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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Hard

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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 1-30 of 37 questions

1 Hard

1 of 37

Analyzing +4 | -1

Let

a,\;b,\;c > 0

and

4{a^2} + 9{b^2} + 16{c^2} - 6ab - 12bc - 8ac = 0,

then

b

View answer and explanation

Correct option: A

\le \sqrt {ac}

Option D (

\le \sqrt {ac}

) is correct because

a, b, c

are in HP, and for positive numbers, the harmonic mean is less than or equal to the geometric mean.

2 Hard

2 of 37

Application +4 | -1

Consider the sequence

1,2,2,3,3,3, \ldots

, where

n

occurs

n

times. The number that occurs as 2011 th terms is

View answer and explanation

Correct option: C

Option C (63) is correct because the 2011th term falls within the range where the number 63 occurs.

3 Hard

3 of 37

Analyzing +4 | -1

\frac{1}{{b - c}},\;\frac{1}{{c - a}},\;\frac{1}{{a - b}}

be consecutive terms of an AP, then

{\left( {b - c} \right)^2},\;{\left( {c - a} \right)^2},\;{\left( {a - b} \right)^2}

will be in

View answer and explanation

Correct option: B

Option B (AP) is correct because if

\frac{1}{{b - c}}, \frac{1}{{c - a}}, \frac{1}{{a - b}}

are in A.P., then

{\left( {b - c} \right)^2},{\left( {c - a} \right)^2},{\left( {a - b} \right)^2}

are also in A.P.

4 Hard

4 of 37

Application +4 | -1

x,y,z

are three consecutive positive integers, then

{\log _e}\sqrt x + {\log _e}\sqrt z + \left( {\frac{1}{{2xz + 1}}} \right) + \frac{1}{3}{\left( {\frac{1}{{2xz + 1}}} \right)^3} + \frac{1}{5}{\left( {\frac{1}{{2xz + 1}}} \right)^5} + \;...

View answer and explanation

Correct option: B

{\log _e}y

Option B (

{\log _e}y

) is correct because the expression simplifies to

{\log _e}\sqrt{xz} + \frac{1}{2}{\log _e}\left[ {\frac{{1 + \frac{1}{{2xz + 1}}}}{{1 - \frac{1}{{2xz + 1}}}}} \right]

which further simplifies to

{\log _e}y

when x, y, and z are consecutive integers.

5 Hard

5 of 37

Application +4 | -1

If the sum of

n

terms of an A.P. is

c n^2

, then the sum of their squares is

View answer and explanation

Correct option: C

\frac{n c^2\left(4 n^2-1\right)}{3}

Option C (

\frac{{n c^2(4 n^2-1)}}{{3}}

) is correct because it is the sum of the squares of the terms of the A.P.

6 Hard

6 of 37

Application +4 | -1

\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots +\infty=\frac{\pi^4}{90}

, then the value of

\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots \infty

View answer and explanation

Correct option: A

\frac{\pi^4}{96}

Option A (

\frac{\pi^4}{96}

) is correct because it is the value of the given infinite series after subtracting the even terms and simplifying.

7 Hard

7 of 37

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}

8 Hard

8 of 37

Analyzing +4 | -1

a,b,c,d

and

p

are distinct real number such that

\left( {{a^2} + {b^2} + {c^2}} \right){p^2} - 2\left( {ab + bc + cd} \right)p + \left( {{b^2} + {c^2} + {d^2}} \right) \le 0

, then

a,b,c,d

View answer and explanation

Correct option: B

are in GP

Option B (are in GP) is correct because the given inequality leads to the condition where the ratio of consecutive terms is constant, indicating a geometric progression.

9 Hard

9 of 37

Understanding +4 | -1

p=\sum\limits_{{n=0}}^{\infty} x^n, q=\sum\limits_{{n=0}}^{\infty} y^n, r=\sum\limits_{{n=0}}^{\infty}(x y)^n

, (where

|x|,|y| < 1

) then which of the following is true?

View answer and explanation

Correct option: B

p q+r=p r+q r

Option B (

p q+r=p r+q r

) is correct because substituting the series representations of p, q, and r and simplifying leads to the equality.

10 Hard

10 of 37

Analyzing +4 | -1

\sum \limits_{r = 1}^\infty \frac{1}{{{{\left( {2r - 1} \right)}^2}}} = \frac{{{\pi ^2}}}{8},

then

\sum \limits_{r = 1}^\infty \frac{1}{{{r^2}}}

is equal to

View answer and explanation

Correct option: C

\frac{{{\pi ^2}}}{6}

Option C (

\frac{{{\pi ^2}}}{6}

) is correct because by expressing the sum of reciprocals of all squares in terms of the sum of reciprocals of odd squares, we can find the value of the desired sum.

