Answer
$
a_5=\dfrac{125}{4}$
$a_n=\dfrac{4}{5}\left(\dfrac{5}{2}\right)^{n-1}$
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.
First we will find the common ratio $r$.The common ratio, $r$ is known as the quotient of a term and the term preceeding it.
Divide the third term by the second term.
That is, $r=\dfrac{a_{3}}{a_2}=\dfrac{5}{2}$
Applying the formula for finding the $n^{th}$ term of a geometric sequence, which can be written as: $a_n=a_1r^{n-1}$
Plug $\frac{4}{5}$ for $a_1$ and $\frac{5}{2}$ for $r$ in the equation above to obtain:
$$ a_n=\frac{4}{5}\left(\frac{5}{2}\right)^{n-1}$$
With $n=5,$ the equation above yields the result :
$$a_5=\frac{4}{5}\left(\frac{5}{2}\right)^{5-1}=\dfrac{125}{4}$$
Hence,
$
a_5=\dfrac{125}{4}$ and $a_n=\dfrac{4}{5}\left(\dfrac{5}{2}\right)^{n-1}$
