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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 391-420 of 564 questions

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391 Medium
391 of 564
Application +4 | -1
Suppose a,b,ca, b, c are distinct positive real numbers such that a,2b,3ca, 2 b, 3 c are in A.P. and a,b,ca, b, c are in G.P. The common ratio of G.P. is
View answer and explanation
Correct option: C
1/31 / 3
Option C (1/3) is correct because by using the conditions for arithmetic and geometric progressions, we can derive a relationship between a and c, leading to the common ratio of 1/3.
392 Medium
392 of 564
Application +4 | -1
If one geometric mean GG and two arithmetic means pp and qq are inserted between two numbers, then G2G^2 is equal to
View answer and explanation
Correct option: B
(2p−q)(2q−p)(2 p-q)(2 q-p)
Option B ((2p-q)(2q-p)) is correct because it represents the square of the geometric mean in terms of the arithmetic means.
393 Medium
393 of 564
Application +4 | -1
The four numbers are in AP, such that their sum is 50 and greatest of them is 4 times the least, are
View answer and explanation
Correct option: C
5,10,15,205,10,15,20
Option C (5,10,15,205,10,15,20) is correct because these four numbers are in AP, their sum is 50, and the greatest is 4 times the least.
394 Medium
394 of 564
Application +4 | -1
If 11 A.M.s are inserted between 28 and 10 , then the middle term in the series is
View answer and explanation
Correct option: B
19
Option B (19) is correct because it is the middle term when 11 arithmetic means are inserted between 28 and 10.
395 Medium
395 of 564
Application +4 | -1
3⋅537‾=3 \cdot \overline{537}=
View answer and explanation
Correct option: C
1178333\frac{1178}{333}
Option C (1178333\frac{{1178}}{{333}}) is correct because it is the result of converting the repeating decimal 3.537(repeating) into a fraction.
396 Medium
396 of 564
Application +4 | -1
Three positive numbers form an increasing GP. If the middle terms in this GP is doubled, the new numbers are in AP. Then, the common ratio of the GP is
View answer and explanation
Correct option: B
2+32+\sqrt{3}
Option B (2+32+\sqrt{3}) is correct because it is the common ratio of the increasing GP, derived from the condition that doubling the middle term results in an AP.
397 Medium
397 of 564
Application +4 | -1
The common difference of an A.P. whose first term is unity and whose second, tenth and thirty fourth terms are in G.P., is
View answer and explanation
Correct option: B
13\frac{1}{3}
The correct answer is B (13\frac{1}{3}) because this common difference makes the second, tenth, and thirty-fourth terms of the A.P. form a G.P.
398 Medium
398 of 564
Analyzing +4 | -1
If the sum of nn terms of the series

1+45+752+1053+…1 + \frac{4}{5} + \frac{7}{{{5^2}}} + \frac{{10}}{{{5^3}}} + \ldots is l+1516(1−15n−1)−(3n−2)4(5n−1)l + \frac{{15}}{{16}}\left( {1 - \frac{1}{{{5^{n - 1}}}}} \right) - \frac{{\left( {3n - 2} \right)}}{{4\left( {{5^{n - 1}}} \right)}}, then ll is
View answer and explanation
Correct option: B
54\frac{5}{4}
Option B (54\frac{5}{4}) is correct because it is the constant term obtained after applying the formula for the sum of an arithmetico-geometric series and comparing coefficients.
399 Medium
399 of 564
Analyzing +4 | -1
The sum of the series 8!+9!2!+10!3!+11!4!+…8 !+\frac{9 !}{2 !}+\frac{10 !}{3 !}+\frac{11 !}{4 !}+\ldots. up to 20 terms
View answer and explanation
Correct option: C
7!(28C20−1)7 !\left({ }^{28} C_{20}-1\right)
Option C (7!(28C20−1)7 !\left({ }^{28} C_{20}-1\right)) is correct because the sum of the series can be simplified using combinatorial identities and adding/subtracting a constant.
400 Medium
400 of 564
Application +4 | -1
If a1,a2,a3,…,a9a_1, a_2, a_3, \ldots, a_9 are in G.P., then ∣log⁡a1log⁡a2log⁡a3log⁡a4log⁡a5log⁡a6log⁡a7log⁡a8log⁡a9∣=\left|\begin{array}{lll}\log a_1 & \log a_2 & \log a_3 \\ \log a_4 & \log a_5 & \log a_6 \\ \log a_7 & \log a_8 & \log a_9\end{array}\right|=
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant of the matrix is zero due to the properties of logarithms and arithmetic progressions in a G.P.
