Question

If S_n denote the sum of first n terms of an A.P. and f(n) = $\frac{S_{3n}}{S_{2n} - S_{n}}$, then $\operatorname { Lim } _ { n \rightarrow \infty }$ $\sum_{r = 1}^{2n}\frac{f(r)}{n}$ is equal to

• A. 6
• B. 3
• C. 2
• D. 0

Answer 1

Answer: (A)

6

Answer 2

S_3n = $\frac{3n}{2}$ [2a + (3n – 1)d]

S_2n = $\frac{2n}{2}$ [2a + (2n – 1) d]

S_n­ = $\frac{n}{2}$ [2a + (n – 1)d]

S_2n – S_n = $\frac{n}{2}$ [2a + (3n – 1)d]

$\frac{S_{3n}}{S_{2n} - S_{n}}$ = 3 = f(n)

S = $\underset{x \rightarrow \infty}{Lim}$ $\frac{1}{n}\sum_{r = 1}^{2n}{f(r)}$ = $\underset{n \rightarrow \infty}{Lim}$ 3$\left( \frac{2n}{n} \right)$ = 6