the common ratio of the infinite geometric series whose 2nd term is 3/2 and its sum is 8?

Answer 1

2nd term is 3/2 --- ar = 3/2
sum of all terms is 8
a/(1-r) = 8

divide the first equation by the 2nd equation:
(ar) / (a/(1-r)) = (3/2) / 8
r(1-r) = 3/16
r - r^2 = 3/16
3 = 16r - 16r^2
16r^2 - 16r + 3 = 0
(4r - 1)(4r - 3) = 0
r = 1/4 or r = 3/4

check for r = 1/4
a(1/4) = 3/2
a = 6

sum(all terms) = 6/(1 - 1/4)
= 6/(3/4) = 8
2nd term = ar = 6(1/4) = 3/2

I will leave the check for the second solution up to you
=

Answer 2

THANKS