401 Medium
401 of 564
Application +4 | -1
The maximum sum of an AP 40,38,36,34,…40,38,36,34, \ldots is
View answer and explanation
Correct option: B
420
Option B (420) is correct because it represents the maximum sum of the arithmetic progression before the terms become negative.
402 Medium
402 of 564
Analyzing +4 | -1
If log⁡xa2+ab+b2=logyb2+bc+c2=log⁡zc2+ca+a2,\frac{{\log x}}{{{a^2} + ab + {b^2}}} = \frac{{\mathrm{log}y}}{{{b^2} + bc + {c^2}}} = \frac{{\log z}}{{{c^2} + ca + {a^2}}}, then xa−b.yb−c.zc−a={x^{a - b}}.{y^{b - c}}.{z^{c - a}} =
View answer and explanation
Correct option: C
1
Option C (1) is correct because when the given ratios are manipulated and combined using logarithm properties, the result simplifies to 1.
403 Medium
403 of 564
Application +4 | -1
a,b,c,da, b, c, d are in GP and are in ascending order such that a+d=112a+d=112 and b+c=48b+c=48. If the GP is continued with aa as the first term, then the sum of the first six terms is
View answer and explanation
Correct option: D
1456
Option D (1456) is correct because solving the equations for the GP yields the sum of the first six terms as 1456.
404 Medium
404 of 564
Analyzing +4 | -1
{an}  and  {bn}\left\{ {{a_n}} \right\}\;and\;\left\{ {{b_n}} \right\} be two sequences given by an=(x)12n+(y)12n{a_n} = {\left( x \right)^{\frac{1}{{{2^n}}}}} + {\left( y \right)^{\frac{1}{{{2^n}}}}} and bn=(x)12n−(y)12n{b_n} = {\left( x \right)^{\frac{1}{{{2^n}}}}} - {\left( y \right)^{\frac{1}{{{2^n}}}}} for all n∈N,n \in N, then a1a2a3…an{a_1}{a_2}{a_3} \ldots {a_n} is equal to
View answer and explanation
Correct option: C
x−ybn\frac{{x - y}}{{{b_n}}}
Option C (x−ybn\frac{{x - y}}{{{b_n}}}) is correct because the product telescopes, simplifying to x−ybn\frac{x-y}{b_n} using the difference of squares repeatedly.
405 Medium
405 of 564
Application +4 | -1
Sum of the series 11+12+14+21+22+24+31+32+34+…\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots up to nn terms is equal to
View answer and explanation
Correct option: B
n2+n2(n2+n+1)\frac{n^2+n}{2\left(n^2+n+1\right)}
Option B (n2+n2(n2+n+1)\frac{n^2+n}{2(n^2+n+1)}) is correct because by finding the general term and using telescoping series, the sum simplifies to the given expression.
406 Medium
406 of 564
Application +4 | -1
If 1,log⁡9(31−x+2),log⁡3(4⋅3x−1)1, \log _9\left(3^{1-x}+2\right), \log _3\left(4 \cdot 3^x-1\right) are in A.P., then x=x=
View answer and explanation
Correct option: D
1−log⁡341-\log _3 4
Option D (1−log⁡341-\log _3 4) is correct because it is the solution to the equation derived from the arithmetic progression condition.
407 Medium
407 of 564
Application +4 | -1
If x=1+y+y2+…x=1+y+y^2+\ldots, then yy is
View answer and explanation
Correct option: C
x−1x\frac{x-1}{x}
Option C (x−1x\frac{x-1}{x}) is correct because the given equation represents an infinite geometric series which can be solved for y.
408 Medium
408 of 564
Analyzing +4 | -1
In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49 , and the sum of the first and the third term is 35 . Then the first term of this geometric progression is:
View answer and explanation
Correct option: C
28
Option C (28) is correct because by using the given conditions to form equations and solving them, we find the first term of the G.P. to be 28.
409 Medium
409 of 564
Application +4 | -1
If a1,a2,…,an+1a_{1}, a_{2}, \ldots, a_{n+1} are in AP, then 1a1a2+1a2a3+⋯+1anan+1\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\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 the given series telescopes, simplifying to this expression.
410 Medium
410 of 564
Understanding +4 | -1
If three numbers be in G.P., then their logarithms will be in
View answer and explanation
Correct option: A
A.P.
Option B (nn+1\frac{n}{n+1}) is correct because the series can be expressed as a telescoping sum, where each term is of the form 1k(k+1)=1k−1k+1\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}, leading to the result nn+1\frac{n}{n+1}.
