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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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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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481 Medium

481 of 564

Application +4 | -1

y = 2{x^2} - 1,

then

\frac{1}{{{x^2}}} + \frac{1}{{2{x^4}}} + \frac{1}{{3{x^6}}} + \ldots \infty

equals to

View answer and explanation

Correct option: A

{\log _e}\left( {\frac{{y + 1}}{{y - 1}}} \right)

Option A (

\log _e\left( {\frac{{y + 1}}{{y - 1}}} \right)

) is correct because it is the result of simplifying the given series using logarithmic properties and the given substitution.

482 Medium

482 of 564

Analyzing +4 | -1

Let

{N}

be any four digit number say

{x}_1 {x}_2 {x}_3 {x}_4

. Then maximum value of

\frac{N}{x_1+x_2+x_3+x_4}

is equal to

View answer and explanation

Correct option: A

1000

Option A (

\frac{1}{2}

) is correct because the sum

\cos \theta + \cos 3\theta + \cos 5\theta

can be simplified using trigonometric identities when

\theta = \frac{\pi}{7}

483 Medium

483 of 564

Analyzing +4 | -1

Let

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

be an A.P. If

a_{7}=3

, the product

a_{1} a_{4}

is minimum and the sum of its first

n

terms is zero, then

n !-4 a_{n}(n+2)

is equal to :

View answer and explanation

Correct option: B

The correct answer is 24 because after finding the values of a and d, and subsequently n, the expression n! - 4*a_n*(n+2) evaluates to 24.

484 Medium

484 of 564

Understanding +4 | -1

3,3^2, 3^3

are in

G P

, then

\log 3, \log 9

and

\log 27

are in

View answer and explanation

Correct option: A

A P

Option A (

A P

) is correct because the logarithms of terms in a geometric progression form an arithmetic progression.

485 Medium

485 of 564

Understanding +4 | -1

If the sum of the first

n

terms of a sequence is of the form

{A} n^2+{B} n

where

{A}, {B}

are constants independent of

n

, then the sequence is

View answer and explanation

Correct option: A

an A.P.

Option D (6) is correct because the number of common roots is equal to the greatest common divisor (GCD) of 12 and 18, which is 6.

486 Medium

486 of 564

Application +4 | -1

a, b

and

c

are in HP, then

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

and

\frac{c}{a+b}

are in

View answer and explanation

Correct option: C

Option C (HP) is correct because if a, b, c are in HP, then their reciprocals are in AP, and manipulating the AP sequence leads to the conclusion that the given terms are also in HP.

487 Medium

487 of 564

Application +4 | -1

2^{10}+2^9 \cdot 3^1+2^8 \cdot 3^2+\ldots+2.3^9+3^{10}

=S-2^{11}

, then

S

is equal to

View answer and explanation

Correct option: A

3^{11}

Option A (

3^{11}

) is correct because it is the result of summing the given series and adding

2^{11}

to it.

488 Medium

488 of 564

Application +4 | -1

If sum of all terms of an infinite GP is

\frac{1}{5}

times the sum of odd terms. Then, common ratio is

View answer and explanation

Correct option: C

-\frac{4}{5}

Option C (

-\frac{{4}}{{5}}

) is correct because by setting up the equation relating the sum of the infinite GP to the sum of its odd terms, we can solve for the common ratio, which is -4/5.

489 Medium

489 of 564

Application +4 | -1

If the

m^{\text {th }}

term of an HP is

n

and

n^{\text {th }}

m

, then the

r^{\text {th }}

term will be

View answer and explanation

Correct option: C

\frac{m n}{r}

Option C (

\frac{m n}{r}

) is correct because the rth term of the corresponding AP is

\frac{r}{mn}

, so the rth term of the HP is its reciprocal.

