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# Integer Answer Type Questions: Sequences and Series | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Q.1. Let Sk, k = 1, 2, ….. , 100, denote the sum of the infinite geometric series whose first term is

and the common ratio is  .  Then the value of

is                         (2010)

Ans. (3)

Sol.   Using   we get

Q.2. Let a1,a2,a3........, a11 be real numbers satisfying

a1=15, 27–2a2> 0 and ak=2ak–1–ak–2 for k = 3, 4,..........11.

if    , then the value of

is equal to (2010)

Ans. (0)

Sol.   Given that  ak = 2ak –1–ak –2

⇒ a, a2 , a3 , ...,a11 are in AP..
If a is the first term and D the commen difference then
a21 + a22 + ... +a211= 990 (
⇒ 11a 2 + d 2 (12 + 22 + ...+ 102 ) + 2ad (1 + 2 + ...+ 10) = 990

⇒ a+ 35d2 + 150d= 90 Using a = 15,
we get 35d2 + 150d +135 = 0 or 7d2 + 30d + 27= 0
⇒ d +3)(7d +9) = 0 ⇒ d= –3 or – 9/7

then a2 = 15 - 3 = 12 or 15

∴ d ≠ – 97

Hence

Q.3. Let a1, a2, a3 .....a100 be an arithmetic progression with  a= 3 and   For any integer n with 1 ≤  n ≤ 20 , let m = 5n. If    does not depend on n, then a2 is              (2011)

Ans. (9)

Sol.  We have

which will be independent of n if d = 6 or d = 0 For a proper A.P. we take d = 6 then a2 = 3 + 6 = 9

Q.4. A pack con tain s n cards n umber ed fr om 1 to n . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k – 20 = (JEE Adv. 2013)

Ans.  (5)

Sol.   Let k, k + 1 be removed from pack.

∴ (1 + 2 + 3 + ... + n) – (k + k + 1) = 1224

for n = 50,  k = 25 ∴ k – 20 = 5

Q.5. Let a, b, c be positive integers such that   is an integer. If a,b, c are in geometric progression and the arithmetic mean of a, b, c is b + 2, then the value of     is

Ans. (4)

Sol.   ∵ a, b, c are in G.P

∴ b = ar and c = ar2

Also    is an integer

⇒ r is an integer

∵  A.M. of a, b, c is b + 2

⇒ a + ar + ar2 = 3ar+6

⇒ a (r- 2r + 1)=6

⇒ a (r - 1)2=6 .

∵  a and r are integers
∴ The only possible values of a and r can be 6 and 2 respectively.

Then

Q.6. Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is (JEE Adv. 2015)

Ans. (9)

Sol.

a7 = a + 6d = 15d

∵  130 < 15d < 140 ⇒ d = 9

(∵ All terms are natural numbers ∴ d ∈N )

Q.7. The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3) ... (1 + x100) is (JEE Adv. 2015)

Ans. (8)

Sol.  In expansion of (1 + x) (1 + x2) (1 + x3) .... (1 + x100) xcan be found in the following ways x9, x1 +2 8, x 2 + 7, x3 + 6, x4 + 5, x1 + 2 + 6, x1 + 3 + 5, x2 + 3 + 4 The coefficient of x9 in each of the above 8 cases is 1. ∴ Required coefficient = 8.

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## Maths 35 Years JEE Main & Advanced Past year Papers

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## Maths 35 Years JEE Main & Advanced Past year Papers

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