Answer
$a_n=-80\left(\frac{1}{4}\right)^{n-1}$
Work Step by Step
$-80,-20,-5,-\frac{5}{4},-\frac{5}{16},-\frac{5}{64},...$
$\frac{a_2}{a_1}=\frac{-20}{-80}=\frac{1}{4}=r$
$ra_2=\frac{1}{4}\times(-20)=-5=a_3$
$ra_3=\frac{1}{4}\times(-5)=-\frac{5}{4}=a_4$
and so on. The sequence is geometric with ratio $r=\frac{1}{4}$. The first term is
$a=-80$ and so the $n^{th}$ term is given by
$a_n=-80\left(\frac{1}{4}\right)^{n-1}$
