Question

What is the sum of an 8-term geometric series if the first term is 12 and the last term is -3,359,232?

Answer 1

We must find the common ratio first...

Since a(n)=ar^(n-1)

-3359232=12r^7

-279936=r^7

-279936^(1/7)=r

-6=r

So our geometric sequence is:

a(n)=12(-6)^(n-1)

And the sum of a geometric sequence is:

s(n)=a(1-r^n)/(1-r), since a=12, r=-6, and n=8 we have:

s(8)=12(1-(-6)^8)/(1--6)

s(8)=12(1-(-6)^8)/7

s(8)= -2,879,340