3n3,3^2, 3^3, \ldots, 3^n is
View answer and explanation
Correct option: D
3(n+1)/23^{(n+1) / 2}
Option D (3(n+1)/23^{(n+1) / 2}) is correct because it is the result of calculating the geometric mean of a geometric progression.
749 Easy
749 of 998
Application +4 | -1
The number of terms of the A.P. 1,4,7,…1, 4, 7, \ldots that must be taken to obtain a sum of 715 is
View answer and explanation
Correct option: C
22
Option C (22) is correct because by using the arithmetic progression sum formula, we can find the number of terms needed to obtain a sum of 715.
750 Easy
750 of 998
Application +4 | -1
The value of 7 log⁡1615+5log⁡2524+3log⁡8180,\log \frac{{16}}{{15}} + 5\log \frac{{25}}{{24}} + 3\log \frac{{81}}{{80}},, is
View answer and explanation
Correct option: A
log⁡2\log 2
Option A log⁡2\log 2 is correct because it is obtained by simplifying the logarithmic expression using logarithmic properties.
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