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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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121 Easy
121 of 998
Application +4 | -1
If 1log⁡x10=2log⁡a10−2,\frac{1}{{{{\log }_x}10}} = \frac{2}{{{{\log }_a}10}} - 2, then x=x =
View answer and explanation
Correct option: D
a2100\frac{{{a^2}}}{{100}}
Option D (a2100\frac{{{a^2}}}{{100}}) is correct because by using the change of base formula and simplifying, we can solve for x and find its value to be a²/100.
122 Easy
122 of 998
Application +4 | -1
If Sn=∑r=0n1nCrS_{n}=\sum_{r=0}^{n} \frac{1}{{ }^{n} C_{r}} and tn=∑r=0nrnCrt_{n}=\sum_{r=0}^{n} \frac{r}{{ }^{n} C_{r}}, then tnSn\frac{t_{n}}{S_{n}} is equal to
View answer and explanation
Correct option: D
12n\frac{1}{2} n
Option C (n2(n+1)2\frac{n^{2}(n+1)}{2}) is correct because it correctly calculates the sum of the series when n is odd, using the given sum for when n is even.
123 Easy
123 of 998
Analyzing +4 | -1
The expansion of log⁡(1+3x+2x2)\log \left( {1 + 3x + 2{x^2}} \right) is
View answer and explanation
Correct option: C
3x−52x2+93x3−174x4+...∞3x - \frac{5}{2}{x^2} + \frac{9}{3}{x^3} - \frac{{17}}{4}{x^4} + ...\infty
Option C (3x−52x2+93x3−174x4+...∞3x - \frac{5}{2}{x^2} + \frac{9}{3}{x^3} - \frac{{17}}{4}{x^4} + ...\infty) is correct because it is the correct expansion of log⁡(1+3x+2x2)\log(1 + 3x + 2x^2) using logarithmic properties and series expansion.
124 Easy
124 of 998
Application +4 | -1
If 3+5+7+…+n  terms5+8+11+…+10  terms=7,\frac{{3 + 5 + 7 + \ldots + n\;\mathrm{terms}}}{{5 + 8 + 11 + \ldots + 10\;\mathrm{terms}}} = 7, then the value of nn is
View answer and explanation
Correct option: A
35
Option A (35) is correct because solving the equation derived from the arithmetic progression sum formula yields n = 35.
125 Easy
125 of 998
Application +4 | -1
If Sn=nP+12n(n−1)Q{S_n} = nP + \frac{1}{2}n\left( {n - 1} \right){\cal Q}, where Sn{S_n} denotes the sum of the first nn terms of an AP, then the common difference is
View answer and explanation
Correct option: D
Q{\cal Q}
Option D (Q{\cal Q}) is correct because the common difference of an AP can be found by subtracting consecutive terms, which simplifies to Q{\cal Q}.
126 Easy
126 of 998
Application +4 | -1
If the sum of the series 2+5x+25x2+125x32+\frac{5}{x}+\frac{25}{x^2}+\frac{125}{x^3} +…+\ldots is finite, then
View answer and explanation
Correct option: A
∣x∣>5|x|>5
Option A (∣x∣>5|x|>5) is correct because the given series is a geometric series, and for its sum to be finite, the absolute value of the common ratio must be less than 1.
127 Easy
127 of 998
Analyzing +4 | -1
The sum of series 1+14.2!+116.4!+164.6!+…∞1 + \frac{1}{{4.2!}} + \frac{1}{{16.4!}} + \frac{1}{{64.6!}} + \ldots \infty is
View answer and explanation
Correct option: A
e+12e\frac{{e + 1}}{{2\sqrt e }}
Option A (e+12e\frac{{e + 1}}{{2\sqrt e }}) is correct because it is derived from the series expansion of ex+e−x2\frac{{{e^x} + {e^{ - x}}}}{2}.
128 Easy
128 of 998
Application +4 | -1
The sum of all the numbers of the form n3n^3 which lie between 100 and 10000 , is
View answer and explanation
Correct option: B
53261
Option B (53261) is correct because the sum of the cubes of integers between the given range is calculated to be 53261.
129 Easy
129 of 998
Application +4 | -1
If log⁡10x=y,{\log _{10}}x = y, then log⁡103x2{\log _{{{10}^3}}}{x^2} equals
View answer and explanation
Correct option: B
23y\frac{2}{3}y
Option B (2/3 y) is correct because it is the simplified form of the logarithm expression using the properties of logarithms.
