Answer
$ a_n =6 \ (-2)^{n-1}$
and $a_5=96$
Work Step by Step
The general formula for nth term of a geometric sequence is: $a_n=a_1 r^{n-1}$
We have : $a_{1}=6 ; \ r=-2$
Thus, the $n^{th}$ term of the sequence can be written as:
$a_n=a_1 r^{n-1} \\ a_n =6 \ (-2)^{n-1}$
Substitute $n=5$ in the above form to obtain:
$a_5=6 \ (-2)^{5-1} \\ =6 \ (-2)^4 \\=96$
