Sum of first $n$ terms in the series ${\cot ^{ - 1}}3 + {\cot ^{ - 1}}7 + {\rm{co}}{{\rm{t}}^{ - 1}}13 + {\cot ^{ - 1}}21 + \ldots $ is given by

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

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

${\tan ^{ - 1}}\left( {n + 1} \right) - {\tan ^{ - 1}}1$

The determinant $\Delta = \left| {\begin{array}{*{20}{c}} a&b&{a\alpha + b}\\ b&c&{b\alpha + c}\\ {a\alpha + b}&{b\alpha + c}&0 \end{array}} \right|$ is equal to zero, if

$\alpha $ is a root of $a{x^2} + bx + c = 0$

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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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61 Easy

61 of 998

Analyzing +4 | -1

\sum_{r=1}^{20}\left(r^2+1\right)(r !)

is equal to

View answer and explanation

Correct option: B

22 !-2(21 !)

Option B (

22 !-2(21 !)

) is the correct answer because the summation telescopes to this result after simplification.

62 Easy

62 of 998

Application +4 | -1

The

(m+n)

th and

(m-n)

th terms of a GP are

p

and

q

, respectively. Then, the

m

th term of the GP is

View answer and explanation

Correct option: B

\sqrt{p q}

Option B (

\sqrt{p q}

) is correct because the

m

th term of the GP can be found by taking the geometric mean of the

(m+n)

th and

(m-n)

th terms.

63 Easy

63 of 998

Application +4 | -1

If the sum of

m

consecutive odd integers is

m^4

, then the first integer is

View answer and explanation

Correct option: D

m^3-m+1

Option D (

m^3-m+1

) is correct because it represents the first term of the arithmetic progression derived from the given sum of consecutive odd integers.

64 Easy

64 of 998

Application +4 | -1

If for every

n \in \mathbf{N}

, sum to

n

terms of an A.P. is

5 n^2+7 n

then its 10th term is

View answer and explanation

Correct option: C

102

Option C (102) is correct because by finding the difference between the sum of the first 10 terms and the sum of the first 9 terms, we can determine the 10th term.

65 Easy

65 of 998

Analyzing +4 | -1

f\left( n \right) = \prod\limits_{i = 2}^{n - 1} {{{\log }_i}\left( {i + 1} \right),}

the value of

\sum\limits_{k = 1}^{100} {f\left( {{2^k}} \right)}

equals

View answer and explanation

Correct option: B

5050

Option B (5050) is correct because the product simplifies to a logarithm, and the sum of the first 100 integers is 5050.

66 Easy

66 of 998

Analyzing +4 | -1

y = {3^{x - 1}} + {3^{ - x - 1}}(x

real), then the least value of

y

View answer and explanation

Correct option: C

2 / 3

Option C (

2/3

) is correct because the least value of

y

\frac{2}{3}

as found using AM-GM inequality.

67 Easy

67 of 998

Application +4 | -1

m

is the A.M. of two distinct real numbers

l

and

n

(l, n>1)

and

G_1 G_2

and

G_3

are three geometric means between

l

and

n

, then

G_1{ }^4+2 G_2{ }^4+G_3{ }^4

equals

View answer and explanation

Correct option: D

4 \operatorname{lm}^2 n

Option D is correct because the expression

G_1{ }^4+2 G_2{ }^4+G_3{ }^4

simplifies to

4 l m^2 n

when

m

is the arithmetic mean and

G_1, G_2, G_3

are geometric means between

l

and

n

68 Easy

68 of 998

Application +4 | -1

If three numbers are in G.P., then the numbers obtained by adding the middle number to each of these numbers are in

View answer and explanation

Correct option: C

H.P.

Option C (H.P.) is correct because the numbers obtained by adding the middle number to each of the terms in a G.P. form a Harmonic Progression.

69 Easy

69 of 998

Application +4 | -1

Let

a_n

be the

n

th term of an AP, if

\sum\limits_{r = 1}^{100} {{a_{2r}}} = \alpha \,\,{\text{and}}\,\,\,\sum\limits_{r = 1}^{100} {{a_{2r - 1}}} = \beta ,

then the common difference of the AP is

View answer and explanation

Correct option: C

\frac{\alpha-\beta}{100}

Option C (

\frac{\alpha-\beta}{100}

) is correct because it represents the common difference of the AP, derived from the given sums of even and odd terms.

