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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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301 Medium
301 of 564
Analyzing +4 | -1
If Sn=∑r=1ntr=16n(2n2+9n+13)S_n=\sum\limits_{{r=1}}^n t_r=\frac{1}{6} n\left(2 n^2+9 n+13\right), then ∑r=1ntr\sum\limits_{{r=1}}^n \sqrt{t_r} is equal to
View answer and explanation
Correct option: C
12n(n+3)\frac{1}{2} n(n+3)
Option C (12n(n+3)\frac{1}{2} n(n+3)) is correct because after finding the general term trt_r and summing the square root of the terms, the result simplifies to 12n(n+3)\frac{1}{2} n(n+3).
302 Medium
302 of 564
Analyzing +4 | -1
The sum of series 11.2.3+13.4.5+15.6.7+…∞\frac{1}{{1.2.3}} + \frac{1}{{3.4.5}} + \frac{1}{{5.6.7}} + \ldots \infty is equal to
View answer and explanation
Correct option: A
log⁡e2−12{\log _e}2 - \frac{1}{2}
Option A (log⁡e2−12{\log _e}2 - \frac{1}{2}) is the correct answer because it is the result of summing the given infinite series using partial fraction decomposition and logarithmic series expansion.
303 Medium
303 of 564
Application +4 | -1
The sum to infinity of the following series 11.2+12.3+13.4+…\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots. shall be-
View answer and explanation
Correct option: B
1
Option B (1) is correct because the series is a telescoping series and its sum converges to 1.
304 Medium
304 of 564
Application +4 | -1
The sum of all two digit numbers which, when divided by 4, yield unity as a remainder is
View answer and explanation
Correct option: C
1210
Option C (1210) is correct because the sum of the arithmetic progression 13, 17, ..., 97 is calculated correctly using the formula for the sum of an AP.
305 Medium
305 of 564
Application +4 | -1
If the equation (x+1)+(x+4)+(x+7)+…+(x+34)=216(x+1)+(x+4)+(x+7)+\ldots+(x+34)=216 represents the sum of the first nn terms of an A.P., then x=x=
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{{1}}{{2}}) is correct because it is the value of x that satisfies the given equation, representing the sum of an arithmetic progression.
306 Medium
306 of 564
Application +4 | -1
The sum of all two digit odd numbers is
View answer and explanation
Correct option: A
2475
Option A (2475) is correct because it is the sum of all two-digit odd numbers, which form an arithmetic progression.
307 Medium
307 of 564
Application +4 | -1
If a,b,ca, b, c are in A.P., then 2ax+1,2bx+1,2cx+1,x≠02^{a x+1}, 2^{b x+1}, 2^{c x+1}, x \neq 0 are in
View answer and explanation
Correct option: D
G.P. for all x≠0x \neq 0
The correct answer is D (G.P. for all x≠0x \neq 0) because the ratio between consecutive terms is constant regardless of the value of x.
308 Medium
308 of 564
Application +4 | -1
The sum of the first nn terms of two arithmetic series are in the ratio (7n+1):(4n+27)(7 n+1):(4 n+27). The ratio of their 11th terms is
View answer and explanation
Correct option: D
4:34: 3
Option D (4:34:3) is correct because it is the ratio of the 11th terms of the two arithmetic series, calculated using the given ratio of their sums.
309 Medium
309 of 564
Understanding +4 | -1
∑i=1n∑j=1i∑k=1j1=\sum\limits_{{i=1}}^n \sum\limits_{{j=1}}^i \sum\limits_{{k=1}}^j 1=
View answer and explanation
Correct option: C
n(n+1)(n+2)6\frac{n(n+1)(n+2)}{6}
Option C (n(n+1)(n+2)6\frac{n(n+1)(n+2)}{6}) is correct because evaluating the triple summation results in the given expression.
310 Medium
310 of 564
Application +4 | -1
If the sum to first nn terms of the AP 2, 4, 6,… is 240, then the value of nn is
View answer and explanation
Correct option: B
15
Option B (15) is correct because the sum of the first n terms of an AP is used to form a quadratic equation, which is then solved for n.
311 Medium
311 of 564
Application +4 | -1
A club consists of members whose ages are in AP, the common difference being 3 months. If the youngest member of the club is just 7 yr old and the sum of the ages of all the members is 250yr250 \mathrm{yr}, then the number of members in the club is
View answer and explanation
Correct option: B
25
Option B (25) is correct because it is the number of members in the club, calculated using the arithmetic progression properties and the given conditions.
312 Medium
312 of 564
Application +4 | -1
If a1,a2,…,an{a_1},{a_2}, \ldots ,{a_n} are in HP, then the expression a1a2+a2a3+..+an−1an{a_1}{a_2} + {a_2}{a_3} + .. + {a_{n - 1}}{a_n} is equal to
View answer and explanation
Correct option: C
(n−1)a1an\left( {n - 1} \right){a_1}{a_n}
Option C ((n−1)a1an\left( {n - 1} \right){a_1}{a_n}) is correct because, using the properties of Harmonic Progression (HP) and manipulating the terms, the expression simplifies to this result.
