If the sum of $n$ terms of an $A.P.$ is $\left(p n+q n^{2}\right),$ where $p$ and $q$ are constants, find the common difference.
If the sum of $n$ terms of an $A.P.$ is $\left(p n+q n^{2}\right),$ where $p$ and $q$ are constants, find the common difference.
It is known that: $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
According to the given condition,
$\frac{n}{2}[2 a+(n-1) d]=p n+q n^{2}$
$\Rightarrow \frac{n}{2}[2 a+n d-d]=p n+q n^{2}$
$\Rightarrow n a+n^{2} \frac{d}{2}-n \cdot \frac{d}{2}=p n+q n^{2}$
Comparing the coefficients of $n^{2}$ on both sides, we obtain
$\frac{d}{2}=q$
$\therefore d=2 q$
Thus, the common difference of the $A.P.$ is $2 q$
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