Answer
$S_{4} =-10$
Work Step by Step
Given geometric sequence
$\frac{-1}{4},\frac{-3}{4},\frac{-9}{4},...$
$a_{1}= \frac{-1}{4}$
Common ratio $r = \frac{a_{n}}{a_{n-1}}$
$r= \frac{a_{2}}{a_{1}} = \frac{-3}{4} \times \frac{-4}{1} =3$
To find sum of first four terms, substitute $a_{1}, r $ and $n=4$ in
$S_{n} =\frac{ a_{1}(1-r^{n})}{1-r}$
$S_{4} =\frac{ \frac{-1}{4}(1-3^{4})}{1-3}$
$S_{4} =\frac{ \frac{-1}{4}(1-81)}{-2}$
$S_{4} =\frac{ \frac{-1}{4}(-80)}{-2}$
$S_{4} = \frac{-1}{4} \times -80 \times \frac{-1}{2}$
$S_{4} =-10$
