Part a.
We have
[tex]a_n = a_1 r^{n-1}[/tex]
[tex]a_3 = a_1 r^2 = -27[/tex]
[tex]a_5 = a_1 r^4 = -162[/tex]
Dividing,
[tex]r^2 = \dfrac{-162}{-27} = 6[/tex]
Ugh. It would have killed them to choose a perfect square?
[tex] r= \pm \sqrt 6[/tex]
We can't really tell the sign on the common ratio; there's not enough information. Let's just pick the positive square root going forward.
Part b
[tex] a_1 = a_3/r^2 = -27/6 = -3/2[/tex]
[tex] a_n = -\frac 3 2 (\sqrt{6})^{n-1} [/tex]
Part c
[tex] a_8 = -\frac 3 2 (\sqrt{6})^{7} = -324 \sqrt{6} [/tex]
Part d
[tex] \displaystyle \sum_{k=1}^{n} a r^{k-1} = \dfrac{ a(1 - r^n)}{1 -r}[/tex]
[tex] \displaystyle s_8 = \sum_{k=1}^{8} a_k = \dfrac{ (-3/2)(1 - (\sqrt{6})^8)}{1 -\sqrt 6}[/tex]
[tex]s_8 = - \dfrac{777}{2} (1 + \sqrt 6) [/tex]
