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Find the reason for the geometric progression in which a1 = 1, n = 3, S3 = 157

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asked by Dany
  1. S3 = 1(1-r^3)/(1-r) = 157
    1+r+r^2 = 157
    r^2+r-156 = 0
    (r-12)(r+13) = 0
    r = 12 or -13
    So, the GPs are

    1,12,144,...
    1,-13,169,...

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    posted by Steve

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