Answer
$
a_5=\dfrac{128}{81}$
$a_n=\dfrac{1}{2}\left(\dfrac{4}{3}\right)^{n-1}$
Work Step by Step
To find $a_5$ and $a_n$ for the given geometric sequence we will apply the formula for finding the $n^{th}$ term of a geometric sequence.
The $n^{th}$ term of a geometric sequence can be found by the formula: $a_n=a_1r^{n-1}....(1)$
where, $r$ is the common ratio, $r$ and can be computed as the quotient of a term and the term preceeding it.
First we will find the common ratio $r$. Divide the third term by the second term. That is, $r=\dfrac{a_{3}}{a_2}=\dfrac{8/9}{3/2}=\dfrac{4}{3}$
Plug $\dfrac{1}{2}$ for $a_1$ and $\dfrac{4}{3}$ for $r$ in the equation (1) to obtain:
$$ a_n=\dfrac{1}{2}\left(\dfrac{4}{3}\right)^{n-1}$$
With $n=5,$ the equation above yields the result :
$a_5=\dfrac{1}{2}\left(\dfrac{4}{3}\right)^{5-1}=\dfrac{128}{81}$
Hence, $
a_5=\dfrac{128}{81}$ and $a_n=\dfrac{1}{2}\left(\dfrac{4}{3}\right)^{n-1}$.
