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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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Showing 181-210 of 998 questions

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181 Easy
181 of 998
Analyzing +4 | -1
If a1,  a2,…,  an{a_1},\;{a_2}, \ldots ,\;{a_n} are in AP with common difference dd, then the sum of the series

sin⁡d  (cosec  a1  cosec  a2+cosec  a2  cosec  a3+...+cosec  an−1  cosec  an)\sin d\;\left( {\mathrm{cosec}\;{a_1}\;\mathrm{cosec\;}{a_2} + \mathrm{cosec}\;{a_2}\;\mathrm{cosec}\;{a_3} + ... + \mathrm{cosec}\;{a_{n - 1}}\;\mathrm{cosec}\;{a_n}} \right) is
View answer and explanation
Correct option: B
cot⁡a1−cot⁡an\cot {a_1} - \cot {a_n}
Option B (cot⁡a1−cot⁡an\cot {a_1} - \cot {a_n}) is correct because the series telescopes, leaving only the first and last terms after simplification.
182 Easy
182 of 998
Application +4 | -1
The H.M. between two numbers is 16/516 / 5, their A.M. is AA and G.M. is GG. If 2A+G2=262 A+G^2=26 then the numbers are
View answer and explanation
Correct option: C
2, 8
Option C (2, 8) is correct because the numbers satisfy the given harmonic mean, arithmetic mean, and geometric mean conditions.
183 Easy
183 of 998
Application +4 | -1
If the sum of the first ten terms of the series (135)2+(225)2+(315)2+42+(445)2+……{\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + {\left( {4\frac{4}{5}} \right)^2} + \ldots \ldots is 165m\frac{16}{5} m, then mm is equal to
View answer and explanation
Correct option: B
101
Option B (101) is correct because after simplifying the series and using summation formulas, the value of m is found to be 101.
184 Easy
184 of 998
Application +4 | -1
If x,2x+2,3x+3,…x,2x + 2,3x + 3, \ldots are in GP, then the fourth term is
View answer and explanation
Correct option: C
−13.5- 13.5
Option C (-13.5) is correct because it is the fourth term of the geometric progression calculated using the given conditions and solving for x.
185 Easy
185 of 998
Analyzing +4 | -1
If S=11⋅2−12⋅3+13⋅4−14⋅5+…+∞,S = \frac{1}{{1 \cdot 2}} - \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} - \frac{1}{{4 \cdot 5}} + \ldots + \infty , then eS{e^S} equals
View answer and explanation
Correct option: B
4e\frac{4}{e}
Option B (4/e) is correct because the given infinite series can be expressed in terms of the logarithm function, allowing for the calculation of e^S.
186 Easy
186 of 998
Application +4 | -1
If a,b,ca,b,c are distinct positive real number and a2+b2+c2=1{a^2} + {b^2} + {c^2} = 1, then 3(a2b2c2)1/33{\left( {{a^2}{b^2}{c^2}} \right)^{1/3}} is
View answer and explanation
Correct option: A
Less than 1
Option A (Less than 1) is correct because, according to AM > GM for distinct positive real numbers, the expression is less than 1.
187 Easy
187 of 998
Application +4 | -1
The product (32)(32)1/6(32)1/36…..to  ∞\left( {32} \right){\left( {32} \right)^{1/6}}{\left( {32} \right)^{1/36}} \ldots ..\mathrm{to}\;\infty is
View answer and explanation
Correct option: C
64
Option C (64) is correct because the product simplifies to 32 raised to the power of an infinite geometric series, which evaluates to 64.
188 Easy
188 of 998
Application +4 | -1
If the number of terms in an AP is 2n+12n + 1, then the ratio of the sum of the odd terms to the sum of even terms is
View answer and explanation
Correct option: A
n+1n\frac{{n + 1}}{n}
Option A is correct because the ratio of the sum of odd terms to the sum of even terms in an AP with 2n+1 terms simplifies to (n+1)/n.
189 Easy
189 of 998
Analyzing +4 | -1
The difference between two numbers is 48 and the difference between their arithmetic mean and their geometric mean is 18. Then, the greater of two numbers is
View answer and explanation
Correct option: D
49
Option D (49) is correct because solving the system of equations derived from the problem statement leads to the greater number being 49.
190 Easy
190 of 998
Application +4 | -1
If a(1b+1c),b(1c+1a),c(1a+1b)a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right) are in A.P., then
View answer and explanation
Correct option: A
a,b,ca, b, c are in A.P.
