An algebra problem by A Former Brilliant Member

The lowest number is 114 and the largest number is 996.

The numbers forms an arithmetic progression.

114,120,126...996

$a_{n}$ $=a_{1}$ $+$ $(n-1)d$

$996$ $=114$ $+$ $(n-1)6$

$n=148$

$S=\frac{n}{2}$ $(a_{1}+a_{n})$

$S=\frac{148}{2}$ $(114+996)$

$S = 82140$