Answer
$S_{4} = \frac{312}{125} $
Work Step by Step
Given geometric sequence
$2,\frac{2}{5},\frac{2}{25},...$
$a_{1} =2$
Common ratio $r = \frac{a_{n}}{a_{n-1}}$
$r= \frac{a_{2}}{a_{1}} = \frac{\frac{2}{5}}{2} = \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}$
To find sum of first four terms, substitute $a_{1}, r $ and $n=4$ in
$S_{n} =\frac{ a_{1}(1-r^{n})}{1-r}$
$S_{4} =\frac{2(1-(\frac{1}{5})^{4})}{1-(\frac{1}{5})}$
$S_{4} =\frac{2(1-\frac{1}{625})}{\frac{5-1}{5}}$
$S_{4} =\frac{2(\frac{625-1}{625})}{\frac{4}{5}}$
$S_{4} = 2 \times \frac{624}{625} \times \frac{5}{4}$
$S_{4} = \frac{312}{125} $