313 Medium
313 of 564
Application +4 | -1
Let 2nd ,8th 2^{\text {nd }}, 8^{\text {th }} and 44th 44^{\text {th }} terms of a non-constant A.P. be respectively the 1st ,2nd 1^{\text {st }}, 2^{\text {nd }} and 3rd 3^{\text {rd }} terms of a G.P. If the first term of the A.P. is 1 , then the sum of its first 20 terms is equal to
View answer and explanation
Correct option: D
970
Option D (970) is correct because the sum of the first 20 terms of the A.P. is calculated correctly using the given conditions.
314 Medium
314 of 564
Analyzing +4 | -1
The sum of the series 34.8−3.54.8.12+3.5.74.8.12.16−  ...\frac{3}{{4.8}} - \frac{{3.5}}{{4.8.12}} + \frac{{3.5.7}}{{4.8.12.16}} - \;...
View answer and explanation
Correct option: B
23−34\sqrt {\frac{2}{3}} - \frac{3}{4}
Option D (23−14\sqrt {\frac{2}{3}} - \frac{1}{4}) is incorrect. The correct answer is B (23−34\sqrt {\frac{2}{3}} - \frac{3}{4}) because it is obtained by recognizing the series as a binomial expansion and simplifying.
315 Medium
315 of 564
Analyzing +4 | -1
If x=∑n=0∞an,y=∑n=0∞bn,z=∑n=0∞cnx=\sum_{n=0}^{\infty} a^{n}, y=\sum_{n=0}^{\infty} b^{n}, z=\sum_{n=0}^{\infty} c^{n} where a,b,ca, b, c are in
View answer and explanation
Correct option: D
H.P.
Option D (H.P.) is correct because the given conditions imply that a, b, and c are in Harmonic Progression.
316 Medium
316 of 564
Application +4 | -1
The sum of the first 10 terms of 1⋅x+4⋅x2+7⋅x3+.....1 \cdot x + 4 \cdot {x^2} + 7 \cdot {x^3} + ..... is
View answer and explanation
Correct option: D
28x11−31x10+2x+1(x−1)2\frac{28 x^{11}-31 x^{10}+2 x+1}{(x-1)^2}
Option D (28x11−31x10+2x+1(x−1)2\frac{{28x^{{11}} - 31x^{{10}} + 2x + 1}}{{(x-1)^2}}) is correct because it represents the sum of the arithmetico-geometric series after applying the appropriate formula and simplifying.
317 Medium
317 of 564
Application +4 | -1
If a1,a2,….ana_1, a_2, \ldots . a_n are in H.P., then a1a2+a3+….+an,a2a1+a3+….+an,…………,ana1+a2+….+an−1\frac{a_1}{a_2+a_3+\ldots .+a_n}, \frac{a_2}{a_1+a_3+\ldots .+a_n}, \ldots \ldots \ldots \ldots, \frac{a_n}{a_1+a_2+\ldots .+a_{n-1}} are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because if a₁, a₂, ..., aₙ are in H.P., then the given expression will also be in H.P.
318 Medium
318 of 564
Analyzing +4 | -1
The sum of the series
x−1x+1+12x2−1(x+1)2+13x3−1(x+1)3+…,\frac{{x - 1}}{{x + 1}} + \frac{1}{2}\frac{{{x^2} - 1}}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{3}\frac{{{x^3} - 1}}{{{{\left( {x + 1} \right)}^3}}} + \ldots , is equal to
View answer and explanation
Correct option: A
log⁡ex{\log _e}x
Option A (log⁡ex{\log _e}x) is correct because the given series can be simplified and expressed as the logarithm of x.
319 Medium
319 of 564
Application +4 | -1
If the arithmetic mean of two numbers aa and b,a>b>0b, a>b>0, is five times their geometric mean, then a+ba−b\frac{a+b}{a-b} is equal to :
View answer and explanation
Correct option: D
5612\frac{5 \sqrt{6}}{12}
The correct answer is D (5612\frac{{5 \sqrt{{6}}}}{{12}}) because it is obtained by using the arithmetic and geometric mean relationship and applying componendo and dividendo.
320 Medium
320 of 564
Analyzing +4 | -1
Given that α,γ\alpha, \gamma are roots of the equation Ax2−4x+1=0A x^2-4 x+1=0 and β,δ\beta, \delta are the roots of the equation Bx2−6x+1=0B x^2-6 x+1=0, then values of AA and BB such that α,β,γ\alpha, \beta, \gamma and δ\delta are in HP, are
View answer and explanation
Correct option: A
A=3,B=8A=3, B=8
Option A (A=3, B=8) is correct because by using the properties of HP and AP, we can find the values of A and B.
321 Medium
321 of 564
Analyzing +4 | -1
If 1,log⁡yx,log⁡zy1, \log _y x, \log _z y and −15log⁡xz-15 \log _x z are in AP\mathrm{AP}, then
View answer and explanation
Correct option: A
z3=xz^3=x
Option A (z3=xz^3=x) is correct because, after solving the given AP conditions, we find the relationship between x and z.
