Answer
$a_5=\dfrac{243}{256}$
$a_n=3\left(\dfrac{-3}{4}\right)^{n-1}$
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,
$$r=\dfrac{a_{3}}{a_2}=\dfrac{27/16}{-9/4}=\dfrac{-3}{4}$$
Plug $3$ for $a_1$ and $\dfrac{-3}{4}$ for $r$ in equation $(1)$ above to obtain:
$$ a_n=3\left(\dfrac{-3}{4}\right)^{n-1}$$
With $n=5,$ the equation above yields the result :
$$a_5=10\left(\dfrac{-1}{2}\right)^{5-1}=\dfrac{243}{256}$$
Hence,
$a_5=\dfrac{243}{256}$ and $a_n=3\left(\dfrac{-3}{4}\right)^{n-1}$.
