Answer
Geometric
$a_5=\frac{4}{27}$
Work Step by Step
We are given the sequence $\{a_n\}$:
$$\dfrac{3}{4}, \dfrac{1}{2},\dfrac{1}{3},\dfrac{2}{9}\dots $$
Check if the sequence is arithmetic:
$$\begin{align*}
a_2-a_1&=\dfrac{1}{2}-\dfrac{3}{4}=-\dfrac{1}{4}\\
a_3-a_2&=\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{1}{6}\\
a_4-a_3&=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}.
\end{align*}$$
Since the differences between consecutive terms are not constant, the sequence is not arithmetic.
Check if the sequence is geometric:
$$\begin{align*}
\dfrac{a_2}{a_1}&=\dfrac{\frac{1}{2}}{\frac{3}{4}}=\dfrac{2}{3}\\
\dfrac{a_3}{a_2}&=\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{2}{3}\\
\dfrac{a_4}{a_3}&=\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{3}.
\end{align*}$$
Since the ratios between consecutive terms are constant, the sequence is geometric.
Its first term is $a_1=\frac{3}{4}$ and its common ratio $r=\frac{2}{3}$.
We calculate the fifth term:
$$a_5=a_4r=\dfrac{2}{9}\cdot\dfrac{2}{3}=\dfrac{4}{27}.$$