Answer
The $n^{th}$ term of the sequence is given by the formula:
$a_n =1 (\dfrac{-1}{3})^{n-1}$
and $a_5=\dfrac{1}{81}$
Work Step by Step
We are given: $a_{1}=1 ; \ r=\dfrac{-1}{3}$
The $n^{th}$ term of the sequence is given by the formula:
$a_n=a_1r^{n-1}$
$a_n =1 \ (\dfrac{-1}{3})^{n-1}$
Now, the 5th term can be computed by substituting $5$ for $n$:
$a_5=1 \ (\dfrac{-1}{3})^{5-1}=\ (\dfrac{-1}{3})^{4}=\dfrac{1}{81}$
