Answer
$422$
Work Step by Step
The sum of the first $n$ terms can be computed as:
$S_n=\dfrac{a_1(1-r^{n})}{1-r}...(1)$
where, $a_1$ is first term and $r$ is the common ratio, $r$ and can be computed as the quotient of a term and the term preceeding it. In a geometric series the ratio between consecutive terms is constant.
We can see from the given series that $a_1=243 \left(\dfrac{2}{3}\right)^{1}=162$ and $r=\dfrac{2}{3}$
Plug $5$ for $n$ and $162$ for $a_1$ and $\dfrac{2}{3}$ for $r$ in the equation (1) to obtain:
$$S_5=\dfrac{162\left[1-\left(\dfrac{2}{3}\right)^{5}\right]}{1-\dfrac{2}{3}}=\dfrac{162\left[1-\left(\dfrac{2}{3}\right)^5\right]}{\dfrac{1}{3}}=422$$
Thus, the sum of the first five terms is: $S_5=422$.
