Let a_1, a_2, a_3, ., a_ 100 be an arithmetic progression with a_1=3 and S_p=∑_ i=1 ^p a_i , 1 ≤ p ≤ 100 . For any integer n with 1 ≤ n ≤ 20 , let m=5 n . If (S_m)/(S_n) does not depend on n , then a_2 is

Correct answer: 3 Given, a_1=3, m=5 n and a_1, a_2, are in AP. (S_m)/(S_n)= S_ 5 n S_n is independent of n . Now, ((5 n)/(2)[2 × 3+(5 n-1) d])/((n)/(2)[2 × 3+(n-1) d]) f\ (6-d)+5 n\ (6-d)+n independent of n , if aligned 6-d= & 0 d=6 a_2= & a_1+d=3+6=9 & Or aligned If d=0 (S_m)/(S_n) is independent of n . a_2=3

JEE Advanced 2011Paper 1

Exam: JEE Advanced
Year: 2011
Paper: Paper 1
Subject: Mathematics
Chapter: Sequences and Series

JEE Advanced 2011 · Paper 1 · Mathematics · Sequences and Series

Let a_1, a_2, a_3, ., a_ 100 be an arithmetic progression with...

Let

a_1, a_2, a_3, \ldots ., a_{100}

be an arithmetic progression with

a_1=3

and

S_p=\sum_{i=1}^p a_i

1 \leq p \leq 100

. For any integer

n

with

1 \leq n \leq 20

, let

m=5 n

. If

\frac{S_m}{S_n}

does not depend on

n

, then

a_2

Correct answer

Step-by-step solution

Given, a_1=3, m=5 n and

a_1, a_2, \ldots

are in AP.

\therefore \frac{S_m}{S_n}=\frac{S_{5 n}}{S_n}

is independent of

n

. Now,

\quad \frac{\frac{5 n}{2}[2 \times 3+(5 n-1) d]}{\frac{n}{2}[2 \times 3+(n-1) d]} \Rightarrow \quad \frac{f\{(6-d)+5 n\}}{(6-d)+n}

independent of

n

, if

\begin{aligned} \quad 6-d= & 0 \Rightarrow d=6 \\ \therefore \quad a_2= & a_1+d=3+6=9 \\ & \text{ Or }\end{aligned}

d=0 \frac{S_m}{S_n}

is independent of

n

. \therefore \quad a_2=3

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