Asked by Tam
The sum to infinity of a convergent geometric progression is S and the sum to infinity of the square of the terms is 2S. Given further that the sum of the first two terms is 25/8, find the value of S.
Answers
Answered by
oobleck
S1 = a(1 + r + r^2 + ...) = a/(1-r)
S2 = a^2(1 + r^2 + r^4 + ...) = a^2/(1-r^2)
a/(1-r) = 25/8
a^2/(1-r^2) = 25/4
divide to get
(1+r)/a = 1/2
a(1+r) = 25/8
r = 1/4 and a = 5/2
S = a/(1-r) = (5/2)/(3/4) = 10/3
r = -9/4 and a = -5/2
S = a/(1-r) = (-5/2)/(13/4) = -13/10
S2 = a^2(1 + r^2 + r^4 + ...) = a^2/(1-r^2)
a/(1-r) = 25/8
a^2/(1-r^2) = 25/4
divide to get
(1+r)/a = 1/2
a(1+r) = 25/8
r = 1/4 and a = 5/2
S = a/(1-r) = (5/2)/(3/4) = 10/3
r = -9/4 and a = -5/2
S = a/(1-r) = (-5/2)/(13/4) = -13/10
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