411 Medium
411 of 564
Application +4 | -1
If log⁡aab=x,{\log _a}ab = x, then the value of log⁡bab{\log _b}ab is
View answer and explanation
Correct option: B
xx−1\frac{x}{{x - 1}}
Option B (xx−1\frac{x}{{x - 1}}) is correct because it is the correct expression for log⁡bab{\log _b}ab in terms of xx, derived from the given equation log⁡aab=x{\log _a}ab = x.
412 Medium
412 of 564
Application +4 | -1
If the interior angles of a convex polygon are in A.P. with the smallest angle measuring 120 (in degrees) and the common difference between measures of angles being 5, then the polygon will have _____ sides.
View answer and explanation
Correct option: D
9
Option D (9) is correct because when n=9, the angles form a valid A.P. within the constraints of a convex polygon, while n=16 results in an angle exceeding 180 degrees.
413 Medium
413 of 564
Analyzing +4 | -1
If (10)9+2(11)1(108)+3(11)2(10)7+…(10)^{9}+2(11)^{1}\left(10^{8}\right)+3(11)^{2}(10)^{7}+\ldots.
+10(11)9=k(10)9+10(11)^{9}=k(10)^{9}, then kk is equal to:
View answer and explanation
Correct option: A
100
Option A (100) is correct because by manipulating the series and subtracting, we can find the value of k.
414 Medium
414 of 564
Application +4 | -1
19 If xx be the AMA M and y,zy, z be two GM's between two positive numbers, then y3+z3xyz\frac{y^3+z^3}{x y z} is equal to-
View answer and explanation
Correct option: B
2
Option B (2) is correct because given x is the A.M. and y, z are two G.M.'s between two positive numbers, the expression simplifies to 2.
415 Medium
415 of 564
Application +4 | -1
If a,ba, b and cc are the sides of a △ABC\triangle A B C are in AP, then cot⁡C2\cot \frac{C}{2} is equal to
View answer and explanation
Correct option: A
3tan⁡A23 \tan \frac{A}{2}
Option A (3tan⁡A23 \tan \frac{A}{2}) is correct because, given that a, b, and c are in AP, we can express cot⁡C2\cot \frac{C}{2} in terms of tan⁡A2\tan \frac{A}{2} using trigonometric identities.
416 Medium
416 of 564
Application +4 | -1
If Sn=∑r=1n1+2+22+… upto r terms 2rS_n=\sum\limits_{{r=1}}^n \frac{1+2+2^2+\ldots \text { upto } r \text { terms }}{2^r}, then SnS_n is equal to
View answer and explanation
Correct option: C
n−1+12nn-1+\frac{1}{2^n}
Option C (n−1+12nn-1+\frac{1}{2^n}) is correct because the sum of the geometric series is calculated and simplified correctly.
417 Medium
417 of 564
Application +4 | -1
nthn^{\text {th}} term of the series 2+4+7+11+…2+4+7+11+\ldots will be
View answer and explanation
Correct option: C
n2+n+22\frac{n^2+n+2}{2}
Option C is correct because the nth term is derived by finding the pattern in the differences between consecutive terms and summing the resulting series.
418 Medium
418 of 564
Application +4 | -1
If the ratio of AM between two positive real numbers aa and bb to their HM is m:n,m:n, then a:ba:b is equal to
View answer and explanation
Correct option: C
m+m−nm−m−n\frac{{\sqrt m + \sqrt {m - n} }}{{\sqrt m - \sqrt {m - n} }}
Option C (m+m−nm−m−n\frac{{\sqrt m + \sqrt {m - n} }}{{\sqrt m - \sqrt {m - n} }}) is correct because it is derived by solving for the two numbers a and b using the given ratio of AM to HM.
419 Medium
419 of 564
Application +4 | -1
If the sum of nn terms of a G.P. is S{S}, their product is pp and the sum of their reciprocals is R,R, then p2=p^2=
View answer and explanation
Correct option: D
(SR)n\left(\frac{S}{R}\right)^n
Option D ((SR)n\left(\frac{{S}}{{R}}\right)^n) is correct because it is derived by relating the sum, product, and sum of reciprocals of a G.P.
420 Medium
420 of 564
Analyzing +4 | -1
If a=log⁡1218a=\log _{12} 18 and b=log⁡2454b=\log _{24} 54, then the value of (a+b)2+a(10−a)−b(10+b)(a+b)^2+a(10-a)-b(10+b) is equal to
View answer and explanation
Correct option: C
2
The correct answer is C (2). The expression simplifies to a2+b2+2ab+10a−a2−10b−b2=2a^2 + b^2 + 2ab + 10a - a^2 - 10b - b^2 = 2 after substituting the given values of aa and bb and simplifying the logarithmic expressions.
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