The sum of $n$ terms of the series $1+2+5+12+25+46+\ldots$ is

$\frac{1}{24} n(n+1)\left(n^2-3 n+8\right)$

If $\left| a \right| < 1,\;b = \mathop \sum \limits_{k = 1}^\infty \frac{{{a^k}}}{k}$, then $a$ is equal to

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{k!}}$

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^k}{b^k}}}{k}$

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^k}{b^k}}}{{\left( {k - 1} \right)!}}$

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{\left( {k + 1} \right)!}}$

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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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Showing 931-960 of 998 questions

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931 Easy

931 of 998

Application +4 | -1

If the second term of a GP is 2 and the sum of its infinite terms is 8 , then its first term is

View answer and explanation

Correct option: D

Option D (4) is correct because by using the given information about the second term and the sum of the infinite terms, we can set up a system of equations and solve for the first term, which equals 4.

932 Easy

932 of 998

Application +4 | -1

Let

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

be a G.P. If

\frac{a_3}{a_1}=25

, then

\frac{a_9}{a_5}

equals

View answer and explanation

Correct option: C

5^4

Option C (

5^4

) is the correct answer because, in a geometric progression, the ratio of terms equidistant from each other is constant, and this ratio can be calculated from the given information.

933 Easy

933 of 998

Application +4 | -1

Sum of the series

\sum\limits_{{r=1}}^{2016}(-1)^r(a+r d)

View answer and explanation

Correct option: D

1008 d

Option D (

1008 d

) is correct because the sum simplifies to

1008d

after evaluating the series.

934 Easy

934 of 998

Application +4 | -1

\log _5 2, \log _5\left(2^x-3\right)

and

\log _5\left(\frac{17}{2}+2^{x-1}\right)

are in

A P

, then the value of

x

View answer and explanation

Correct option: C

Option C (3) is correct because it is the solution obtained by setting up and solving the equation derived from the arithmetic progression property.

935 Easy

935 of 998

Application +4 | -1

a,b,c,d,e,f

are A.M.’s between 2 and 12, then

a + b + c + d + e + f

is equal to

View answer and explanation

Correct option: B

Option B (42) is correct because the sum of the arithmetic means is calculated using the average of the first and last term.

936 Easy

936 of 998

Application +4 | -1

Let

a_1, a_2, a_3, \ldots, a_n \ldots

be an A.P. whose common difference is an integer and

S_{n}

denote the sum of its

n

terms. If

a_1=1, a_n=300

and

15 \leq n \leq 50

, then the ordered pair

\left(S_{n-4}, a_{n-4}\right)

is equal to

View answer and explanation

Correct option: A

(2490,248)

Option A (2490, 248) is correct because it is derived from finding the common difference and then calculating

S_{n-4}

and

a_{n-4}

937 Easy

937 of 998

Application +4 | -1

The number which should be added to the numbers 2, 14, 62, so that the resulting numbers may be in GP, is

View answer and explanation

Correct option: B

Option B (2) is correct because adding 2 to the numbers 2, 14, and 62 results in a geometric progression (4, 16, 64).

938 Medium

938 of 998

Application +4 | -1

{2^x}{.9^{2x + 3}} = {7^{x + 5}},

then

x =

View answer and explanation

Correct option: C

\frac{{5\log 7 - 6\log 3}}{{\log 162 - \log 7}}

Option C (

\frac{{5\log 7 - 6\log 3}}{{\log 162 - \log 7}}

) is correct because taking the logarithm of both sides and solving for x results in the given expression.

939 Easy

939 of 998

Application +4 | -1

Three numbers are in GP such that their sum is 38 and their product is 1728. The greatest number among them is

View answer and explanation

Correct option: A

Option A (18) is correct because it is the largest number in the geometric progression that satisfies the given sum and product conditions.

940 Easy

940 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

\frac{16}{5} \mathrm{~m}

, then

m

equal to

View answer and explanation

Correct option: D

101

Option D (101) is correct because the sum of the given series can be simplified and calculated using the formula for the sum of squares of natural numbers.

941 Easy

941 of 998

Application +4 | -1

\log \left( {x + y} \right) = \log 2 + \frac{1}{2}\log x + \frac{1}{2}\log y,\;\mathrm{then}

View answer and explanation

Correct option: B

x - y = 0

Option B (

x - y = 0

) is correct because it is the simplified form of the given logarithmic equation, indicating that x and y are equal.

942 Medium

942 of 998

Application +4 | -1

The

\operatorname{sum} \frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+\ldots

upto 11 -terms is:

View answer and explanation

Correct option: C

\frac{11}{2}

Option C (

\frac{11}{2}

) is correct because the series can be expressed as a telescoping series, allowing for simplification and direct calculation.

943 Medium

943 of 998

Application +4 | -1

Sum of the series

0.5+0.55+0.555+\ldots

upto

n

terms is

View answer and explanation

Correct option: D

\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right]

Option D is correct because the sum of the series can be expressed and simplified to

\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right]

944 Easy

944 of 998

Application +4 | -1

{a^{1/x}} = {b^{1/y}} = {c^{1/z}}

and

a,b,c

are in geometrical progression, then

x,y,z

are in

View answer and explanation

Correct option: A

Option A is correct because if

{a^{1/x}} = {b^{1/y}} = {c^{1/z}}

and

a,b,c

are in GP, then

x, y, z

must be in AP.

945 Medium

945 of 998

Application +4 | -1

The sum of

n

terms of the series

1+2+5+12+25+46+\ldots

View answer and explanation

Correct option: C

\frac{1}{12} n(n+1)\left(n^2-3 n+8\right)

Option C (

\frac{1}{12} n(n+1)(n^2-3 n+8)

) is correct because the sum of n terms of the series is derived to be in this form by finding the nth term and summing it.