130 Easy
130 of 998
Application +4 | -1
Let a1,a2,…a_1, a_2, \ldots be in AP and q1,q2,…q_1, q_2, \ldots be in GP. If a1=q1=2a_1=q_1=2 and a10=q10=3a_{10}=q_{10}=3, then
View answer and explanation
Correct option: C
a7q19=a19q10a_7 q_{19}=a_{19} q_{10}
Option C (a7q19=a19q10a_7 q_{19}=a_{19} q_{10}) is correct because by calculating the values of a7a_7, a19a_{19}, q7q_7, q10q_{10}, and q19q_{19}, we can verify that a7q19=a19q10a_7 q_{19}=a_{19} q_{10}.
131 Easy
131 of 998
Application +4 | -1
Sum of the series S=1+12(1+2)+S=1+\frac{1}{2}(1+2)+ 13(1+2+3)+14(1+2+3+4)+…\frac{1}{3}(1+2+3)+\frac{1}{4}(1+2+3+4)+\ldots upto 20 terms is
View answer and explanation
Correct option: C
115
Option C (115) is correct because the sum of the series can be simplified by finding a general term and then summing the resulting arithmetic series.
132 Easy
132 of 998
Application +4 | -1
The sum to nn terms of the series 312+512+22+712+22+32+…\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots is
View answer and explanation
Correct option: A
6nn+1\frac{6 n}{n+1}
Option A (6nn+1\frac{{6 n}}{{n+1}}) is correct because the series can be simplified to a telescoping series, and the sum to nn terms is 6nn+1\frac{{6n}}{{n+1}}.
133 Easy
133 of 998
Application +4 | -1
If a+b2,b,b+c2\frac{{a + b}}{2},b,\frac{{b + c}}{2} are in HP, then a,b,ca,b,c are in
View answer and explanation
Correct option: C
GP
Option C (GP) is correct because the given condition leads to b2 = ac, which is the condition for a, b, and c to be in GP.
134 Easy
134 of 998
Application +4 | -1
If a>0a>0, then ∑n=1∞(aa+1)n\sum\limits_{n=1}^{\infty}\left(\frac{a}{a+1}\right)^n equals
View answer and explanation
Correct option: D
aa
Option D (aa) is correct because the given expression is an infinite geometric series whose sum can be directly calculated using the formula a1−r\frac{a}{1-r}.
135 Easy
135 of 998
Analyzing +4 | -1
If a1,a2,…a_1, a_2, \ldots are in HP and fk=∑r=1nar−akf_k=\sum\limits_{r=1}^n a_r-a_k, then 2α1,2α2,2α3,2α4,…2^{\alpha_1}, 2^{\alpha_2}, 2^{\alpha_3}, 2^{\alpha_4}, \ldots are in
{Where,   α1=a1f1,   α2=a2f2,   α3=a3f3,…}\left\{ {{\text{Where, }}\,{\text{ }}{\alpha _1} = \frac{{{a_1}}}{{{f_1}}},\,\,\,{\alpha _2} = \frac{{{a_2}}}{{{f_2}}},\,\,\,{\alpha _3} = \frac{{{a_3}}}{{{f_3}}}, \ldots } \right\}
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because while the reciprocals of α_1, α_2, α_3 are in AP, it doesn't necessarily imply that 2α_1, 2α_2, 2α_3 are in AP, GP, or HP.
136 Easy
136 of 998
Analyzing +4 | -1
Let a1=0a_1=0 and a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n be real numbers such that ∣ai∣=∣ai−1+1∣\left|a_i\right|=\left|a_{i-1}+1\right| for all ii, then the A.M. of the numbers a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n has the value AA where
View answer and explanation
Correct option: C
A≥−12A \geq-\frac{1}{2}
Option C (A≥−12A \geq-\frac{1}{2}) is correct because, by summing the given recurrence relation and using the definition of the arithmetic mean, we can derive that A≥−12A \geq-\frac{1}{2}.
137 Easy
137 of 998
Application +4 | -1
Given sum of the first nn terms of an A.P. is 2n+3n22 n+3 n^{2}. Another
View answer and explanation
Correct option: B
6n2−n6 n^{2}-n
Option B (6n2 - n) is correct because it produces an A.P. where the first term and the difference between the second and first term align with the given series.
138 Easy
138 of 998
Application +4 | -1
The sum of 100 terms of the series 0.9+0.09+0.009…0.9 + 0.09 + 0.009 \ldots will be
View answer and explanation
Correct option: A
1−(110)1001 - {\left( {\frac{1}{{10}}} \right)^{100}}
Option A (1−(110)1001 - {\left( {\frac{1}{{10}}} \right)^{100}}) is correct because it represents the sum of the given series, which is a geometric progression, up to 100 terms.
