Answer
$n=13$ or $n=27$
Work Step by Step
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 $n=27$