The sum of first $n$ terms of the series ${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + \ldots $ is $\frac{{n{{\left( {n + 1} \right)}^2}}}{2}$ when $n$ is even. When $n$ is odd the sum is

${\left\{ {\frac{{n\left( {n + 1} \right)}}{2}} \right\}^2}$

If ${a_1},\;{a_2}, \ldots ,\;{a_n}$ are in AP with common difference $d$, then the sum of the series $\sin d\;\left( {{\rm{cosec}}\;{a_1}\;{\rm{cosec\;}}{a_2} + {\rm{cosec}}\;{a_2}\;{\rm{cosec}}\;{a_3} + ... + {\rm{cosec}}\;{a_{n - 1}}\;{\rm{cosec}}\;{a_n}} \right)$ is

${\rm{cosec}}\;{a_1} = {\rm{cosec}}\;{a_n}$

The sum of $n$ terms of the series $5+7+13+31+85+\ldots$ is

$\frac{1}{2}\left(3^n+7 n+1\right)$

$\frac{1}{2}\left(3^n+5 n\right)$

$\frac{1}{2}\left(3^n+7 n\right)$

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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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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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181 Hard

181 of 998

Application +4 | -1

The

20^{\text {th }}

term of the series

2+3+5+9+16

+\ldots

.... is

View answer and explanation

Correct option: A

990

Option C (

{I}+{P}

) is correct because after performing the matrix multiplication and simplification using trigonometric identities, the result is equal to

{I}+{P}

182 Easy

182 of 998

Application +4 | -1

The value of 0.234 is

View answer and explanation

Correct option: A

\frac{{232}}{{990}}

Option A (

\frac{{232}}{{990}}

) is correct because the repeating decimal 0.234 (with 34 repeating) can be expressed as the fraction

\frac{{232}}{{990}}

183 Medium

183 of 998

Application +4 | -1

The sum of first

n

terms of the series

{1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + \ldots

\frac{{n{{\left( {n + 1} \right)}^2}}}{2}

when

n

is even. When

n

is odd the sum is

View answer and explanation

Correct option: B

\frac{{{n^2}\left( {n + 1} \right)}}{2}

Option B (

\frac{{{n^2}\left( {n + 1} \right)}}{2}

) is correct because when n is odd, the last term is n² and the sum can be calculated by adding n² to the sum of the previous (n-1) terms.

184 Easy

184 of 998

Application +4 | -1

If 100 times the 100 th term of an

\mathrm{AP}

with non-zero common difference equals the 50 times its 50th term, then the 150th term of this AP is

View answer and explanation

Correct option: C

zero

Option C (zero) is correct because the 150th term of the AP is zero, derived from the given relationship between the 100th and 50th terms.

185 Easy

185 of 998

Application +4 | -1

Given,

\log 2=0.301

and

\log 3=0.477,

then the number of digits before decimal in

3^{12} \times 2^8

View answer and explanation

Correct option: C

Option C (9) is correct because the number of digits is found by calculating the logarithm and adding 1 to the integer part.

186 Easy

186 of 998

Application +4 | -1

The product of three geometric means between

4

and

\frac{1}{4}

will be

View answer and explanation

Correct option: D

Option D (1) is correct because the product of geometric means between two numbers is equal to the geometric mean of the two numbers raised to the power of the number of means.

187 Easy

187 of 998

Analyzing +4 | -1

{a_1},\;{a_2}, \ldots ,\;{a_n}

are in AP with common difference

d

, then the sum of the series

\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)

View answer and explanation

Correct option: B

\cot {a_1} - \cot {a_n}

Option B (

\cot {a_1} - \cot {a_n}

) is correct because the series telescopes, leaving only the first and last terms after simplification.

188 Easy

188 of 998

Application +4 | -1

The H.M. between two numbers is

16 / 5

, their A.M. is

A

and G.M. is

G

. If

2 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.

189 Easy

189 of 998

Application +4 | -1

If the sum of the first ten terms of the series

{\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

\frac{16}{5} m

, then

m

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.

190 Easy

190 of 998

Application +4 | -1

x,2x + 2,3x + 3, \ldots

are in GP, then the fourth term is

View answer and explanation

Correct option: C

- 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.

191 Easy

191 of 998

Analyzing +4 | -1

S = \frac{1}{{1 \cdot 2}} - \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} - \frac{1}{{4 \cdot 5}} + \ldots + \infty ,

then

{e^S}

equals

View answer and explanation

Correct option: B

\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.

192 Easy

192 of 998

Application +4 | -1

a,b,c

are distinct positive real number and

{a^2} + {b^2} + {c^2} = 1

, then

3{\left( {{a^2}{b^2}{c^2}} \right)^{1/3}}

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.

193 Easy

193 of 998

Application +4 | -1

The product

\left( {32} \right){\left( {32} \right)^{1/6}}{\left( {32} \right)^{1/36}} \ldots ..\mathrm{to}\;\infty

View answer and explanation

Correct option: C

Option C (64) is correct because the product simplifies to 32 raised to the power of an infinite geometric series, which evaluates to 64.

