Answer
$\dfrac{13}{36}$
Work Step by Step
The sum of the first $n$ terms of a geometric sequence can be calculated using the formula:
$$S_n=\dfrac{a_1(1-r^n)}{1-r}; r \ne 1$$
where, $r$ is the common ratio, $r$ and can be computed as the quotient of a term and the term preceding it.
The given sequence has:
$a_1=\dfrac{3}{4}$
$r=\dfrac{a_2}{a_1}=\dfrac{-\frac{1}{2}}{\frac{3}{4}}=-\dfrac{1}{2} \cdot \dfrac{4}{3}=-\dfrac{2}{3}$
Plug these values in the formula above to obtain:
\begin{align*}S_n&=\dfrac{\frac{3}{4}\left(1-\left(\frac{-2}{3}\right)^n\right)}{1-\left(\frac{-2}{3}\right)}\\
\\S_n&=\dfrac{\frac{3}{4}\left(1-\left(\frac{-2}{3}\right)^n\right)}{\frac{5}{3}}
\end{align*}
With $n=4$ the equation above yields:
\begin{align*}S_4&=\dfrac{\frac{3}{4}\left(1-\left(\frac{-2}{3}\right)^4\right)}{\frac{5}{3}}\\
\\S_4&=\dfrac{\frac{3}{4}\left(1-\frac{16}{81}\right)}{\frac{5}{3}}\\
\\S_4&=\dfrac{3}{4}\left(\dfrac{65}{81}\right)\cdot \dfrac{3}{5}\\
\\S_4&=\dfrac{13}{36}
\end{align*}
