# Arithmetic

The first, second and third terms of a geometric progression are 2k+3, k+6 and k, respectively. Given that all the terms of geometric progression are positive, calculate

(a) the value of the constant k
(b) the sum to infinity of the progression.

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1. since the ratio between terms is constant,

(k+6)/(2k+3) = k/(k+6)
(k+6)^2 = k(2k+3)
k^2+12k+36 = 2k^2+3k
k^2 - 9k - 36 = 0
(k-12)(k+3) = 0
k = 12

and the sequence is
27,18,12,...

so the ratio is 2/3 and S = 27/(1/3) = 81

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posted by Steve

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