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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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571 Easy

571 of 998

Application +4 | -1

a_1, a_2, a_3, \ldots, a_n

are in H.P. then the expression

a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n

is equal to

View answer and explanation

Correct option: D

(n-1) a_1 a_n

Option D (

(n-1) a_1 a_n

) is correct because it is derived by manipulating the relationship between consecutive terms in a harmonic progression and summing them.

572 Medium

572 of 998

Application +4 | -1

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

are in HP, then the expression

{a_1}{a_2} + {a_2}{a_3} + .. + {a_{n - 1}}{a_n}

is equal to

View answer and explanation

Correct option: C

\left( {n - 1} \right){a_1}{a_n}

Option C (

\left( {n - 1} \right){a_1}{a_n}

) is correct because, using the properties of Harmonic Progression (HP) and manipulating the terms, the expression simplifies to this result.

573 Easy

573 of 998

Analyzing +4 | -1

\frac{1}{a} + \frac{1}{c} + \frac{1}{{a - b}} + \frac{1}{{c - b}} = 0

and

b \ne a + c,

are in

View answer and explanation

Correct option: B

H.P.

Option B (H.P.) is correct because the given condition leads to the relationship

2ac = ab + bc

, which is the condition for

a, b, c

to be in Harmonic Progression.

574 Easy

574 of 998

Application +4 | -1

The sum of first three terms of an increasing A.P. is 9 and sum of their squares is 35. The sum to

n

terms of the series is

View answer and explanation

Correct option: D

n^2

Option D (

n^2

) is correct because, given the sum of the first three terms and their squares, the sum to n terms of the A.P. is

n^2

575 Easy

575 of 998

Application +4 | -1

\cos (x-y), \cos x

and

\cos (x+y)

are in HP, the

\cos x \sec \left(\frac{y}{2}\right)

is equal to

View answer and explanation

Correct option: A

\pm \sqrt{2}

Option A (

\frac{1}{2}

) is correct because the ratio of the first term to the common difference is

\frac{1}{2}

, derived from the given condition that the sum of the first ten terms is four times the sum of the first five terms.

576 Medium

576 of 998

Application +4 | -1

If the arithmetic mean of two numbers

a

and

b, a>b>0

, is five times their geometric mean, then

\frac{a+b}{a-b}

is equal to :

View answer and explanation

Correct option: D

\frac{5 \sqrt{6}}{12}

The correct answer is D (

\frac{{5 \sqrt{{6}}}}{{12}}

) because it is obtained by using the arithmetic and geometric mean relationship and applying componendo and dividendo.

577 Easy

577 of 998

Application +4 | -1

If log 2,

\log \left( {{2^n} - 1} \right)\;\mathrm{and}\log \left( {{2^n} + 3} \right)\;

are in AP, then

n

is equal to

View answer and explanation

Correct option: B

{\log _2}5

Option B (

\log _2 5

) is correct because using the properties of arithmetic progressions and logarithms, we find that n = log base 2 of 5.

578 Medium

578 of 998

Application +4 | -1

T_1, T_2, T_3, T_4, T_5, T_6, T_7, \ldots

are in AP with common difference 2 . Then, the difference of any two consecutive terms of the sequence

T_1, T_4, T_7, \ldots

View answer and explanation

Correct option: C

Option C (6) is correct because the difference between consecutive terms in the sequence

T_1, T_4, T_7, \ldots

is 3 times the common difference of the original AP.

579 Easy

579 of 998

Application +4 | -1

123. If

\sum_{n=1}^{5} \frac{1}{n(n+1)(n+2)(n+3)}=\frac{k}{3}

, then

\mathrm{k}

is equal to

View answer and explanation

Correct option: C

\frac{55}{336}

Option C (

\frac{55}{336}

) is correct because it results from correctly applying partial fraction decomposition and telescoping the series.

580 Medium

580 of 998

Application +4 | -1

If the AM of two numbers be

A

and GM be

G

, then the numbers will be

View answer and explanation

Correct option: C

A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)}

Option C (

A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)}

) is correct because solving the equations for AM and GM leads to this expression for the two numbers.

