#### Answer

$r = \dfrac{1}{4}$
$a_5=-\dfrac{1}{32}$
The $n^{th}$ term of the geometric sequence is: $a_n=-8 \cdot \left(\dfrac{1}{4}\right)^{n-1}$

#### Work Step by Step

RECALL:
(1) The common ratio of a geometric sequence is equal to the quotient of any two consecutive terms:
$r =\dfrac{a_n}{a_{n-1}}$
(2) The $n^{th}$ term of a geometric sequence is given by the formula:
$a_n = a\cdot r^{n-1}$
where
$a$ = first term
$r$ = common ratio
The sequence is said to be geometric.
Thus, we can proceed to solving for the common ratio:
$r=\dfrac{-2}{-8}
\\r = \dfrac{1}{4}$
The fifth term can be found by multiplying the common ratio to the fourth term.
The fourth term is $-\dfrac{1}{8}$.
Thus, the fifth term is:
$a_5=-\dfrac{1}{8} \cdot \left(\dfrac{1}{4}\right)
\\a_5=-\dfrac{1}{32}$
With a first term of $-8$ and a common ratio of $r=\dfrac{1}{4}$, the $n^{th}$ term of the geometric sequence is:
$a_n=a \cdot r^{n-1}
\\a_n=-8 \cdot \left(\dfrac{1}{4}\right)^{n-1}$