490 Medium

490 of 564

Application +4 | -1

In the sequence

1,2,2,4,4,4

4,8,8,8,8,8,8, \ldots

, where

n

consecutive terms have the value

n

, then 1025 th term is

View answer and explanation

Correct option: B

2^{10}

Option B (

2^{10}

) is correct because the 1025th term falls within the block of

2^{10}

consecutive terms having the value

2^{10}

491 Medium

491 of 564

Application +4 | -1

The value of

x

, satisfying

\log _{a}\left(1+\log _{b}\left\{1+\log _{c}\left(1+\log _{p} x\right)\right\}\right)

=0

, is

View answer and explanation

Correct option: D

Option D (1) is correct because solving the logarithmic equation step-by-step leads to x = 1.

492 Medium

492 of 564

Analyzing +4 | -1

a, b, c

are positive integers forming an increasing G.P. whose common ratio is a natural number.

b-a

is cube of a natural number and

\log _6 a+\log _6 b+\log _6 c=6

, then

a+b+c=

View answer and explanation

Correct option: D

189

Option D (189) is correct because it satisfies all the given conditions: a, b, c are positive integers in increasing G.P., b-a is a cube of a natural number, and the sum of the logarithms is 6.

493 Medium

493 of 564

Analyzing +4 | -1

Let

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

. be a G.P. of increasing positive numbers. Let the sum of its

6^{\text {th }}

and

8^{\text {th }}

terms be 2 and the product of its

3^{\text {rd }}

and

5^{\text {th }}

terms be

\frac{1}{9}

. Then

6\left(a_{2}+a_{4}\right)\left(a_{4}+a_{6}\right)

is equal to

View answer and explanation

Correct option: A

Option A, 3, is correct because solving the geometric progression equations and substituting the values gives 3.

494 Medium

494 of 564

Application +4 | -1

Let

x, y, z \in R^{+}

such that

x+y+z=27

. If maximum value of

x^2 y^3 z^4

\lambda \cdot 6^{10}

, then find the value of

\lambda

View answer and explanation

Correct option: A

Option A (9) is correct because the maximum value of the expression is found using AM-GM inequality, resulting in

\lambda = 9

495 Medium

495 of 564

Analyzing +4 | -1

{5^{3{x^2}{{\log }_{10}}2}} = {2^{\left( {x + \frac{1}{2}} \right){{\log }_{10}}25}},

then

x

equals to

View answer and explanation

Correct option: A

1,\; - \frac{1}{3}

Option A (

1, -\frac{1}{3}

) is correct because these are the solutions obtained by equating the exponents after taking logarithms and solving the resulting quadratic equation.

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496 Medium

496 of 564

Analyzing +4 | -1

a, b, c

are real numbers such that

3\left(a^2+b^2+c^2+1\right)=2(a+b+c+a b+b c+c a)

, then

a, b, c

are in

View answer and explanation

Correct option: C

GP and AP

Option C (GP and AP) is correct because the given condition leads to a = b = c = 1, which satisfies both the properties of a Geometric Progression and an Arithmetic Progression.

497 Medium

497 of 564

Application +4 | -1

If 6 th term of a GP is 32 and its 8 th term is 128 , then the value of the common ratio is

View answer and explanation

Correct option: B

Option B (2) is correct because the common ratio can be found by dividing the 8th term by the 6th term and taking the square root.

498 Medium

498 of 564

Application +4 | -1

The sum of all odd numbers of four digits which are divisible by 9 , is

View answer and explanation

Correct option: A

2754000

Option A (2754000) is correct because it is the sum of the arithmetic progression of four-digit odd numbers divisible by 9.

499 Medium

499 of 564

Application +4 | -1

\lim\limits_{x\, \to \,2} \left( {\sum\limits_{n\, = \,1}^9 {\frac{x}{{n\left( {n + 1} \right){x^2} + 2\left( {2n + 1} \right)x + 4}}} } \right)

is equal to

View answer and explanation

Correct option: A

\frac{9}{{44}}

Option A (

\frac{9}{{44}}

) is correct because it is the value of the limit after simplifying the summation using partial fractions.

500 Medium

500 of 564

Analyzing +4 | -1

The sum of the series

\frac{1}{{2!}} + \frac{1}{{4!}} + \frac{1}{{6!}} + \ldots

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the series can be expressed in terms of e and simplified to this expression.

