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 four times the ratio $\frac{a}{d}.$ ?

Answer 1

$\frac{S_{pn}}{S_n}=\frac{\frac{pn}{2}[2a+(pn-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{p[(2a-d)+npd]}{(2a-d)+nd}$

Above will be independent of n only when

$2a-d=0$ as in that case n will cancel from both $N^{r}and{D}^r$

$\frac{a}{d}=\frac{1}{2}$