# JEE Main

Explore popular questions from Sequence and Series for JEE Main. This collection covers Sequence and Series previous year JEE Main questions hand picked by experienced teachers.

## Mathematics

Sequence and Series

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Q 1. If {tex} ( 10 ) ^ { 9 } + 2 ( 11 ) ^ { 1 } ( 10 ) ^ { 8 } + 3 ( 11 ) ^ { 2 } ( 10 ) ^ { 7 } + \cdots + 10 ( 11 ) ^ { 9 } = k ( 10 ) ^ { 9 } , {/tex} then {tex} k {/tex} is equal to

100

B

110

C

{tex} \frac { 121 } { 10 } {/tex}

D

{tex} \frac { 441 } { 100 } {/tex}

##### Explanation

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Q 2. If the {tex} 2 ^ { \text { nd } } , 5 ^ { \text { th } } {/tex} and {tex} 9 ^ { \text { th } } {/tex} terms of a non-constant AP are in GP, then the common ratio of this GP is

A

{tex} \frac { 7 } { 4 } {/tex}

B

{tex} \frac { 8 } { 5 } {/tex}

{tex} \frac { 4 } { 3 } {/tex}

D

{tex} 1 {/tex}

##### Explanation

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Q 3. Let {tex} a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots , a _ { n } {/tex} be in AP. If {tex} a _ { 3 } + a _ { 7 } + a _ { 11 } + a _ { 15 } = 72 , {/tex} then the sum of its first {tex}17{/tex} terms is equal to

306

B

204

C

153

D

612

##### Explanation

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Q 4. If {tex} \tan n \theta = \tan m \theta , {/tex} then the different values of {tex} \theta {/tex} will be in

{tex} A P {/tex}

B

{tex} G P {/tex}

C

{tex} \mathrm { HP } {/tex}

D

None of these

##### Explanation

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Q 5. If the sum of {tex} n {/tex} terms of an AP is {tex} n A + n ^ { 2 } B , {/tex} where {tex} A , B {/tex} are constants, then its common difference will be

A

{tex} A - B {/tex}

B

{tex} A + B {/tex}

C

{tex}2 A {/tex}

{tex}2 B {/tex}

##### Explanation

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Q 6. The {tex} 9 ^ { \text { th } } {/tex} term of the series {tex} 27 + 9 + 5 \frac { 2 } { 5 } + 3 \frac { 6 } { 7 } + \cdots {/tex} will be

1{tex} \frac { 10 } { 17 } {/tex}

B

{tex} \frac { 10 } { 17 } {/tex}

C

{tex} \frac { 16 } { 27 } {/tex}

D

{tex} \frac { 17 } { 27 } {/tex}

##### Explanation

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Q 7. If {tex} \log _ { 3 } 2 , \log _ { 3 } \left( 2 ^ { x } - 5 \right) {/tex} and {tex} \log _ { 3 } \left( 2 ^ { x } - \frac { 7 } { 2 } \right) {/tex} are in AP, then {tex} x {/tex} is equal to

A

{tex} 1 , \frac { 1 } { 2 } {/tex}

B

{tex} 1 , \frac { 1 } { 3 } {/tex}

C

{tex} 1 , \frac { 3 } { 2 } {/tex}

None of these

##### Explanation

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Q 8. Let {tex} T _ { r } {/tex} be the {tex} r ^ { \text { th } } {/tex} term of an AP for {tex} r = 1,2,3 , \cdots . {/tex} If for some positive integers {tex} m , n , {/tex} we have {tex} T _ { m } = \frac { 1 } { n } {/tex} and {tex} T _ { n } = \frac { 1 } { m } , {/tex} then {tex} T _ { m n } {/tex} equals

A

{tex}\large \frac { 1 } { m n } {/tex}

B

{tex}\large \frac { 1 } { m } + \frac { 1 } { n } {/tex}

1

D

0

##### Explanation

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Q 9. If the ratio of the sum of {tex} n {/tex} terms of two APs is {tex} ( 7 n + 1 ) : ( 4 n + 27 ) {/tex} , then the ratio of their {tex} 11 ^ { \text { th } } {/tex} terms will be