11 Hard

11 of 37

Understanding +4 | -1

S = \sum \limits_{n = 0}^\infty \frac{{{{\left( {\log x} \right)}^{2n}}}}{{\left( {2\;n} \right)\;!}},

then

S

equals

View answer and explanation

Correct option: C

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

Option C is correct because the given series is the Taylor series expansion of cosh(log x), which simplifies to (x + x^(-1))/2.

12 Hard

12 of 37

Analyzing +4 | -1

\frac{1}{{{e^{3x}}}}\left( {{e^x} + {e^{5x}}} \right) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots

, then

2{a_1} + {2^3}{a_3} + {2^5}{a_5} + \ldots

is equal to

View answer and explanation

Correct option: D

Option D (0) is correct because when the given expression is simplified, the coefficients of the odd powers of x are zero, resulting in the sum being zero.

13 Hard

13 of 37

Application +4 | -1

The coefficient of

{x^n}

in the expansion of

\frac{{{e^{7\;x}} + {e^x}}}{{{e^{3\;x}}}},

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the coefficient of x^n in the expansion is

\frac{{{{4^n}} + {{\left( { - 2} \right)}^n}}}{{n\\;!}}

, which doesn't match any of the given options.

14 Hard

14 of 37

Analyzing +4 | -1

a,\;b,\;c,\;d

are in HP, then

View answer and explanation

Correct option: C

Both (a) and (b)

Option D (None of these) is correct because the correct sum of the series is

e^2 - 1

, which is not among the given options.

15 Hard

15 of 37

Analyzing +4 | -1

a = \sum \limits_{n = 0}^\infty \frac{{{x^{3\;n}}}}{{\left( {3\;n} \right)\;!}},b = \sum \limits_{n = 1}^\infty \frac{{{x^{3\;n - 2}}}}{{\left( {3\;n - 2} \right)\;!}}

and

c = \sum \limits_{n = 1}^\infty \frac{{{x^{3\;n - 1}}}}{{\left( {3n - 1} \right)\;!}},

then the value of

{a^3} + {b^3} + {c^3} - 3\;abc

View answer and explanation

Correct option: A

Option A (1) is correct because the expression

{a^3} + {b^3} + {c^3} - 3abc

can be factored, and using properties of complex numbers and exponential functions, it simplifies to 1.

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16 Hard

16 of 37

Analyzing +4 | -1

The sum of the series

1 + \frac{{{1^2} + {2^2}}}{{2\;!}} + \frac{{{1^2} + {2^2} + {3^2}}}{{3\;!}} + \frac{{{1^2} + {2^2} + {3^2} + {4^2}}}{{4\;!}} + \ldots ,

View answer and explanation

Correct option: B

\frac{{17}}{6}e

Option B (

\frac{{17}}{6}e

) is correct because it is found by expressing the general term of the series, and then using the power series expansion of

e^x

17 Hard

17 of 37

Application +4 | -1

In an AP, if

s_n

is the sum of the first

n

terms (

n

is odd) and

s_n^{\prime}

is the sum of the first

n

odd terms, then

s_n / s_n^{\prime}=

View answer and explanation

Correct option: A

2 n / n+1

Option A (

2 n / n+1

) is correct because it is derived by finding the ratio of the sum of the first n terms to the sum of the first n odd terms of the AP.

18 Hard

18 of 37

Analyzing +4 | -1

The correct statement is

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because each of the given series summations are correctly derived and stated.

19 Hard

19 of 37

Analyzing +4 | -1

0 < \phi < \frac{\pi }{2},\;x = \sum \limits_{n = 0}^\infty {\cos ^{2n}}\phi ,\;y = \sum \limits_{n = 0}^\infty {\sin ^{2n}}\phi

and

z = \sum \limits_{n = 0}^\infty {\cos ^{2n}}\phi {\sin ^{2n}}\phi

, then

View answer and explanation

Correct option: B

xyz = xy + z

Option B (

xyz = xy + z

) is correct because it is derived by expressing

x

y

, and

z

in terms of

\sin^2{\phi}

and

\cos^2{\phi}

and then simplifying the equation.

20 Hard

20 of 37

Analyzing +4 | -1

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

are in AP with common difference

d \ne 0

, then

\left( {\sin d} \right)\left[ {\sec {a_1}\sec {a_2} + \sec {a_2}\sec {a_3} + ... + \sec {a_{n - 1}}\sec {a_n}} \right]

is equal to

View answer and explanation

Correct option: C

\tan {a_n} - \tan {a_1}

Option C (

\tan {a_n} - \tan {a_1}

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

21 Hard

21 of 37

Application +4 | -1

If the numbers

{3^{2a - 1}},14,{3^{4 - 2a}}(0 < a < 1)

are the first three terms of an AP, then its fifth term is equal to

View answer and explanation

Correct option: C

Option C (53) is correct because by solving for 'a' and finding the common difference, the fifth term of the AP can be determined.

