Question

If the sum of $n$ terms of an $A.P.$ is $(pn+{qn}^2)$, where $p$ and $q$ are constants, find the common difference.

Answer 1

It is known that, for an A.P with first term $a$ and common difference $d$
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

According to the given condition,
$\frac{n}{2}[2a+(n-1\left)d\right]=pn+{qn}^2\Rightarrow \frac{n}{2}[2a+nd-d]=pn+{qn}^2\Rightarrow na+n^2\frac{d}{2}-n.\frac{d}{2}=pn+{qn}^2$

Comparing the coefficient of $n^2$ on both sides,we obtain
$\frac{d}{2}=q$

$\therefore d=2q$

Thus, the common difference of the A.P. is $2q$.