Question

If $S_n$ be the sum of first n terms of an A.P. and if $\frac{S_{pn}}{S_n}$ is independent of n, then find the ratio $\frac{a}{d}$.

Answer 1

$\frac{S_{pn}}{S_n}=\frac{\frac{pn}{2}[2a+(pn-1\left)d\right]}{\frac{n}{2}[2a+(n-1\left)d\right]}=\frac{p\left[\right(2a-d)+npd]}{(2a-d)+nd}$
Above will be independent of n only when $2a-d=0$ as in that case n will cancel from both $N^r$ and $D^r$
$\therefore \frac{a}{d}=\frac{1}{2}$.