22 Hard

22 of 37

Application +4 | -1

If the sum of first

n

terms of an AP is

c n^2

, then the sum of the squares of these terms is

View answer and explanation

Correct option: C

\left[n\left(4 n^2-1\right) c^2\right] / 3

Option C (

\left[n\left(4 n^2-1\right) c^2\right] / 3

) is correct because it is derived by finding the nth term of the AP and then summing the squares of the terms.

23 Hard

23 of 37

Analyzing +4 | -1

a,\;b,\;c

are in GP and

\log a - \log 2b,\;\log 2b,\log 2b - \log 3c

and

\log 3c - \log a\;

are in AP, then

a,\;b,\;c

are the length of the sides of a triangle which is

View answer and explanation

Correct option: B

Obtuse angled

Option B (Obtuse angled) is correct because the cosine of the largest angle is negative, indicating an obtuse angle.

24 Hard

24 of 37

Analyzing +4 | -1

\log \left( {1 - x + {x^2}} \right) = {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \ldots ,

and

n

is not a multiple of 3 then

{a_n}

is equal to

View answer and explanation

Correct option: B

\frac{{{{\left( { - 1} \right)}^n}}}{n}

Option B (

\frac{{{{\left( { - 1} \right)}^n}}}{n}

) is correct because it represents the coefficient of

x^n

in the logarithmic expansion when n is not a multiple of 3.

25 Hard

25 of 37

Application +4 | -1

If the set of natural numbers is partitioned into subsets

{S_1} = \left\{ 1 \right\},\;\;{S_2} = \{ 2,\;3),\;\;{S_3} = \left\{ {4,\;5,\;6} \right\}

and so on. Then, the sum of the terms in

{S_{50}}

View answer and explanation

Correct option: A

62525

Option A (62525) is correct because the sum of the terms in

S_{50}

is calculated by finding the first term and using the arithmetic series formula.

26 Hard

26 of 37

Analyzing +4 | -1

A natural number has prime factorisation given by

n = {2^x}{3^y}{5^z}

, where

y

and

z

are such that

y + z = 5

and

{y^{ - 1}} + {z^{ - 1}} = \frac{5}{6}\,

y > z

. Then, the number of odd divisors of

n

, including 1, is

View answer and explanation

Correct option: D

Option D (12) is correct because the number of odd divisors is calculated by (y+1)(z+1) where y and z are the powers of the odd prime factors 3 and 5 respectively.

27 Hard

27 of 37

Analyzing +4 | -1

The sum of the products of the numbers

\pm 1, \pm 2, \ldots \pm n

, taken two at a time is

View answer and explanation

Correct option: C

\frac{{ - n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}

Option C (

\frac{{ - n(n + 1)(2n + 1)}}{6}

) is correct because the sum of the products of the numbers is derived from the square of the sum of the numbers, accounting for the squares of individual numbers.

28 Hard

28 of 37

Application +4 | -1

x,y,z

are three consecutive positive integers, then

\frac{1}{2}{\log _e}x + \frac{1}{2}{\log _e}z + \frac{1}{{2\;x\;z + 1}} + \frac{1}{3}{\left( {\frac{1}{{2\;x\;z + 1}}} \right)^3} + \ldots

is equal to

View answer and explanation

Correct option: B

{\log _e}y

Option B (

{\log _e}y

) is correct because by using the properties of logarithms and the given condition that x, y, and z are consecutive integers, the expression simplifies to

{\log _e}y

29 Hard

29 of 37

Application +4 | -1

The

m

th term of an

\mathrm{AP}

a

and its

n

th term is

b

, and

m \neq n

. Then the sum of the first

(m+n)

terms of the

\mathrm{AP}

View answer and explanation

Correct option: A

\frac{m+n}{2}\left[a+b+\frac{a-b}{m-n}\right]

Option A is correct because the sum of the first (m+n) terms of an AP is derived using the formulas for the nth term and the sum of n terms of an AP.

30 Hard

30 of 37

Application +4 | -1

The sum of

1 \cdot n+2 \cdot(n-1)+3 \cdot(n-2)+\ldots n \cdot(1) \, \forall \, n \in \mathrm{N}

View answer and explanation

Correct option: C

\frac{n(n+1)(n+2)}{6}

Option C (

\frac{{n(n+1)(n+2)}}{{6}}

) is correct because it represents the sum of the series after simplification of the rth term and summation.

Revision notes and key takeaways

Review SEQUENCE OF SERIES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 37 single-correct-answer MCQs on SEQUENCE OF SERIES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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