Answer
$a_n=-\dfrac{1}{4}(4)^{n-1}$
Work Step by Step
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1)
When $n=4$, then we have
Equation (1) gives: $a_4=a_1(r)^{4-1}=a_1r^3$
Plug in $a_4=-16$ and $a_1=-\dfrac{1}{4}$
$-16=-\dfrac{1}{4}r^3$
This gives: $r=4$
Hence, $a_n=-\dfrac{1}{4}(4)^{n-1}$
