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

ar^2 = 7/4
a/(1r) = 14
a = 14(1r)
back into the 1st...
14(1r)r^2 = 7/4
8(1r)r^2 = 1
8r^2  8r^3  1 = 0
8r^3  8r^2 + 1 = 0
after a few trialanderror attempts, I found
f(1/2) = 0 ,so 2r1 was a factor
(2r1)(4r^2 2r1) = 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, ....
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