Question

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

Answer 1

`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.