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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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601 Easy

601 of 998

Application +4 | -1

a

is the A.M. of

b

and

c

and the two geometric means are

G_1

and

G_2

are inserted between

b

and

c

such that

G_1^3+G_2^3=\lambda a b c

, then

\lambda=

View answer and explanation

Correct option: B

Option B (2) is correct because, after inserting the geometric means and using the given condition, the value of lambda is found to be 2.

602 Easy

602 of 998

Application +4 | -1

Three numbers form an increasing G.P. If the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P. is

View answer and explanation

Correct option: B

2+\sqrt{3}

Option D (-13.5) is correct because solving for x and finding the common ratio leads to the fourth term being -13.5.

603 Easy

603 of 998

Application +4 | -1

There are four numbers of which the first three are in G.P. and the last three are in A.P. with common difference 9. If the first and last number are equal, the sum of the four numbers is

View answer and explanation

Correct option: B

Option B (21) is correct because by setting up the G.P. and A.P. conditions, we find the value of 'a' and then sum the four numbers.

604 Easy

604 of 998

Analyzing +4 | -1

The sum of the series

\frac{{{1^2}}}{{3!}} + \frac{{{2^2}}}{{4!}} + \frac{{{3^2}}}{{5!}} + \ldots \mathrm{to}\infty

equals

View answer and explanation

Correct option: C

2e - 5

Option C (

2e - 5

) is correct because the series can be manipulated to relate to the exponential function's expansion.

605 Easy

605 of 998

Application +4 | -1

If the third term of a GP is

P

. Then, the product of the first 5 terms of the GP is

View answer and explanation

Correct option: D

{P^5}

Option D (

{P^5}

) is correct because the product of the first 5 terms of a GP with the third term as P is equal to

{P^5}

606 Easy

606 of 998

Application +4 | -1

a_1, a_2, a_3, \ldots .

. is an A.P. such that

a_{10}=\frac{1}{20}

and

a_{20}=\frac{1}{10}

, then the sum of first 200 terms is

View answer and explanation

Correct option: D

100 \frac{1}{2}

Option D is correct because by finding the common difference and the first term of the A.P., the sum of the first 200 terms can be calculated, resulting in

100 \frac{1}{2}

607 Easy

607 of 998

Application +4 | -1

If an infinite geometric series the first term is a and common ratio is

r

. If the sum of the series is 4 and the second term is 3/4, then

\left( {a,r} \right)

View answer and explanation

Correct option: D

\left( {3,\;1/4} \right)

Option D ((3, 1/4)) is correct because the values of a and r satisfy both the sum of the infinite geometric series and the value of the second term.

608 Easy

608 of 998

Application +4 | -1

If 3 arithmetic means, 3 geometric means and 3 harmonic means are inserted between 1 and 5 , then the cubic equation whose roots are first A.M., second G.M. and third H.M. between 1 and 5 , is

View answer and explanation

Correct option: A

x^3-\left(\frac{9}{2}+\sqrt{5}\right) x^2+\left(\frac{9 \sqrt{5}}{2}+5\right) x-5 \sqrt{5}=0

Option A (

x^3-(\frac{9}{2}+\sqrt{5}) x^2+(\frac{9 \sqrt{5}}{2}+5) x-5 \sqrt{5}=0

) is correct because it is the cubic equation formed with the first A.M., second G.M., and third H.M. between 1 and 5 as its roots.

609 Easy

609 of 998

Application +4 | -1

The value of

{e^{{{\log }_{10}}\,\,\tan \,\,1^\circ \,\, + \,\,{{\log }_{10}}\,\,\tan \,\,2^\circ \, + \,\,{{\log }_{10}}\,\,3^\circ + \ldots + \,\,{{\log }_{10}}\,\,\tan \,\,89^\circ }}

, is

View answer and explanation

Correct option: D

\mathrm{1}

Option D (

\mathrm{1}

) is correct because the expression simplifies to

e^0

, which equals 1.

