## Intermediate Algebra (6th Edition)

$S_{4} =-10$
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 $S_{n} =\frac{ a_{1}(1-r^{n})}{1-r}$ $S_{4} =\frac{ \frac{-1}{4}(1-3^{4})}{1-3}$ $S_{4} =\frac{ \frac{-1}{4}(1-81)}{-2}$ $S_{4} =\frac{ \frac{-1}{4}(-80)}{-2}$ $S_{4} = \frac{-1}{4} \times -80 \times \frac{-1}{2}$ $S_{4} =-10$