Answer
$S_{\infty}=\dfrac{18}{5}$
Work Step by Step
Using $S_{\infty}=\dfrac{a_1}{1-r}$, the sum of the infinite geometric sequence $
6,-4,\dfrac{8}{3},...
$ is
\begin{array}{l}
S_{\infty}=\dfrac{6}{1-\left( -\dfrac{2}{3} \right)}\\\\
S_{\infty}=\dfrac{6}{\dfrac{5}{3}}\\\\
S_{\infty}=\dfrac{18}{5}
.\end{array}