70 Easy

70 of 998

Analyzing +4 | -1

Sum of first

n

terms in the series

{\cot ^{ - 1}}3 + {\cot ^{ - 1}}7 + \mathrm{co}{\mathrm{t}^{ - 1}}13 + {\cot ^{ - 1}}21 + \ldots

is given by

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the sum of the series can be expressed in multiple equivalent forms using trigonometric identities.

71 Easy

71 of 998

Application +4 | -1

An infinite G.P. has first term '

x

' and sum '5', then

x

belongs to

View answer and explanation

Correct option: C

0 < x< 10

Option C (

0 < x< 10

) is correct because the first term 'x' must lie within this range for the infinite G.P. to have a sum of 5.

72 Easy

72 of 998

Application +4 | -1

p, x_1, x_2 \ldots x_i, \ldots

and

q, y_1, y_2 \ldots, y_i, \ldots

are in A.P., with common difference

a

and

b

respectively, then the centre of mean position of the points

A_i\left(x_i, y_i\right)

i=1,2 \ldots n

lies on the line

View answer and explanation

Correct option: C

b x-a y=b p-a q

Option C (

b x-a y=b p-a q

) is correct because the center of the mean position lies on this line, derived from the properties of arithmetic progressions.

73 Easy

73 of 998

Application +4 | -1

If the first two terms of a H.P. are

2 / 5

and

12 / 23

respectively. Then, the largest term is

View answer and explanation

Correct option: A

5 th term

Option A (5 th term) is correct because the largest term in the H.P. corresponds to the least value of 37 - 7n, which occurs at n=5.

74 Easy

74 of 998

Analyzing +4 | -1

Let

\lambda

be the greatest integer for which

5 p^2-16,2 p \lambda, \lambda^2

are distinct consecutive terms of an

A P

, where

p \in R

. If the common difference of the

A P

\left(\frac{m}{n}\right), m, n \in N

and

m, n

are relative prime, the value of

m+n

View answer and explanation

Correct option: C

143

Option C (30) is correct because it is the minimum value of the expression, obtained by applying the AM-GM inequality and completing the square.

75 Easy

75 of 998

Application +4 | -1

|\alpha|<1,|\beta|<1

1-\alpha+\alpha^2-\alpha^3+\ldots

. to

\infty=s_1

1-\beta+\beta^2-\beta^3+\ldots

\infty=s_2

, then

1-\alpha \beta+\alpha^2 \beta^2-\alpha^3 \beta^3+\ldots

\infty

equals

View answer and explanation

Correct option: C

\frac{s_1 s_2}{1-s_1-s_2+2 s_1 s_2}

Option C (

\frac{s_1 s_2}{1-s_1-s_2+2 s_1 s_2}

) is correct because it correctly expresses the sum of the series in terms of

s_1

and

s_2

after substituting

\alpha

and

\beta

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76 Easy

76 of 998

Application +4 | -1

{\log _6}\left\{ {{{\log }_4}\left( {\sqrt {x + 4} + \sqrt x } \right)} \right\} = 0,

then

x =

View answer and explanation

Correct option: D

\frac{9}{4}

Option D (

\frac{9}{4}

) is correct because solving the logarithmic equation by converting it to exponential form and simplifying leads to x = 9/4.

77 Easy

77 of 998

Application +4 | -1

Fifth term of an GP is 2, then the product of its 9 term is

View answer and explanation

Correct option: B

512

Option B (512) is correct because the product of the 9 terms of the GP, given the 5th term is 2, simplifies to 2 raised to the power of 9, which equals 512.

78 Easy

78 of 998

Application +4 | -1

A student read common difference of an AP as -3 instead of 3 and obtained the sum of first 10 terms as -30. Then, the actual sum of first 10 terms is equal to

View answer and explanation

Correct option: A

240

Option A (240) is correct because by using the incorrect common difference to find the first term and then calculating the sum with the correct common difference, we obtain 240.

79 Easy

79 of 998

Application +4 | -1

Sum of the series

\sum\limits_{{r=1}}^n r \log \frac{r+1}{r}

View answer and explanation

Correct option: A

\log \frac{(n+1)^n}{n !}

Option A (

\log \frac{(n+1)^n}{n !}

) is correct because it is the simplified form of the sum of the given series, derived using logarithmic properties and telescoping.

