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

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

$\frac{{n{{\left( {n + 1} \right)}^2}}}{4}$

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

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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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571 Easy

571 of 998

Application +4 | -1

The sum of

1^3+2^3+3^3+4^3+\ldots+15^3

, is

View answer and explanation

Correct option: C

14400

Option C (14400) is correct because it is the direct application of the formula for the sum of the cubes of the first n natural numbers.

572 Easy

572 of 998

Analyzing +4 | -1

a_1, a_2, a_3, \ldots ., a_{2 n+1}

are in A.P., then

\frac{a_{2 n+1}-a_1}{a_{2 n+1}+a_1}+\frac{a_{2 n}-a_2}{a_{2 n}+a_2}+\ldots .+\frac{a_{n+2}-a_n}{a_{n+2}+a_n}

is equal to

View answer and explanation

Correct option: A

\frac{n(n+1)}{2} \cdot \frac{a_2-a_1}{a_{n+1}}

Option A (

\frac{n(n+1)}{2} \cdot \frac{a_2-a_1}{a_{n+1}}

) is correct because the given expression simplifies to this form using the properties of an arithmetic progression.

573 Easy

573 of 998

Analyzing +4 | -1

Let

V_r

denote the sum of first

r

terms of an A.P. whose first term is

r

and common difference is

(2 r-1)

. Then, the sum

V_1+V_2+V_3+\ldots+V_n

, is

View answer and explanation

Correct option: B

\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)

Option B (

\frac{1}{12} n(n+1)(3 n^2+n+2)

) is correct because it is the simplified expression for the sum of the first n terms of the series, derived by summing the individual V_r terms.

574 Easy

574 of 998

Analyzing +4 | -1

Let

\left[ x \right]

denote greatest integer less than or equal to

x

. If for

n \in N,\,{\left( {1 - x + {x^3}} \right)^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}}

, then

\sum\limits_{j = 0}^{\left[ {\frac{{3n}}{2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {\frac{{3n - 1}}{2}} \right]} {{a_{2j + 1}}}

is equal to:

View answer and explanation

Correct option: A

Option A (1) is correct because by substituting x=1 and x=-1 into the given equation and adding/subtracting, we can isolate the required summation and find its value.

575 Easy

575 of 998

Application +4 | -1

The eighth term of a geometric progression is 128 and common ratio is 2. The product of the first five terms is

View answer and explanation

Correct option: B

4^5

Option B (

4^5

) is correct because, with a=1 and r=2, the product of the first five terms of the geometric progression is

2^{10}

, which equals

4^5

576 Easy

576 of 998

Analyzing +4 | -1

\left| x \right| < 1

and

\left| y \right| < 1,

the sum to infinity of the sequence

x + y,\left( {{x^2} + xy + {y^2}} \right),\left( {{x^3} + {x^2}y + {y^3}} \right), \ldots ,

View answer and explanation

Correct option: A

\frac{{x + y - xy}}{{1 - x - y + xy}}

Option A (

\frac{{x + y - xy}}{{1 - x - y + xy}}

) is correct because it represents the sum to infinity of the given sequence under the conditions

|x| < 1

and

|y| < 1

577 Easy

577 of 998

Analyzing +4 | -1

The sum of series

\frac{1}{{1.2}} - \frac{1}{{2.3}} + \frac{1}{{3.4}} - \ldots \infty

is equal to

View answer and explanation

Correct option: D

{\log _e}\left( {\frac{4}{e}} \right)

Option D (

{\log _e}\left( {\frac{4}{e}} \right)

) is correct because the sum of the given series can be simplified to this logarithmic expression.

578 Easy

578 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.

579 Easy

579 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.

580 Easy

580 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

581 Easy

581 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.

582 Easy

582 of 998

Analyzing +4 | -1

The sum of the first

n

terms of the series

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

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

where 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 it represents the sum of the series when n is odd, derived by adding the (2m+1)th term to the sum of 2m terms.

583 Easy

583 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.

584 Easy

584 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.

585 Easy

585 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.

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586 Easy

586 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.

587 Easy

587 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.

588 Easy

588 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.

589 Easy

589 of 998

Application +4 | -1

If the 3rd term of a G.P. is 4, then the product of its first 5 terms is

View answer and explanation

Correct option: B

1024

Option B (1024) is correct because the product of the first 5 terms of a G.P. with a given 3rd term can be expressed in terms of that term.

590 Easy

590 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.

591 Easy

591 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.

592 Easy

592 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.

593 Easy

593 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.

594 Easy

594 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

595 Easy

595 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.

596 Easy

596 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.

597 Easy

597 of 998

Analyzing +4 | -1

Consider the pattern shown below:

\begin{array}{cccccccccccccccccccc} {{\text{Row}}}&{\text{1}}&{\text{1}}&{}&{}&{} \\ {{\text{Row}}}&{\text{2}}&{\text{3}}&{\text{5}}&{}&{} \\ {{\text{Row}}}&{\text{3}}&{\text{7}}&{\text{9}}&{{\text{11}}}&{} \\ {{\text{Row}}}&{\text{4}}&{{\text{13}}}&{{\text{15}}}&{{\text{17}}}&{{\text{19,}}\;\,{\text{etc}}{\text{.}}} \end{array}

The number at end of row 60 is

View answer and explanation

Correct option: A

3659

Option A (3659) is correct because the number at the end of row 60 is calculated using the formula derived from the pattern:

T_n = n^2 + n - 1

598 Easy

598 of 998

Analyzing +4 | -1

a, b, c

are three distinct real numbers in G.P. If

a+b+c=x b

, then

View answer and explanation

Correct option: A

x < -1

or,

x > 3

Option A (

x < -1

or,

x > 3

) is correct because the condition for

a, b, c

to be distinct real numbers in G.P. leads to the inequality

(x-3)(x+1) > 0

, which gives

x < -1

x > 3

599 Easy

599 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.

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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