Question

If $S_n$ denotes the sum of first n terms of an A.P. and $f\left(n\right)=\frac{S_{3n}}{S_{2n}-S_n}$,then $\underset{n\to \infty }{lim}\sum _{r=1}^{2n}\frac{f\left(r\right)}{n}$ is equal to
(Note : $S_{2n}\ne S_n$)

• A. $6$
• B. $3$
• C. $2$
• D. $0$

Answer 1

The correct option is A $6$

$S_{3n}=\frac{3n}{2}[2a+(3n-1\left)d\right]$

$S_{2n}=\frac{2n}{2}[2a+(2n-1\left)d\right]$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{2n}-S_n=\frac{n}{2}[2a+(3n-1\left)d\right]$

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

$S={lim}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^{2n}f\left(r\right)$

$={lim}_{n\to \infty }\frac{1}{n}\left[f\right(1)+f(2)+f(3)+.....+f(n)+f(n+1)+......+f(2n\left)\right]$

$={lim}_{n\to \infty }3\left(\frac{2n}{n}\right)=6$