Option A is correct because if the given terms are in A.P., then a, b, and c are also in A.P.
191 Easy
191 of 998
Understanding +4 | -1
2{m−nm+n+13(m−nm+n)3+15(m−nm+n)5+…}  2\left\{ {\frac{{m - n}}{{m + n}} + \frac{1}{3}{{\left( {\frac{{m - n}}{{m + n}}} \right)}^3} + \frac{1}{5}{{\left( {\frac{{m - n}}{{m + n}}} \right)}^5} + \ldots } \right\}\; is equal to
View answer and explanation
Correct option: A
log⁡(mn)\log \left( {\frac{m}{n}} \right)
Option A (log⁡(mn)\log \left( {\frac{m}{n}} \right)) is correct because the given expression is the expansion of log⁡(mn)\log \left( {\frac{m}{n}} \right).
192 Easy
192 of 998
Application +4 | -1
If sum of the first 21 terms of the series log⁡91/2x+log⁡91/3x+log⁡91/4x+……\log _{9^{1 / 2}} x+\log _{9^{1 / 3}} x+\log _{9^{1 / 4}} x+\ldots \ldots, where x>0x>0 is 504 , then x\mathrm{x} is equal to:
View answer and explanation
Correct option: A
81
Option A (81) is correct because solving the logarithmic series and simplifying the equation leads to x = 81.
193 Easy
193 of 998
Remembering +4 | -1
Third term of the sequence whose nth n^{\text {th }} term is an=2na_n=2^n, is
View answer and explanation
Correct option: A
8
Option A (8) is correct because substituting n=3 into the given formula an=2na_n = 2^n yields a3=23=8a_3 = 2^3 = 8.
194 Easy
194 of 998
Application +4 | -1
If log⁡102,log⁡10(2x+1)\log _{10} 2, \log _{10}\left(2^x+1\right) and log⁡10(2x+3)\log _{10}\left(2^x+3\right) are in A.P., then
View answer and explanation
Correct option: D
x=12log⁡25x=\frac{1}{2} \log _2 5
Option D (x=12log⁡25x=\frac{{1}}{{2}} \log _2 5) is correct. By using the properties of arithmetic progressions and logarithms, we can solve for x and get the result.
195 Easy
195 of 998
Application +4 | -1
If the sum of first 11 terms of an AP.,  a1,a2,a3.....AP.,\,\,{a_1},{a_2},{a_3}..... is 0 (a1≠0),\left( {{a_1} \ne 0} \right), then the sum of the AP.,  a1,a3,a5..... a23AP.,\,\,{a_1},{a_3},{a_5}.....\,{a_{23}} is ka1,k{a_1}, where kk is equal to
View answer and explanation
Correct option: D
−725- \frac{{72}}{5}
Option D (−725\frac{{-72}}{5}) is correct because it is the value of k obtained by using the given condition about the sum of the first 11 terms of the AP being 0.
196 Easy
196 of 998
Application +4 | -1
The arithmetic mean of first nn odd natural number is
View answer and explanation
Correct option: C
nn
Option C (n) is correct because the arithmetic mean of the first n odd natural numbers is n.
197 Easy
197 of 998
Analyzing +4 | -1
If x,y,zx, y, z are in GP⁡(x,y,z>1)\operatorname{GP}(x, y, z>1), then 12x+ln⁡x,14x+ln⁡y,16x+ln⁡z\frac{1}{2 x+\ln x}, \frac{1}{4 x+\ln y}, \frac{1}{6 x+\ln z} are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because since x, y, z are in GP, their logarithms are in AP. Also, 2x, 4x, 6x are in AP. Therefore, 2x + ln(x), 4x + ln(y), 6x + ln(z) are in AP, and their reciprocals are in HP.
198 Easy
198 of 998
Application +4 | -1
The number of pairs of consecutive odd natural numbers both of which are larger than 10, such that their sum is less than 40, is
View answer and explanation
Correct option: A
4
The correct answer is A (4) because there are four pairs of consecutive odd natural numbers greater than 10 whose sum is less than 40.
199 Easy
199 of 998
Application +4 | -1
The HM of two numbers is 4. Their AM is AA and GM is GG. If 2A+G2=272A + {G^2} = 27, then AA is equal to
View answer and explanation
Correct option: B
92\frac{9}{2}
Option B is correct because by using the relationships between AM, GM, and HM, and the given equation, we can solve for A.