80 Easy

80 of 998

Application +4 | -1

The value of

1 - {\log _e}2 + \frac{{{{\left( {{{\log }_e}2} \right)}^2}}}{{2\;!}} - \frac{{{{\left( {{{\log }_e}2} \right)}^3}}}{{3\;!}} + \ldots ,

View answer and explanation

Correct option: B

\frac{1}{2}

Option B is correct because the given expression is the Taylor series expansion of e^(-log(2)), which simplifies to 1/2.

81 Easy

81 of 998

Analyzing +4 | -1

The determinant

\Delta = \left| {\begin{array}{cccccccccccccccccccc} a&b&{a\alpha + b}\\ b&c&{b\alpha + c}\\ {a\alpha + b}&{b\alpha + c}&0 \end{array}} \right|

is equal to zero, if

View answer and explanation

Correct option: B

a,b,c

are in G.P.

Option B (

a,b,c

are in G.P.) is correct because the determinant simplifies to zero when

b^2 = ac

, which is the condition for a geometric progression.

82 Easy

82 of 998

Analyzing +4 | -1

a,b,c

are in A.P. as well as in G.P. then

View answer and explanation

Correct option: D

a = b = c

Option D (

a = b = c

) is correct because if

a, b, c

are in both A.P. and G.P., then they must be equal.

83 Easy

83 of 998

Application +4 | -1

Let

S_n

denote the sum of first

n

terms of an A.P.If

S_{2 n}=3 S_n

then the ratio

S_{3 n} / S_n

is equal to

View answer and explanation

Correct option: B

The correct answer is 6 because the ratio

S_{3n} / S_n

is calculated using the given condition

S_{2n} = 3S_n

and the formula for the sum of an A.P.

84 Easy

84 of 998

Application +4 | -1

{a_1},{a_2},{a_3}, \ldots ,{a_n}

be an AP of non-zero terms, then

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

is equal to

View answer and explanation

Correct option: A

\frac{{n - 1}}{{{a_1}{a_n}}}

Option A,

\frac{{n - 1}}{{{a_1}{a_n}}}

, is correct because it represents the sum of the series of reciprocals of products of consecutive terms in an arithmetic progression.

85 Easy

85 of 998

Application +4 | -1

If the fourth power of the common difference of an A.P. with integer entries is added to the product of any four consecutive terms of it, then the resulting sum is

View answer and explanation

Correct option: C

the square of an integer

Option C (the square of an integer) is correct because the resulting sum can be expressed as a perfect square.

86 Easy

86 of 998

Analyzing +4 | -1

Let

\alpha ,\beta

are the roots of

f\left( x \right) = a{x^2} + bx + c,\;a \ne 0

and

\Delta = {b^2} - 4ac.\;

\alpha + \beta ,\;{\alpha ^2} + {\beta ^2}\;\mathrm{and}\;{\alpha ^3} + {\beta ^3}

are in GP, then

View answer and explanation

Correct option: C

c\Delta = 0

Option C (

c\Delta = 0

) is correct because it follows from the condition that

\alpha + \beta

{\alpha ^2} + {\beta ^2}

, and

{\alpha ^3} + {\beta ^3}

are in GP, leading to either

\alpha \beta = 0

\alpha = \beta

87 Easy

87 of 998

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: D

The correct answer is B (12) because the given series is an arithmetico-geometric progression, and its sum to infinity can be calculated using the appropriate formula.

88 Easy

88 of 998

Application +4 | -1

The sum to

n

terms of the series

\left( {{n^2} - {1^2}} \right) + 2\left( {{n^2} - {2^2}} \right) + 3\left( {{n^2} - {3^2}} \right) + \ldots ,

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is derived by expressing the sum as a difference of two summations and using standard summation formulas.

89 Easy

89 of 998

Application +4 | -1

p

and

q

are the two arithmetic means and

\mathrm{G}

is the geometric mean between two numbers

a

and

b

, then

\mathrm{G}^2=

View answer and explanation

Correct option: D

(2 p-q)(2 q-p)

Option D ((2p-q)(2q-p)) is correct because it accurately represents G² in terms of the arithmetic means p and q.

90 Easy

90 of 998

Application +4 | -1

b-c, 2 b-\lambda, b-a

are in

H P

, then

a-\frac{\lambda}{2}

b-\frac{\lambda}{2}, c-\frac{\lambda}{2}

are is

View answer and explanation

Correct option: B

The correct answer is GP because after the given harmonic progression is manipulated, it results in the condition for a geometric progression.

Revision notes and key takeaways

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