501 Medium

501 of 564

Application +4 | -1

if the first term k terms of an

AP

is an integer with common difference 2 , then the number of possible values of

\mathrm{k}

for which the sum of first k terms is 153 is

(\mathrm{k}>1)

View answer and explanation

Correct option: C

Option C (5) is correct because there are 5 possible values of k for which the sum of the first k terms of the AP is 153, derived from the factors of 153.

502 Medium

502 of 564

Application +4 | -1

If the

7^{\text {th }}

term of a harmonic progression is 8 and the

8^{\text {th }}

term is 7 , then its

15^{\text {th }}

term is

View answer and explanation

Correct option: D

\frac{56}{15}

Option D (

\frac{56}{15}

) is correct because the 15th term of the HP is the reciprocal of the 15th term of the corresponding AP.

503 Medium

503 of 564

Analyzing +4 | -1

In the expansion of

2{\log _e}x - {\log _e}\left( {x + 1} \right) - {\log _e}\left( {x - 1} \right)

the coefficient of

{x^{ - 4}}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because by expanding the logarithmic expression and identifying the coefficient of

x^{-4}

, we find it to be

\frac{1}{2}

504 Medium

504 of 564

Application +4 | -1

The sum of the products of the numbers

\pm 1, \pm 2, \pm 3, \pm 4, \pm 5

taking two at a time is

View answer and explanation

Correct option: B

-55

Option B (-55) is correct because the sum of the products of the numbers

\pm 1, \pm 2, \pm 3, \pm 4, \pm 5

taken two at a time is -55.

505 Medium

505 of 564

Analyzing +4 | -1

\mathrm{A}>0, \mathrm{B}>0

and

\mathrm{A}+\mathrm{B}=\frac{\pi}{6}

, then the minimum value of

\tan \mathrm{A}+\tan \mathrm{B}

is :

View answer and explanation

Correct option: B

4-2 \sqrt{3}

Option B (

4-2 \sqrt{3}

) is correct because by using the tangent addition formula and AM-GM inequality, we find the minimum value of

\tan A + \tan B

to be

4-2\sqrt{3}

506 Medium

506 of 564

Analyzing +4 | -1

Let

a_1, a_2, a_3, a_4

and

a_5

be such that

a_1, a_2

and

a_3

are in A.P.,

a_2, a_3

and

a_4

are in G.P.; and

a_3, a_4

and

a_5

are in H.P. Then,

a_1, a_3

and

a_5

are in

View answer and explanation

Correct option: A

G.P.

Option A (G.P.) is correct because the given relationships between the terms lead to the conclusion that

a_1, a_3,

and

a_5

are in geometric progression.

507 Medium

507 of 564

Analyzing +4 | -1

a,b,c,d,e

are five numbers in which the first three are in A.P. and the last three are in H.P. If the three numbers in the middle are in G.P., then the numbers in the odd places are in

View answer and explanation

Correct option: B

G.P.

Option B (G.P.) is correct because the numbers in the odd places satisfy the condition for a geometric progression.

508 Medium

508 of 564

Application +4 | -1

2^{x}=4^{y}=8^{z}

and

^{x y z}=288

, then

\frac{1}{2 x}+\frac{1}{4 y}+\frac{1}{8 z}=

View answer and explanation

Correct option: D

11 / 96

Option D (11/96) is correct because solving for x, y, and z and then substituting into the expression yields 11/96.

509 Medium

509 of 564

Application +4 | -1

a

is the

A M

b

and

c

and the two geometric means are

G_1

and

G_2

, then

G_1^3+G_2^3

is equal to

View answer and explanation

Correct option: B

2 a b c

Option B (2abc) is correct because, given a is the AM of b and c, and G1 and G2 are the two geometric means, G1³ + G2³ simplifies to 2abc.

510 Medium

510 of 564

Application +4 | -1

If the A.M. between

p^{\text {th }}

and

q^{\text {th }}

terms of an A.P. is equal to the A.M. between

r^{\text {th }}

and

s^{\text {th }}

terms of the same A.P., then

p+q

is equal to

View answer and explanation

Correct option: D

r+s

Option D (r+s) is correct because, in an A.P., if the A.M. of two terms is equal, their indices must sum to the same value.

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