Question

Given a Geometric sequence where: a3 = –27 and a5 = –162

Answer 1

Part a.

We have

$a_n = a_1 r^{n-1}$

$a_3 = a_1 r^2 = -27$

$a_5 = a_1 r^4 = -162$

Dividing,

$r^2 = \dfrac{-162}{-27} = 6$

Ugh.  It would have killed them to choose a perfect square?

$r= \pm \sqrt 6$

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

$a_1 = a_3/r^2 = -27/6 = -3/2$

$a_n = -\frac 3 2 (\sqrt{6})^{n-1}$

Part c

$a_8 = -\frac 3 2 (\sqrt{6})^{7} = -324 \sqrt{6}$

Part d

$\displaystyle \sum_{k=1}^{n} a r^{k-1} = \dfrac{ a(1 - r^n)}{1 -r}$

$\displaystyle s_8 = \sum_{k=1}^{8} a_k = \dfrac{ (-3/2)(1 - (\sqrt{6})^8)}{1 -\sqrt 6}$

$s_8 = - \dfrac{777}{2} (1 + \sqrt 6)$