My answer is wrong... I realized that... The 2nd term on both of those was not 8. Sorry.
Hello, Mr_Green!
The sum of an infinite geometric series is 36.
The second term of the series is 8.
Find two possibilities for combinations of a (the first term) and r (the common ratio).
Given the first term a and the common ratio r,
. . the nth term is: .a_n .= .ar^{n-1}
. . and the sum is: .S .= .a/(1 - r)
We are told the second term is 8: .a_2 .= .ar .= .8 . → . r = 8/a . [1]
We are told the sum is 36: .S .= .a/(1 - r) = 36 . → . a .= .36 - 36r . [2]
Substitute [1] into [2]: .a .= .36 - 36(8/a)
We have a quadratic: .aČ - 36a + 288 .= .0
. . which factors: .(a - 12)(a - 24) .= .0
. . and has roots: .a .= .12, 24
Substitute into [1] and we get: .r .= .2/3, 1/3
Therefore, the series are: .a = 12, r = 2/3 .and .a = 24, r = 1/3