Integer Answer Type Questions: Sequences and Series | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q.1. Let S_k, 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 a_k=2a_k–1–a_k–2 for k = 3, 4,..........11.

 if    , then the value of

is equal to (2010)

Ans. (0)

Sol.   Given that  a_k = 2a_{k –1}–a_{k –2}

⇒ a₁ , 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

⇒ a² + 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  a₁ = 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
 (JEE Adv. 2014)

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 (r² - 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) x⁹ can be found in the following ways x⁹, x^{1 +2} 8, x ^{2 + 7}, x^{3 + 6}, x^{4 + 5}, x^{1 + 2 + 6}, x^{1 + 3 + 5}, x^{2 + 3 + 4} The coefficient of x9 in each of the above 8 cases is 1. ∴ Required coefficient = 8.