## College Algebra (6th Edition)

$S_{50}= 6600$
$-15,-9,-3,3,...$ 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 $n=50$ $S_{50}= \frac{50}{2}(2(-15)+(50-1)6)$ $S_{50}=25(-30+(49)6)$ $S_{50}=25(-30+294)$ $S_{50}=25(264)$ $S_{50}= 6600$