Question

Find the sum of a geometric series if a1=8, and an=0.394 and r=9/11 (the fraction 9/11) ?

Answer 1

Is this an infinite series... see below

Explanation:

`a_n= 8*(9/11)^(n-1)`

Given that `a_n=0.394`
Let `x` be `n-1`

Solve for x:
`0.394= 8*(9/11)^(x)`

`0.04925= (9/11)^x`

`log0.04925= log(9/11)^x`
`log0.04925= xlog(9/11)`

`x= log0.04925/log(9/11)`

`x=15`

`n=16`

Using that the sum would be:

`S_(16)= 8((1-(9/11)^(16))/(1-(9/11)))=42.226`

If the series you provided was an infinite series then:

`S= 8/(1-9/11)= 44`