Question

If the sum of the first $n$ terms of an arithmetic progression, whose first term is the sum of the first $n$ positive integers and whose common difference is $n$ , is $\left(8 n^{2} - 11 n - 20\right)$ , then the sum of all the possible values of $n$ is

Answer 1

$a_{1}=\frac{n \left(n + 1\right)}{2},d=n$
$\Rightarrow S_{n}=\frac{n}{2}\left(n \left(n + 1\right) + \left(n - 1\right) n\right)$
$=\frac{n}{2}\left(n^{2} + n + n^{2} - n\right)=n^{3}$
But, given that $S_{n}=8n^{2}-11n-20.$
$\Rightarrow 8n^{2}-11n-20=n^{3}$
$\Rightarrow n^{3}-8n^{2}+11n+20=0$
$\Rightarrow \left(n + 1\right)\left(n - 4\right)\left(n - 5\right)=0$
$\Rightarrow n=4$ or $5 \, \left(n \neq - 1\right)$
$\Rightarrow $ sum of all possible values $=9$