How do you find the sum of the infinite geometric series 15 - 3 + 0.6 - 0.12 + . . .?

1 Answer
Feb 3, 2016

S_oo=12.5

Explanation:

Given is the infinite geometric series 15 - 3 + 0.6 - 0.12 + . . .
The first term is T_1=a=15, second term T_2=-3

Since it's a geometric series, we have to divide the second term by the first term, so T_2/T_1=-cancel{3}^1/cancel{15}^5=-1/5
So, T_2/T_1=r=(-1)/5

The formula for the infinite geometric series is
S_oo=a/(1-r) for |r|<1
We have all the required terms, so by replacing them for their respective values we have
S_oo=15/(1-(-1/5))=15/(1+1/5)=15/(6/5)=cancel{15}^5*5/cancel{6}^2=5*5/2=25/2

So now you know how I got that answer up there.