Explanation
A geometric serie has a common ratio, the formula for the nth term is
[tex]\begin{gathered} a_n=a\cdot r^{n-1} \\ where \\ a_n=n^{th}\text{ term of the sequ}ence \\ a=\text{ first term of the sequence} \\ r\text{ is the common ratio} \end{gathered}[/tex]so
Step 1
find the geometric serie
a) find the common ratio
let
[tex]\begin{gathered} \text{first term= 120, so} \\ a_n=a\cdot r^{n-1} \\ 120=a\cdot r^{1-1} \\ 120=a\cdot r^0=a\cdot1 \\ 120=a \end{gathered}[/tex]so
a=120
and
[tex]\begin{gathered} \text{second term} \\ -80=a\cdot r^{2-1} \\ -80=120\cdot r^1 \\ -\frac{80}{120}=r \\ r=-\frac{2}{3} \\ \text{hence} \\ a_n=\text{ 120(}\frac{-2}{3})^{n-1} \end{gathered}[/tex]so, the sequence is
[tex]a_n=\text{ 120(}\frac{-2}{3})^{n-1}[/tex]let's check
[tex]\begin{gathered} a_n=\text{ 120(}\frac{-2}{3})^{n-1} \\ a=1 \\ a_1=\text{ 120(}\frac{-2}{3})^{1-1}=\text{120(}\frac{-2}{3})^0=120\rightarrow ok \\ a=2 \\ a_2=\text{120(}\frac{-2}{3})^{2-1}=120\cdot(-\frac{2}{3})=-80\rightarrow ok \\ a=3 \\ a_3=\text{ 120(}\frac{-2}{3})^{3-1}=120(\frac{-2}{3})^2=53.33=\frac{160}{3}\rightarrow ok \\ a=\text{ 4} \\ a_4=\text{ 120(}\frac{-2}{3})^{4-1}=\text{120(}\frac{-2}{3})^3=-35.55\rightarrow ok \end{gathered}[/tex]Step 2
now, let's find the sum:
the sum of geometric serie is give by
[tex]\begin{gathered} S_n=\frac{a(1-r^n)}{1-r} \\ \end{gathered}[/tex]let n= 8,
replace
[tex]\begin{gathered} S_n=\frac{120(1-(-\frac{2}{3})^8)}{1-(-\frac{2}{3})} \\ S_n=\frac{120(1-(-\frac{2}{3})^8)}{1+\frac{2}{3}} \\ S_n=\frac{120(1-(\frac{256}{6561})^{})}{\frac{5}{3}} \\ S_n=\frac{120(1-(\frac{256}{6561})^{})}{\frac{5}{3}} \\ S_n=\frac{120(\frac{6305}{6561})}{\frac{5}{3}} \\ S_n=\frac{120(\frac{6305}{6561})}{\frac{5}{3}}=\frac{120\cdot6305\cdot3}{6561\cdot5}=\frac{2269800}{32805}=\frac{50440}{729} \end{gathered}[/tex]therefore, the answer is
[tex]\frac{50440}{729}[/tex]I hope this helps you