139 Easy
139 of 998
Application +4 | -1
The sum to nn terms of the series 43+109+2827+...\frac{4}{3} + \frac{{10}}{9} + \frac{{28}}{{27}} + ... is
View answer and explanation
Correct option: B
3n(2n+1)−12(3n)\frac{{{3^n}\left( {2n + 1} \right) - 1}}{{2\left( {{3^n}} \right)}}
Option B is correct because the sum is calculated by separating the series into two parts: a constant series and a geometric series, and then summing each part.
140 Easy
140 of 998
Application +4 | -1
If the sum of the series 3+14(3+d)+142(3+2d)+…3+\frac{1}{4}(3+d)+\frac{1}{4^2}(3+2 d)+\ldots to ∞\infty is 8 , then the value of dd, is
View answer and explanation
Correct option: A
9
Option A (9) is correct because it is the value of d that satisfies the given infinite series equation.
141 Easy
141 of 998
Analyzing +4 | -1
If 4a2+9b2+16c2=2(3ab+6bc+4ca)4{a^2} + 9{b^2} + 16{c^2} = 2\left( {3ab + 6bc + 4ca} \right), where a,b,ca,b,c are non-zero numbers, then a,b,ca,b,c are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the given equation can be rearranged to show that 2a=3b=4c2a = 3b = 4c, which implies that a,b,ca, b, c are in Harmonic Progression.
142 Easy
142 of 998
Application +4 | -1
The coefficient of x4{x^4} in the expansion of 1−2x−x2e−x\frac{{1 - 2x - {x^2}}}{{{e^{ - x}}}} is
View answer and explanation
Correct option: A
1−k−k2k!\frac{{1 - k - {k^2}}}{{k!}}
Option C is correct because by substituting the given value of x and applying the properties of logarithms, the expression simplifies to -6.
143 Easy
143 of 998
Application +4 | -1
A man saves ₹200₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹40₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after
View answer and explanation
Correct option: C
21 months
Option C (21 months) is correct because it is the total number of months required to reach the total savings of ₹ 11040, considering the initial constant savings and the subsequent increasing savings.
144 Easy
144 of 998
Analyzing +4 | -1
If mm is a root of the given equation (1−ab)x2−(a2+b2)x−(1+ab)=0\left( {1 - ab} \right){x^2} - \left( {{a^2} + {b^2}} \right)x - \left( {1 + ab} \right) = 0 and mm harmonic means are inserted between aa and bb, then the difference between the last and the first of the means equals
View answer and explanation
Correct option: B
ab(b−a)ab\left( {b - a} \right)
Option B (ab(b−a)ab(b - a)) is correct because the difference between the last and first harmonic means inserted between a and b is ab(b−a)ab(b-a).
145 Easy
145 of 998
Application +4 | -1
If the 9th 9^{\text {th }} term of an AP is 35 and 19th 19^{\text {th }} term is 75 , then its 20th 20^{\text {th }} term will be
View answer and explanation
Correct option: B
79
Option B (79) is correct because by finding the common difference and the first term of the AP, the 20th20^{th} term can be calculated as 79.
146 Easy
146 of 998
Application +4 | -1
The minimum value of 3tan⁡2θ+12cot⁡2θ3{\tan ^2}\theta + 12{\cot ^2}\theta is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the minimum value of the expression 3tan⁡2θ+12cot⁡2θ3{\tan ^2}\theta + 12{\cot ^2}\theta is 12, which is not among the given options. This is found using AM-GM inequality.
147 Easy
147 of 998
Application +4 | -1
If the first, second and the last terms of an A.P. are a,ba, b, c respectively, then the sum of the A.P. is
View answer and explanation
Correct option: C
(a+c)(b+c−2a)2(b−a)\frac{(a+c)(b+c-2 a)}{2(b-a)}
Option C is correct because the sum of the A.P. is calculated using the first, second, and last terms.
148 Easy
148 of 998
Application +4 | -1
The maximum value of the sum of the AP 50,48,46,44,…50,48,46,44, \ldots is
View answer and explanation
Correct option: D
650
Option D (650) is correct because it represents the sum of the AP up to the point where the terms become non-positive, maximizing the sum.
149 Easy
149 of 998
Application +4 | -1
The value of (0.16)log⁡2.5(13+132+133+…∞){\left( {0.16} \right)^{{{\log }_{2.5}}\left( {\frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + \ldots \infty } \right)}} , is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the expression simplifies to 4 after evaluating the infinite geometric series and applying logarithmic properties.
150 Easy
150 of 998
Application +4 | -1
If the (p+q)\left( {p + q} \right)th term of a geometric series is mm and the (p−q)\left( {p - q} \right)th term is n, then the ppth term is
View answer and explanation
Correct option: A
(mn)1/2{\left( {mn} \right)^{1/2}}
Option A ((mn)1/2{( {mn} )}^{1/2}) is correct because it is derived by multiplying the two given terms and taking the square root.
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