Question

What is the sum of an 8–term geometric series if the first term is 13 and the last term is –3,639,168?

Answer 1

We know a formula to find any term in a geometric sequence, \(\large x_{n} = ar^{n-1}\)
We know that the last term is -3629168 and the first term is 13, well, \(\large 3639168 = 13r^{8-1}\)
\(\large 3639168 = 13r^{7}\)
With that, you'll be able to find r, the common ratio, and then use the formula \[\large \sum_{k=0}^{n-1}(ar^k) = a \frac{1-r^n}{1-r}\]

Answer 2

I forgot the '-' sign before 3639168, my bad.

Answer 3

im getting -279936 = r^7 and r= -6 but i dont know what to do after

Answer 4

Okay, you common ratio is -6
You first term is 13
And by using the formula I showed about, plug everything in
\[\huge \sum_{k=0}^{8-1}=(ar^k)=13 (\frac{1-(-6)^8}{1-(-6)})\]