581 Easy

581 of 998

Application +4 | -1

The value of

{\log _2}\left[ {{{\log }_2}\left\{ {{{\log }_3}\left( {{{\log }_3}{{27}^3}} \right)} \right\}} \right]

View answer and explanation

Correct option: B

Option B (0) is correct because after simplifying the nested logarithms, the expression evaluates to 0.

582 Medium

582 of 998

Application +4 | -1

\sum \limits_{n = 1}^\infty \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots \left( {n + k} \right)}}

is equal to

View answer and explanation

Correct option: C

\frac{1}{{\left( {k - 1} \right)k!}}

Option A (

\frac{1}{{\left( {k - 1} \right)\left( k! \right)}}

) is correct because it is the result of the infinite summation of the given series using partial fraction decomposition.

583 Medium

583 of 998

Application +4 | -1

The set of real values of

x

satisfying

|x-1| \leq 3

and

|x-1| \geq 1

View answer and explanation

Correct option: C

[-2,0] \cup[2,4]

Option C ([-2,0]

\cup

[2,4]) is correct because it represents the intersection of the solution sets of the two inequalities.

584 Easy

584 of 998

Application +4 | -1

If the ratio of the sum of n term of two AP’s be

\left( {7n + 1} \right):\left( {4n + 27} \right),

then the ratio of their 11th term will be

View answer and explanation

Correct option: C

4\mathrm{ }:\mathrm{ }3

Option C (4 : 3) is correct because substituting n = 21 into the ratio of the sums and simplifying gives the ratio of their 11th terms as 4/3.

585 Medium

585 of 998

Application +4 | -1

The sum of the series

1+2 \cdot 2+3 \cdot 2^2+4 \cdot 2^3+\ldots+100 \cdot 2^{99}

View answer and explanation

Correct option: B

99 \cdot 2^{100}+1

Option B,

99 \cdot 2^{100}+1

, is the correct answer because it represents the sum of the given series, which is found by manipulating the series and using the formula for the sum of a geometric progression.

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586 Easy

586 of 998

Application +4 | -1

\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \ldots \mathrm{to\;}\infty = \frac{{{\mathrm{\pi }^2}}}{6},

then

\frac{1}{{{1^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \ldots

equals

View answer and explanation

Correct option: A

{\mathrm{\pi }^2}/8

Option A (

\frac{{\pi ^2}}{8}

) is correct because it is the result of subtracting the sum of the reciprocals of even squares from the sum of the reciprocals of all squares.

587 Easy

587 of 998

Application +4 | -1

Let

A_r, r=1,2, \ldots .29

be arithmetic means between

303

and

-57

where

A_r>A_{r+1} \, \forall \,r=1,2, \ldots 28

. If

S

be the sum of these means, then the value of

\left[\frac{S}{\left(A_{14}-12\right)\left|A_r\right|_{\text {min. }}}\right]

\left[[k]\right.

denotes greatest integer less than or equal to

k

and

\left|A_r\right|_{\min }

. denotes the minimum value of

\left.\left|A_r\right| \cdot\right.

View answer and explanation

Correct option: A

Option A (9) is correct because the arithmetic means are calculated, and the expression involving the sum and minimum absolute value is evaluated to find the greatest integer less than or equal to the result.

588 Easy

588 of 998

Application +4 | -1

If three successive terms of a G.P. with common ratio

r(r>1)

from the sides of a

\triangle A B C

and

[r]

denotes greatest integer function, then

[r]+[-r]=

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because the greatest integer function values of r and -r sum to -1 under the given conditions.

589 Easy

589 of 998

Application +4 | -1

The sum of all odd numbers between

1

and

1000

which are divisible by

3

, is

View answer and explanation

Correct option: A

83667

Option A (83667) is correct because the sum of the arithmetic series 3 + 9 + 15 + ... + 999 is calculated to be 83667.

590 Easy

590 of 998

Application +4 | -1

a,b,c

are in AP,

b - a,c - b

and a are in GP, then

a:b:c

View answer and explanation

Correct option: A

1:2:3

Option A (

1:2:3

) is correct because it satisfies both the arithmetic progression and geometric progression conditions.

591 Easy

591 of 998

Application +4 | -1

The harmonic mean and geometric mean of two positive numbers are in the ratio

4: 5

, then two numbers are in the ratio

View answer and explanation

Correct option: A

4: 1

Option A (

n+1-2^{-n}

) is correct. By rewriting the series and summing the geometric progression, we get the result.

