Question

If $S_n$ be the sum of first n terms of an A.P. and if $\frac{S_{pn}}{S_n}$ in independent of n, then find the second term . Assume a=1.

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 expression will be independent of $n$ only when $2a-d=0$
as in that case $n$ will cancel from both $NumeratorandDenominator$.
$\Rightarrow \frac{a}{d}=\frac{1}{2}$

Then, sum of first term and the common difference $3$