#### Answer

$S_{50}= 6600$

#### Work Step by Step

$-15,-9,-3,3,...$
Sum of $n$ terms,
$S_{n}= \frac{n}{2}(a_{1}+a_{n})$
Substituting $a_{n}= a_{1}+(n-1)d$
$S_{n}= \frac{n}{2}(a_{1}+a_{1}+(n-1)d)$
$S_{n}= \frac{n}{2}(2a_{1}+(n-1)d)$
Substituting $a_{1}=-15, d= -9+15 = 6$ and $n=50$
$S_{50}= \frac{50}{2}(2(-15)+(50-1)6)$
$S_{50}=25(-30+(49)6)$
$S_{50}=25(-30+294)$
$S_{50}=25(264)$
$S_{50}= 6600$