200 Easy
200 of 998
Application +4 | -1
If the sum of the first nn natural numbers is 1/51/5 times the sum of their squares, then the value of nn is
View answer and explanation
Correct option: C
7
Option C is correct because by setting up the equation based on the given condition and solving for n, we find that n = 7.
201 Easy
201 of 998
Application +4 | -1
The value of 91/3×91/9×91/27×…∞{9^{1/3}} \times {9^{1/9}} \times {9^{1/27}} \times \ldots \infty is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the expression simplifies to 9(1/2) which equals 3.
202 Easy
202 of 998
Application +4 | -1
If log⁡52,log⁡5(2x−5)\log _5 2, \log _5\left(2^x-5\right) and log⁡5(2x−7/2)\log _5\left(2^x-7 / 2\right) are in A.P., then xx is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because it is the only value of x that satisfies the given arithmetic progression condition and the logarithmic domain restrictions.
203 Easy
203 of 998
Analyzing +4 | -1
If A, G and H are respectively A.M., G.M., and H.M. of the positive integers a,ba, b and cc, then which of the following is not true?
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Correct option: A
a2=AHa^2=A H
Option A (a2=AHa^2=A H) is not necessarily true for positive integers a, b, and c and their A.M., G.M., and H.M., while the other options hold under the given conditions.
204 Easy
204 of 998
Analyzing +4 | -1
Let the positive numbers a,b,ca, b, c, dd be in A.P. Then, abc,abd,acd,bcda b c, a b d, a c d, b c d are
View answer and explanation
Correct option: D
in H.P.
Option D (in H.P.) is correct because if a, b, c, d are in AP, then abc, abd, acd, bcd are in HP.
205 Easy
205 of 998
Application +4 | -1
If −7<x<18-7 < x < 18 and 9<y<209 < y < 20, then the range of x+yx+y is:
View answer and explanation
Correct option: D
(2,38)(2,38)
Option D ((2,38)(2,38)) is correct because it represents the range of x+yx+y obtained by adding the inequalities −7<x<18-7 < x < 18 and 9<y<209 < y < 20.
206 Easy
206 of 998
Application +4 | -1
If the sum
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 \ldots+ up to 20 terms is equal
to k21\frac{\mathrm{k}}{21}, then k\mathrm{k} is equal to:
View answer and explanation
Correct option: A
120
Option A (120) is correct because the sum of the series is calculated correctly, and k is found to be 120.
207 Easy
207 of 998
Application +4 | -1
If a(b−c)x2+b(c−a)xy+c(a−b)y2a(b-c) x^2+b(c-a) x y+c(a-b) y^2 is a perfect square, the quantities a,b,ca, b, c are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the given condition for the expression to be a perfect square leads to the relationship that a,b,ca, b, c are in Harmonic Progression.
208 Easy
208 of 998
Application +4 | -1
For x∈Rx \in \mathbf{R}, let [x][x] denote the greatest integer ≤x\leq x. Largest natural number nn for which

E=[π2]+[1100+π2]+[2100+π2]…+[n100+π2]<43,E=\left[\frac{\pi}{2}\right]+\left[\frac{1}{100}+\frac{\pi}{2}\right]+\left[\frac{2}{100}+\frac{\pi}{2}\right] \ldots+\left[\frac{n}{100}+\frac{\pi}{2}\right]<43, is
View answer and explanation
Correct option: A
41
Option A (41) is correct because it is the largest integer 'n' such that the sum of the greatest integer functions is less than 43.
209 Easy
209 of 998
Analyzing +4 | -1
The sixth term of an AP is 2, and its common difference is greater than one. The value of the common difference of the progression so that the product of the first, fourth and fifth terms is greatest is
View answer and explanation
Correct option: A
8/58 / 5
Option A (8/5) is correct because it maximizes the product of the first, fourth, and fifth terms of the arithmetic progression, given the constraints.
210 Easy
210 of 998
Application +4 | -1
If mm is the AM of two distinct real numbers ll and nn (l,n>1)(l, n>1) and G1,G2G_1, G_2 and G3G_3 are three geometric means between ll and nn, then G14+2G24+G34G_1^4+2 G_2^4+G_3^4 equals
View answer and explanation
Correct option: C
4 lm2n4 \mathrm{~lm}^2 n
Option C (4 lm2n4 \mathrm{~lm}^2 n) is correct because it is the value of the expression G14+2G24+G34G_1^4+2 G_2^4+G_3^4, derived from the given conditions of AM and geometric means.
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