610 Easy

610 of 998

Application +4 | -1

Let

T_r

be the

r

th term of an AP, for

r=1

2, \ldots

. If for some positive integers

m

and

n

, we have

T_m=\frac{1}{n}

and

T_n=\frac{1}{m}

, then

T_{m+n}

equals

View answer and explanation

Correct option: C

\frac{1}{m}

Option C (

\frac{1}{m}

) is correct because by using the properties of arithmetic progressions, we can determine that

T_{m+n} = \frac{1}{m}

611 Hard

611 of 998

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.

612 Easy

612 of 998

Understanding +4 | -1

1+a+a^2+a^3+\ldots+a^{x-1}=(1+a)\left(1+a^2\right)\left(1+a^4\right)

, then

x=

View answer and explanation

Correct option: C

Option C (8) is correct because by multiplying (1-a) on both sides and using the difference of squares repeatedly, we can simplify the right side to 1 - a⁸, which implies x = 8.

613 Medium

613 of 998

Application +4 | -1

In a certain AP, 5 times the 5 th term is equal to 8 times the 8 th term, then 13 th term is

View answer and explanation

Correct option: D

Option D (0) is correct because if 5 times the 5th term equals 8 times the 8th term in an AP, the 13th term must be 0.

614 Easy

614 of 998

Application +4 | -1

{2^{{{\log }_{10}}3\sqrt 3 }} = {3^{k\log _{10}^2\;}},

then

k =

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{3}{2}

) is correct because, by taking logarithms and simplifying, the value of k is found to be

\frac{3}{2}

615 Easy

615 of 998

Application +4 | -1

The sum of

n

terms of three AP’s whose first term is 1 and common differences are 1, 2, 3 respectively are

{S_1},\;{S_2},\;{S_3}

respectively. The true relation is

View answer and explanation

Correct option: B

{S_1} + {S_3} = 2\;{S_2}

Option B (

{S_1} + {S_3} = 2\;{S_2}

) is correct because the sum of the first and third AP's is twice the sum of the second AP.

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616 Easy

616 of 998

Application +4 | -1

The value of

n

for which

\frac{{{x^{n + 1}} + {y^{n + 1}}}}{{{x^n} + {y^n}}}

is the geometric mean of

x

and

y

View answer and explanation

Correct option: A

n = - \frac{1}{2}

Option A (

n = - \frac{1}{2}

) is correct because substituting this value into the given expression and simplifying leads to the geometric mean of x and y.

617 Medium

617 of 998

Application +4 | -1

Let

\alpha, \beta

and

\gamma

be three positive real numbers. Let

f(x)=\alpha x^5+\beta x^3+\gamma x, x \in \mathbb{R}

and

g: \mathbb{R} \rightarrow \mathbb{R}

be such that

g(f(x))=x

for all

x \in \mathbb{R}

. If

a_1, a_2, a_3, \ldots, a_n

be in arithmetic progression mean is zero, then the value of

f\left(g\left(\frac{1}{n} \sum\limits_{i=1}^n f\left(a_i\right)\right)\right)

is equal to

View answer and explanation

Correct option: A

The correct answer is A (0) because the function

f(x)

is odd, and the sum of an arithmetic progression with a mean of zero is zero, leading to

f(g(0)) = 0

618 Easy

618 of 998

Application +4 | -1

The value of

{2^{{{\log }_3}7}} - {7^{{{\log }_3}2}}

View answer and explanation

Correct option: C

Option C (0) is correct because

{2^{{{\log }_3}7}}

is equal to

{7^{{{\log }_3}2}}

, so their difference is zero.

619 Easy

619 of 998

Application +4 | -1

a,b

and

c

are in AP, then

{2^{ax + 1}},{2^{bx + 1}},{2^{cx + 1}},x \ne 0

are in

View answer and explanation

Correct option: D

The correct answer is D (GP) because if a, b, and c are in AP, then the given exponential terms form a GP regardless of the value of x.

620 Medium

620 of 998

Application +4 | -1

An infinite GP has the first term

'x'

and sum 5, then

x

belongs to

View answer and explanation

Correct option: C

0 < x < 10

Option C is correct because the condition for the sum of an infinite GP to exist requires the common ratio to be between -1 and 1, leading to the inequality 0 < x < 10.