A

{tex} 2 : 3 {/tex}

B

{tex} 3 : 4 {/tex}

{tex} 4 : 3 {/tex}

D

{tex} 5 : 6 {/tex}

##### Explanation

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Q 10. The sum of the series {tex} \frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 6 } + \dots {/tex} to 9 terms is

A

{tex} - \frac { 5 } { 6 } {/tex}

B

{tex} - \frac { 1 } { 2 } {/tex}

C

1

{tex} - \frac { 3 } { 2 } {/tex}

##### Explanation

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Q 11. If {tex} a _ { 1 } , a _ { 2 } , \ldots , a _ { n } {/tex} are in AP with common difference {tex} d , {/tex} then the sum of the following series is{tex} \sin d \left( \cosec a _ { 1 } \cosec a _ { 2 } + \cosec a _ { 2 } \cosec a _ { 3 } + \cdots + \cosec a _ { n - 1 } \right. {/tex} {tex} \cosec a _ { n } ) {/tex}

A

{tex} \sec a _ { 1 } - \sec a _ { n } {/tex}

{tex} \cot a _ { 1 } - \cot a _ { n } {/tex}

C

{tex} \tan a _ { 1 } - \tan a _ { n } {/tex}

D

{tex} \cosec a _ { 1 } - \cosec a _ { n } {/tex}

##### Explanation

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Q 12. The sum of the integers from 1 to 100 which are not divisible by 3 or 5 is

A

2489

B

4735

C

2317

2632

##### Explanation

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Q 13. The sum of numbers from 250 to 1000 which are divisible by 3 is

A

135657

B

136557

C

161575

156375

##### Explanation

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Q 14. The number of terms of the {tex} A P\,\, 3,7,11,15 , \ldots {/tex} to be taken so that the sum is 406 is

A

5

B

10

C

12

14

##### Explanation

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Q 15. If {tex} a , b , c , d , e , f {/tex} are in AP, then the value of {tex} e - c {/tex} will be

A

2{tex} ( c - a ) {/tex}

B

2{tex} ( f - d ) {/tex}

2{tex} ( d - c ) {/tex}

D

{tex} d - c {/tex}

##### Explanation

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Q 16. If the sum of three numbers of an arithmetic sequence is {tex}15{/tex} and the sum of their squares is {tex} 83 , {/tex} then the numbers are

A

{tex} 4,5,6 {/tex}

{tex} 3,5,7 {/tex}

C

{tex} 1,5,9 {/tex}

D

{tex} 2,5,8 {/tex}

##### Explanation

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Q 17. If the sum of three consecutive terms of an AP is {tex}51{/tex} and the product of last and first terms is {tex} 273 , {/tex} then the numbers are

{tex} 21,17,13 {/tex}

B

{tex} 20,16,12 {/tex}

C

{tex} 22,18,14 {/tex}

D

{tex} 24,20,16 {/tex}

##### Explanation

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Q 18. If {tex} a , b , c {/tex} are in {tex} A P , {/tex} then {tex} ( a + 2 b - c ) ( 2 b + c - a ) ( c + a - b ) {/tex} equals

A

{tex} \frac { 1 } { 2 } a b c {/tex}

B

{tex} a b c {/tex}

C

2{tex} a b c {/tex}

4{tex} a b c {/tex}

##### Explanation

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Q 19. If twice the {tex} 11 ^ { \text { th } } {/tex} term of an AP is equal to 7 times of its {tex} 21 ^ { \text { st } } {/tex} term, then its {tex} 25 ^ { \text { th } } {/tex} term is equal to

A

24

B

120

0

D

None of these

##### Explanation

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Q 20. If {tex} x , y , z {/tex} are in {tex} G P {/tex} and {tex} a ^ { x } = b ^ { y } = c ^ { z } , {/tex} then