322 Medium
322 of 564
Analyzing +4 | -1
The consecutive terms 11+x,11−x,11−x\frac{1}{1+\sqrt{x}}, \frac{1}{1-x}, \frac{1}{1-\sqrt{x}} of a series are in
View answer and explanation
Correct option: C
AP\mathrm{AP}
Option C (AP) is correct because the difference between consecutive terms is constant, indicating an arithmetic progression.
323 Medium
323 of 564
Application +4 | -1
If T1,T2,T3,T4,T5,T6,T7,…T_1, T_2, T_3, T_4, T_5, T_6, T_7, \ldots are in AP with common difference 2 . Then, the difference of any two consecutive terms of the sequence T1,T4,T7,…T_1, T_4, T_7, \ldots is
View answer and explanation
Correct option: C
6
Option C (6) is correct because the difference between consecutive terms in the sequence T1,T4,T7,…T_1, T_4, T_7, \ldots is 3 times the common difference of the original AP.
324 Medium
324 of 564
Application +4 | -1
If the AM of two numbers be AA and GM be GG, then the numbers will be
View answer and explanation
Correct option: C
A±(A+G)(A−G)A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)}
Option C (A±(A+G)(A−G)A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)}) is correct because solving the equations for AM and GM leads to this expression for the two numbers.
325 Medium
325 of 564
Application +4 | -1
∑n=1∞1(n+1)(n+2)…(n+k)\sum \limits_{n = 1}^\infty \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots \left( {n + k} \right)}} is equal to
View answer and explanation
Correct option: C
1(k−1)k!\frac{1}{{\left( {k - 1} \right)k!}}
Option A (1(k−1)(k!)\frac{1}{{\left( {k - 1} \right)\left( k! \right)}}) is correct because it is the result of the infinite summation of the given series using partial fraction decomposition.
326 Medium
326 of 564
Application +4 | -1
The set of real values of xx satisfying ∣x−1∣≤3|x-1| \leq 3 and ∣x−1∣≥1|x-1| \geq 1 is
View answer and explanation
Correct option: C
[−2,0]∪[2,4][-2,0] \cup[2,4]
Option C ([-2,0] ∪\cup[2,4]) is correct because it represents the intersection of the solution sets of the two inequalities.
327 Medium
327 of 564
Application +4 | -1
The sum of the series 1+2⋅2+3⋅22+4⋅23+…+100⋅2991+2 \cdot 2+3 \cdot 2^2+4 \cdot 2^3+\ldots+100 \cdot 2^{99} is
View answer and explanation
Correct option: B
99⋅2100+199 \cdot 2^{100}+1
Option B, 99⋅2100+199 \cdot 2^{100}+1, is the correct answer because it represents the sum of the given series, which is found by manipulating the series and using the formula for the sum of a geometric progression.
328 Medium
328 of 564
Application +4 | -1
Let a three term strictly increasing arithmetic sequence of integers are one such that the sum of the squares of the three terms equals the product of the middle term and the square of the common difference, then the sum of third term of such sequence is pp, then (p−25)(p-25) is
View answer and explanation
Correct option: A
6
Option A (6) is correct because by setting up the equation based on the given condition and solving for the common difference, we find the third term of the sequence and then calculate p-25.
329 Medium
329 of 564
Analyzing +4 | -1
The value of log⁡2+2(15+13⋅153+15⋅155+…+∞)\log 2 + 2\left( {\frac{1}{5} + \frac{1}{3} \cdot \frac{1}{{{5^3}}} + \frac{1}{5} \cdot \frac{1}{{{5^5}}} + \ldots + \infty } \right) is
View answer and explanation
Correct option: D
log  3\mathrm{log\;}3
Option D (log  3\mathrm{log\;}3) is correct because the expression simplifies to log⁡3\log 3 after using the logarithmic identity log⁡(1+x1−x)=2(x+x33+x55−…)\log \left( {\frac{{1 + x}}{{1 - x}}} \right) = 2\left( {x + \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \ldots } \right).
330 Medium
330 of 564
Analyzing +4 | -1
The sum of nn terms of the series 51.2⋅13+72⋅3⋅132+93.4⋅133+114.5⋅134+…\frac{5}{1.2} \cdot \frac{1}{3}+\frac{7}{2 \cdot 3} \cdot \frac{1}{3^2}+\frac{9}{3.4} \cdot \frac{1}{3^3}+\frac{11}{4.5} \cdot \frac{1}{3^4}+\ldots. is:
View answer and explanation
Correct option: A
1−1n+1⋅13n1-\frac{1}{n+1} \cdot \frac{1}{3^n}
The correct answer is A (1−1n+1⋅13n1-\frac{1}{n+1} \cdot \frac{1}{3^n}) because the sum of the series can be simplified and expressed in this form by decomposing the general term.
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