592 Easy

592 of 998

Application +4 | -1

The value of

\sqrt e

rounded off to three decimal places, is

View answer and explanation

Correct option: A

1.648

Option A (1.648) is correct because it is the value of

\sqrt{e}

rounded off to three decimal places, calculated using the series expansion of

e^x

593 Easy

593 of 998

Analyzing +4 | -1

The sum of the series

\frac{{{1^2} \cdot 2}}{{1\;!}} + \frac{{{2^2} \cdot 3}}{{2\;!}} + \frac{{{3^2} \cdot 4}}{{3\;!}} + \frac{{{4^2} \cdot 5}}{{4\;!}} + \ldots ,

View answer and explanation

Correct option: C

7e

Option C (

7e

) is correct because the series can be decomposed and simplified to a sum involving the exponential constant e.

594 Medium

594 of 998

Analyzing +4 | -1

The value of

\log 2 + 2\left( {\frac{1}{5} + \frac{1}{3} \cdot \frac{1}{{{5^3}}} + \frac{1}{5} \cdot \frac{1}{{{5^5}}} + \ldots + \infty } \right)

View answer and explanation

Correct option: D

\mathrm{log\;}3

Option D (

\mathrm{log\;}3

) is correct because the expression simplifies to

\log 3

after using the logarithmic identity

\log \left( {\frac{{1 + x}}{{1 - x}}} \right) = 2\left( {x + \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \ldots } \right)

595 Easy

595 of 998

Application +4 | -1

p,q,r,s \in N

and they are four consecutive terms of an A.P., then

p\mathrm{th},q\mathrm{th},r\mathrm{th}

and sth terms of a G.P. are in

View answer and explanation

Correct option: B

G.P.

Option B (G.P.) is correct because if p, q, r, s are in A.P., then the pth, qth, rth, and sth terms of a G.P. are also in G.P.

596 Easy

596 of 998

Application +4 | -1

Suppose

m

th term of an A.P. is

1 / n

and

n

th term of the A.P. is

1 / m

. If

r

th term of the A.P. is 1, then

r

is equal to

View answer and explanation

Correct option: A

m n

Option A (

mn

) is correct because by using the given information about the mth and nth terms, we can find the common difference and then the value of r for which the rth term is 1.

597 Medium

597 of 998

Application +4 | -1

The sum of numbers of three digits which are divisible by 7 , is

View answer and explanation

Correct option: A

70336

Option A (70336) is correct because it is the sum of all three-digit numbers divisible by 7, calculated using the arithmetic series formula.

598 Medium

598 of 998

Application +4 | -1

Let

a

and

b

be two positive real numbers. Suppose

A_1

A_2

are two arithmetic means;

G_1, G_2

are two geometric means and

{H}_1, {H}_2

are two harmonic means between

a

and

b

, then

\frac{A_1+A_2}{H_1+H_2}

is equal to

View answer and explanation

Correct option: A

\frac{2\left(a^2+b^2\right)+5 a b}{9 a b}

Option A is correct because the expression can be simplified using the formulas for arithmetic, geometric, and harmonic means.

599 Medium

599 of 998

Application +4 | -1

A person purchases

1 \mathrm{~kg}

tomatoes from each of the 4 places at the rate of

1 \mathrm{~kg}, 2 \mathrm{~kg}, 3 \mathrm{~kg}

and

4 \mathrm{~kg}

per rupee respectively. On the average, he has purchased

x \mathrm{~kg}

tomatoes per rupee, then the value of

x

View answer and explanation

Correct option: C

1.92

Option C (1.92) is correct because the average rate is best represented by the Harmonic Mean when dealing with rates per unit.

600 Easy

600 of 998

Application +4 | -1

The sum to infinity of the series

1 + \frac{4}{5} + \frac{7}{{{5^2}}} + \frac{{10}}{{{5^3}}} + \ldots ,

View answer and explanation

Correct option: C

\frac{{35}}{{16}}

Option C (35/16) is correct because it is the sum to infinity of the given arithmetico-geometric progression (AGP).

Revision notes and key takeaways

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

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