## College Algebra (10th Edition)

$n=13$ or $n=27$
The sum $S_{n}$ of the first $n$ terms of $\left\{a_{n}\right\}$: $S_{n}=\displaystyle \frac{n}{2}\left[2a_{1}+(n-1)d\right]$ $702=\displaystyle \frac{n}{2}\left[2(78)-4(n-1)\right]\quad$ ... solve for n $1404=n[156-4n+4]$ $1404=n[160-4n]$ $4n^{2}-160n+1404=0$ $n^{2}-40n+351=0$ ... use the quadratic formula $n=\displaystyle \frac{40\pm\sqrt{40^{2}-4(1)(351)}}{2(1)}$ $n=\displaystyle \frac{40\pm 14}{2}$ $n=13$ or $n=27$