194 Easy

194 of 998

Application +4 | -1

If the number of terms in an AP is

2n + 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

\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.

195 Medium

195 of 998

Application +4 | -1

x_1, x_2, \ldots, x_n

are

n

non-zero real numbers such that

\begin{aligned}\left(x_1^2+x_2^2+\ldots+x_{N-1}^2\right) & \left(x_2^2+x_3^2+\ldots+x_N^2\right) \\ & \leq\left(x_1 x_2+x_2 x_3+\ldots+x_{n-1} x_n\right)^2,\end{aligned}

then

x_1, x_2, \ldots, x_n

are in

View answer and explanation

Correct option: B

Option B (GP) is correct because the given inequality implies that the numbers must be in a geometric progression.

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196 Medium

196 of 998

Understanding +4 | -1

5,25,125, \ldots

are in

G P

, then

1,5,25, \ldots

are in

View answer and explanation

Correct option: B

G P

Option B (

G P

) is correct because dividing each term of a geometric progression by a constant results in another geometric progression.

197 Easy

197 of 998

Application +4 | -1

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, 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.

198 Easy

198 of 998

Understanding +4 | -1

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 \left( {\frac{m}{n}} \right)

Option A (

\log \left( {\frac{m}{n}} \right)

) is correct because the given expression is the expansion of

\log \left( {\frac{m}{n}} \right)

199 Easy

199 of 998

Application +4 | -1

If sum of the first 21 terms of the series

\log _{9^{1 / 2}} x+\log _{9^{1 / 3}} x+\log _{9^{1 / 4}} x+\ldots \ldots

, where

x>0

is 504 , then

\mathrm{x}

is equal to:

View answer and explanation

Correct option: A

Option A (81) is correct because solving the logarithmic series and simplifying the equation leads to x = 81.

200 Easy

200 of 998

Remembering +4 | -1

Third term of the sequence whose

n^{\text {th }}

term is

a_n=2^n

, is

View answer and explanation

Correct option: A

Option A (8) is correct because substituting n=3 into the given formula

a_n = 2^n

yields

a_3 = 2^3 = 8

201 Easy

201 of 998

Application +4 | -1

\log _{10} 2, \log _{10}\left(2^x+1\right)

and

\log _{10}\left(2^x+3\right)

are in A.P., then

View answer and explanation

Correct option: D

x=\frac{1}{2} \log _2 5

Option D (

x=\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.

202 Easy

202 of 998

Application +4 | -1

If the sum of first 11 terms of an

AP.,\,\,{a_1},{a_2},{a_3}.....

is 0

\left( {{a_1} \ne 0} \right),

then the sum of the

AP.,\,\,{a_1},{a_3},{a_5}.....\,{a_{23}}

k{a_1},

where

k

is equal to

View answer and explanation

Correct option: D

- \frac{{72}}{5}

Option D (

\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.

203 Medium

203 of 998

Application +4 | -1

In the four numbers first three are in GP and last three are in AP whose common difference is 6. If the first and last numbers are same, then first number will be

View answer and explanation

Correct option: D

Option D (8) is correct because it is the first number of the sequence that satisfies the given conditions of the first three numbers being in GP and the last three in AP.

204 Easy

204 of 998

Application +4 | -1

The arithmetic mean of first

n

odd natural number is

View answer and explanation

Correct option: C

n

Option C (n) is correct because the arithmetic mean of the first n odd natural numbers is n.

205 Medium

205 of 998

Application +4 | -1

The sum of

n

terms of the series

5+7+13+31+85+\ldots

View answer and explanation

Correct option: A

\frac{1}{2}\left(3^n+8 n-1\right)

Option A (

\frac{1}{2}(3^n+8n-1)

) is correct because the sum of n terms of the series is derived to be in this form by identifying the pattern in the differences of successive terms.

206 Easy

206 of 998

Application +4 | -1

The HM of two numbers is 4. Their AM is

A

and GM is

G

. If

2A + {G^2} = 27

, then

A

is equal to

View answer and explanation

Correct option: B

\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.

207 Easy

207 of 998

Application +4 | -1

If the sum of the first

n

natural numbers is

1/5

times the sum of their squares, then the value of

n

View answer and explanation

Correct option: C

Option C is correct because by setting up the equation based on the given condition and solving for n, we find that n = 7.

208 Easy

208 of 998

Application +4 | -1

The value of

{9^{1/3}} \times {9^{1/9}} \times {9^{1/27}} \times \ldots \infty

View answer and explanation

Correct option: C

Option C (3) is correct because the expression simplifies to 9^(1/2) which equals 3.

209 Easy

209 of 998

Application +4 | -1

\log _5 2, \log _5\left(2^x-5\right)

and

\log _5\left(2^x-7 / 2\right)

are in A.P., then

x

is equal to

View answer and explanation

Correct option: B

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.

210 Medium

210 of 998

Application +4 | -1

If the numbers

a, b, c, d, e

form an A.P., then the value of

a-4 b+6 c-4 d+e

, is

View answer and explanation

Correct option: C

Option C (0) is correct because the expression simplifies to zero due to the properties of arithmetic progressions.

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