Answer
The equation for the $n$th term is
$a_n=-\frac{1}{8}(2)^{n-1}$
and $a_6=-4$
Work Step by Step
The given series is
$-\frac{1}{8},-\frac{1}{4},-\frac{1}{2},-1,...$
First term $a_1=-\frac{1}{8}$.
Common ratio $r=\frac{-1/4}{-1/8}=2$.
Equation for a geometric sequence is
$\Rightarrow a_n=a_1r^{n-1}$
Substitute $-\frac{1}{8}$ for $a_1$ and $2$ for $r$.
$\Rightarrow a_n=-\frac{1}{8}(2)^{n-1}$
Substitute $6$ for $n$.
$\Rightarrow a_6=-\frac{1}{8}(2)^{6-1}$
Simplify.
$\Rightarrow a_6=-\frac{1}{8}(2)^{5}$
$\Rightarrow a_6=-4$.
