SOLUTION: The 3rd and the 4th terms of a geometric progression are 12 and 8 respectively. Find the sum to infinity of the progression.

Question 1083080: The 3rd and the 4th terms of a geometric progression are 12 and 8 respectively. Find the sum to infinity of the progression.

Found 2 solutions by amfagge92, natolino_2017:
Answer by amfagge92(93) (Show Source): You can put this solution on YOUR website!
81
Answer by natolino_2017(77) (Show Source): You can put this solution on YOUR website!
a(n) = a(r^(n-1))
a(3) = a(r^2) = 12.
a(4) = a(r^3) = 8.
a(4)/(a(3)) = r = 8/12 = 2/3.
a(4) = a(2/3)^3 = 8.
a = 27.
So a(n) = 27*(2/3)^(n-1).
using 1/(1-r) =lim n--->inf( 1 +r^1 +r^2 + r^3 + ... r^n), if |r|<1
Sum(a(n))) = 27/(1-2/3)) = 27/(1/3) = 81.
@natolino_

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