Answer
$n=27$ or, $n= 13$.
Work Step by Step
We can see that the sum of $n$ terms the arithmetic sequence is $S_n=702$ and the first term is $a_1=78$
There is a constant difference between the terms of: $d=-4$
The sum of the first $n$ terms of the arithmetic sequence is given by:
$S_{n}= \dfrac{n}{2}\left[2a_{1}+(n-1) d\right] ..(1)$
Now, we plug in the above data into Equation-1 to find the number of terms, $n$.
$702= \dfrac{n}{2}[(2)(78)+(-4)(n-1)] \\ 1404=n [156-4n+4]\\ n^2 -40n +351=0$
Use the quadratic formula to find the roots of $n$.
$n=\dfrac{40 \pm\sqrt {(40)^2 -(4)(1) (351)}}{(2)(1)}=\dfrac{40 \pm 14 }{2}$
or, $n=\dfrac{40 +14 }{2}= 27$ and $n=\dfrac{40 -14 }{2}= 13$
Therefore, the we have $n=27$ or $n= 13$ terms.
