Answer
$a_n = 1 \cdot (-\frac{1}{3})^{n-1}$
$a_5=\frac{1}{81}$
Work Step by Step
RECALL:
The $n^{th}$ term $a_n$ of a geometric sequence is given by the formula:
$a_n=a_1 \cdot r^{n-1}$
where
$a_1$ = first term
$r$ = common ratio
The given geometric sequence has:
$a_1=1$
$r=-\frac{1}{3}$
Thus, the $n^{th}$ term of the sequence is given by the formula:
$a_n = 1 \cdot (-\frac{1}{3})^{n-1}$
The 5th term can be found by substituting $5$ for $n$:
$a_5=1 \cdot (-\frac{1}{3})^{5-1}
\\a_5=1 \cdot (-\frac{1}{3})^4
\\a_5 = 1 \cdot \frac{1}{81}
\\a_5=\frac{1}{81}$