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946 Easy

946 of 998

Application +4 | -1

The sum of

{15^2} + {16^2} + {17^2} + ... + {30^2}

is equal to

View answer and explanation

Correct option: B

8440

Option B (8440) is correct because it is the result of calculating the sum of squares from 15 to 30 by subtracting the sum of squares from 1 to 14 from the sum of squares from 1 to 30.

947 Easy

947 of 998

Application +4 | -1

An AP consists of 23 terms. If the sum of the three terms in the middle is 141 and the sum of the last three terms is 261, then the first term is

View answer and explanation

Correct option: D

Option D is correct because by setting up equations based on the given sums of terms in the AP and solving the system of equations, the first term can be determined to be 3.

948 Easy

948 of 998

Application +4 | -1

{A_1},{A_2}

are two A ∙ M’s between

a

and

b

and

{G_1},{G_2}

are two GM’s between

a

and

b

then

\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}}

View answer and explanation

Correct option: A

\frac{{a + b}}{{ab}}

Option A (

\frac{{a + b}}{{ab}}

) is correct because it is derived from the properties of Arithmetic and Geometric Progressions.

949 Easy

949 of 998

Application +4 | -1

H

be the H.M. between

a

and

b

, then the value of

\frac{H}{a} + \frac{H}{b},

View answer and explanation

Correct option: A

Option A (2) is correct because if

H

is the H.M. between

a

and

b

, then

\frac{H}{a} + \frac{H}{b} = 2

950 Easy

950 of 998

Application +4 | -1

Geometric mean of

7,{7^2},{7^3}, \ldots {7^n}

View answer and explanation

Correct option: A

{7^{\frac{{n + 1}}{2}}}

Option A (

{7^{\frac{{n + 1}}{2}}}

) is correct because the geometric mean of the given series is calculated by taking the nth root of the product of the terms.

951 Easy

951 of 998

Analyzing +4 | -1

\left| a \right| < 1,\;b = \sum \limits_{k = 1}^\infty \frac{{{a^k}}}{k}

, then

a

is equal to

View answer and explanation

Correct option: B

\sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{k!}}

Option B (

\sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{k!}}

) is correct because it represents the Maclaurin series expansion of

a

in terms of

b

, derived from

b = -\log(1-a)

952 Medium

952 of 998

Application +4 | -1

The lengths of three unequal edges of a rectangular solid block are in GP. The volume of the block is

216 \mathrm{~cm}^3

and the total surface area is

252 \mathrm{~cm}^2

. The length of the longest edge is

View answer and explanation

Correct option: A

12 \mathrm{~cm}

Option A (

12 \mathrm{~cm}

) is correct because, given the volume and surface area, the longest edge of the rectangular solid block in GP is calculated to be 12 cm.

953 Easy

953 of 998

Application +4 | -1

Let

a_{1}, a_{2}, a_{3}, \ldots

be an A.P, such that

\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+a_{3}+\ldots+a_{q}}=\frac{p^{3}}{q^{3}} ; p \neq q

. Then

\frac{a_{6}}{a_{21}}

is equal to:

View answer and explanation

Correct option: B

\frac{31}{121}

Option B (31/121) is correct because the ratio a6/a21 is calculated by using the given ratio of sums and properties of arithmetic progressions.

954 Easy

954 of 998

Application +4 | -1

If the 7th term of HP is

\frac{1}{{10}}

and the 12th term is

\frac{1}{{25}}

, then the 20th term is

View answer and explanation

Correct option: C

\frac{1}{{49}}

Option C (

\frac{1}{{49}}

) is correct because the 20th term of the Harmonic Progression is found by first finding the corresponding Arithmetic Progression and then inverting the result.

955 Medium

955 of 998

Application +4 | -1

The sum of the series

S=1^2-2^2+3^2-4^2+\ldots-2002^2+2003^2

View answer and explanation

Correct option: A

2007006

Option A (2007006) is correct because the sum of the series can be simplified by pairing terms and using the formula for the sum of integers.

956 Medium

956 of 998

Application +4 | -1

Let the first term

a

and the common ratio

r

of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033 , then the sum of these three terms is equal to

View answer and explanation

Correct option: B

231

Option B (231) is correct because by factoring the given sum of squares and solving for the first term and common ratio, the sum of the first three terms of the GP is found to be 231.

957 Medium

957 of 998

Application +4 | -1

Sum to 30 terms of the series

\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots

View answer and explanation

Correct option: B

\frac{495}{1984}

Option B (

\frac{{n}}{{2}} \log \left(\frac{{a^{{n+3}}}}{{b^{{n-1}}}}\right)

) is correct. By using the properties of logarithms and summing the series, we get the result.

958 Easy

958 of 998

Application +4 | -1

{a_1} = {a^2} = 2,\;{a_n} = {a_{n - 1}} - 1\left( {n > 2} \right),\;\mathrm{then}\;{a_5}

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because, given the recursive formula,

a_5

is calculated to be -1.

959 Easy

959 of 998

Application +4 | -1

{\log _e}2.{\log _x}27 = {\log _{10}}8.{\log _e}10,

then

x =

View answer and explanation

Correct option: B

Option B (3) is correct because by simplifying the given equation using logarithm properties, we find that x = 3.

960 Easy

960 of 998

Application +4 | -1

{a_1},\;{a_2}, \ldots ,\;{a_{n + 1}}

are in AP, then

\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ... + \frac{1}{{{a_n}{a_{n + 1}}}}

View answer and explanation

Correct option: D

\frac{n}{{{a_1}{a_{n + 1}}}}

Option D is correct because the given expression can be simplified using the properties of arithmetic progressions, resulting in the formula n/(a₁aₙ₊₁).

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