A

{tex} \log _ { a } c = \log _ { b } a {/tex}

{tex} \log _ { b } a = \log _ { c } b {/tex}

C

{tex} \log _ { c } b = \log _ { a } c {/tex}

D

None of these

##### Explanation

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Q 21. The value of 7 $\log\frac{16}{15} + 5\log\frac{25}{24} + 3\log\frac{81}{80},$ is

log 2

B

log 3

C

log 5

D

None of these

##### Explanation

We have,

$7\log{\left( \frac{16}{15} \right) + 5\log\left( \frac{25}{24} \right) + 3\log\frac{81}{80}}$

$= \log\left\{ \left( \frac{16}{15} \right)^{7} \times \left( \frac{25}{24} \right)^{5} \times \left( \frac{81}{80} \right)^{3} \right\}$

$= \log\left\{ \frac{2^{28}}{3^{7} \times 5^{7}} \times \frac{5^{10}}{2^{15} \times 3^{5}} \times \frac{3^{12}}{2^{12} \times 5^{3}} \right\} = \log 2$

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Q 22. The sum of first two terms of an infinite G.P. is 1 and every term is twice the sum of the successive terms. Its first term is

A

1/3

B

2/3

3/4

D

1/4

##### Explanation

Let a be the first term and r be the common ration.

We have,

a + ar = 1 ….(i)

and,

arn − 1 = 2(arn + arn + 1 + …)

$\Rightarrow ar^{n - 1} = 2\frac{ar^{n}}{1 - r}$

⇒ rn − 1 − rn = 2rn

⇒ rn − 1 = 3 rn

⇒ 3 r = 1 ⇒ r = 1/3

Putting r = 1/3 in (i), we get a = 3/4

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Q 23. If $\log_{8}\left\{ \log_{2}\left( \log_{3}\left( x^{2} - 4x + 85 \right) \right) \right\} = \frac{1}{3},$ then x equals to

A

5

B

4

C

3

2

##### Explanation

We have,

$\log_{8}{\left\lbrack \log_{2}\left\{ \log_{3}\left( x^{2} - 4x + 85 \right) \right\} \right\rbrack = \frac{1}{3}}$

⇒ log2{log3(x2−4x+85)} = 81/3 = 2

⇒ log3(x2−4x+85) = 22

⇒ x2 − 4x + 85 = 34

⇒ x2 − 4x + 4 = 0 ⇒ (x−2)2 = 0 ⇒ x = 2

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Q 24. If the sum of first n natural numbers is $\frac{1}{78}$ times the sum of their cubes, then the value of n is

A

11

12

C

13

D

14

##### Explanation

Since, $\Sigma n = \frac{1}{78}\Sigma n^{3}$

$\Rightarrow \frac{n(n + 1)}{2} = \frac{1}{78} \times \frac{n^{2}\left( n + 1 \right)^{2}}{4}$

⇒ n2 + n − 156 = 0

⇒ (n+13)(n−12) = 0

⇒ n = 12 [∵n ≠ − 13]

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Q 25. If n geometric means between a and b be G1, G2, …Gn and a geometric mean be G, then the true relation is

A

G1G2…Gn = G

B

G1G2…Gn = G1/n

G1G2…Gn =Gn

D

G1G2…Gn = G2/n

##### Explanation

Since, G is the geometric mean between a and b.

∴ G = (ab)1/2 and n geometric means G1, G2, …, Gn be inserted between a and b.

∴ G1 = ar1, G2 = ar2, …, Gn = arn

Now, G1 • G2 • G3…Gn = anr1 + 2 + ... + n

= anrn(n + 1)/2

But $b = ar^{n + 1}\ \Rightarrow \ \bullet r = \left( \frac{b}{a} \right)^{\frac{1}{n + 1}}$

Therefore, the required product is

$a^{n}\left( \frac{b}{a} \right)^{\left\lbrack \frac{1}{n + 1} \right\rbrack\frac{n\left( + 1 \right)}{2}} = \left( \text{ab} \right)^{\frac{n}{2}}$

$= \left\{ \left( \text{ab} \right)^{\frac{1}{2}} \right\}^{n} = G^{n}$