Find a geometric series which has a third term of 7/4 and a sum to infinity of 14.

Answer 1

ar^2 = 7/4

a/(1-r) = 14
a = 14(1-r)

back into the 1st...

14(1-r)r^2 = 7/4
8(1-r)r^2 = 1
8r^2 - 8r^3 - 1 = 0
8r^3 - 8r^2 + 1 = 0

after a few trial-and-error attempts, I found
f(1/2) = 0 ,so 2r-1 was a factor

(2r-1)(4r^2 -2r-1) = 0

r = 1/2 or r = (1 ± √5)/4

if r = 1/2 , then a = 14(1/2) = 7
GS is 7, 7/2, 7/4 , 7,8 , ....

if r = (1 + √5)/4
we have a contradiction with the sum of the infinitite series
we can only have such a sum if |r| < 1
and both of the irrational answers fall outside of that domain.

so the GS is 7, 7/2, 7/4, ....