621 Medium

621 of 998

Application +4 | -1

the sum of

n

terms of the infinite series

1 \cdot {3^2} + 2 \cdot {5^2} + 3 \cdot {7^2} + ...\infty

View answer and explanation

Correct option: A

\frac{n}{6}\left( {n + 1} \right)\left( {6{n^2} + 14n + 7} \right)

Option A (

\frac{n}{6}

\left( {n + 1} \right)\left( {6{n²} + 14n + 7} \right)) is correct because it is the correct formula for the sum of the given arithmetico-geometric series.

622 Medium

622 of 998

Application +4 | -1

S_n=n P+\frac{n}{2}(n-1) Q

, where

S_n

denotes the sum of first

n

terms of an AP, then the common difference is

View answer and explanation

Correct option: D

Q

Option D (

Q

) is correct because the common difference in an AP is the coefficient of the

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

term in the sum formula.

623 Medium

623 of 998

Application +4 | -1

Given two numbers

a

and

b

. Let

A

denote the single A.M. and

S

denote the sum of

n

A.M.’s between

a

and

b

, then

S/A

depends on

View answer and explanation

Correct option: D

n

Option D (

n

) is correct because the ratio S/A simplifies to n, independent of a and b.

624 Easy

624 of 998

Application +4 | -1

The number of divisors of

3 \times {7^3},\;7 \times {11^2}

and

2 \times 61

are in

View answer and explanation

Correct option: A

Option A (AP) is correct because the number of divisors of the given numbers forms an arithmetic progression with a common difference of -2.

625 Medium

625 of 998

Application +4 | -1

x>0

and

\log _2 x+\log _2(\sqrt{x})+\log _2(\sqrt[4]{x})

+\log _2(\sqrt[8]{x})+\log _2(\sqrt[16]{x})+\ldots=4

, then

x

equals

View answer and explanation

Correct option: C

Option C (4) is correct because simplifying the logarithmic expression and solving for x gives x = 4.

626 Medium

626 of 998

Application +4 | -1

Let

{a_1},{a_2}, \ldots ,{a_{10}}

be in AP and

{h_1},{h_2}, \ldots ,{h_{10}}

be in HP. If

{a_1} = {h_1} = 2\mathrm{\;and\;}{a_{10}} = {h_{10}} = 3

then

{a_4}{h_7}

View answer and explanation

Correct option: D

Option D (6) is correct because it is the product of the 4th term of the AP and the 7th term of the HP, calculated using the properties of AP and HP.

627 Easy

627 of 998

Application +4 | -1

In a geometric progression (GP) the ratio of the sum of the first three terms and first six terms is 125:152 the common ratio is

View answer and explanation

Correct option: D

\frac{3}{5}

Option D (

\frac{3}{5}

) is correct because it is derived by setting up a ratio of sums and solving for the common ratio of the geometric progression.

628 Easy

628 of 998

Application +4 | -1

x=111 \ldots 1

(20 digits),

y=333 \ldots 3(10

digits) and

z=222 \ldots 2\left(10\right.

digits), then

\frac{x-y^2}{z}

equals

View answer and explanation

Correct option: B

Option B (1) is correct because after substituting the given values of x, y, and z and simplifying the expression, the result is 1.

629 Easy

629 of 998

Application +4 | -1

{2^{10}} = {2^9} \cdot {3^1} + {2^8} \cdot {3^2} + ..... + 2 \cdot {3^9} + {3^{10}} = S - {2^{11}},

then

S

is equal to

View answer and explanation

Correct option: B

{3^{11}}

The correct answer is B (

3^{11}

) because it is the value of S, calculated by recognizing the given expression as a geometric progression and using the formula for the sum of a GP.

630 Easy

630 of 998

Application +4 | -1

The sum of the

3^{\text {rd }}

and the

4^{\text {th }}

terms of a G.P. is 60 and the product of its first three terms is 1000 . If the first term of this G.P. is positive, then its

7^{\text {th }}

term is :

View answer and explanation

Correct option: D

320

Option D (320) is correct because after finding the first term and common ratio of the G.P., the 7th term can be calculated as 320.

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