Since it is a geometric sequence is given by:
[tex]\begin{gathered} a_n=a1\cdot r^{n-1} \\ _{\text{ }}where\colon \\ a1=81 \\ a2=27 \\ so\colon \\ 27=81r \\ r=\frac{27}{81} \\ r=\frac{1}{3} \\ _{\text{ }}hence\colon \\ a_n=81(\frac{1}{3})^{n-1} \end{gathered}[/tex]
The sum is given by:
[tex]S_{180}=\sum ^{180}_{n\mathop=1}81(\frac{1}{3})^{n-1}=\frac{a1(r^n^{}-1)}{r^{}-1}=\frac{81(\frac{1}{3}^{180}-1)}{\frac{1}{3}-1}=121.5[/tex]
Answer:
121.5
