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 $\left(8n^2-11n-20\right)$ , then the sum of all the possible values of $n$ is

Answer 1

Answer: 9

$a_1=\frac{n\left(n+1\right)}{2},d=n$
$\Rightarrow S_n=\frac{n}{2}\left(n\left(n+1\right)+\left(n-1\right)n\right)$
$=\frac{n}{2}\left(n^2+n+n^2-n\right)=n^3$
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 $5\left(n\ne -1\right)$
$\Rightarrow$ sum of all possible values $=9$