The given geometric series is
[tex]120-80+\frac{160}{3}-\frac{320}{9}+\cdots[/tex]In a geometric series, there is a common ratio between consecutive terms defined as
[tex]r=\frac{-80_{}}{120_{}}=-\frac{2}{3}[/tex]The sum of the first n terms of a geometric series is given by
[tex]S_n=\frac{a(1-r^n)}{1-r},r<1[/tex]Where a is the first term.
From the given series
a = 120
Hence, the sum of the first 8 terms is
[tex]S_8=\frac{120(1-(-\frac{2}{3})^8)}{1-(-\frac{2}{3})}[/tex]Simplify the brackets
[tex]S_8=\frac{120(1-\frac{2^8}{3^8}^{})}{1+\frac{2}{3}}[/tex]Simplify further
[tex]\begin{gathered} S_8=\frac{120(1-\frac{256}{6561})}{\frac{3+2}{3}} \\ S_8=\frac{120(\frac{6561-256}{6561})}{\frac{5}{3}} \\ S_8=\frac{120(\frac{6305}{6561})}{\frac{5}{3}} \\ S_8=\frac{120\times6305}{6561}\div\frac{5}{3} \\ S_8=\frac{120\times6305}{6561}\times\frac{3}{5} \\ S_8=\frac{120\times6305}{6561}\times\frac{3}{5} \\ S_8=\frac{8\times6305}{729} \\ S_8=\frac{50440}{729} \end{gathered}[/tex]Therefore, the sum of the first 8 terms is
[tex]\frac{50440}{729}[/tex]