mathematics

in a geometric progression, the first term is a and the common ratio is r. the sum of the first two terms is 12 and the third term is 16.
a) determine the ratio (ar^2)/a+ar( aready found the answer as 4/3)
b) if the first term is larger than the second term, find the value of r.

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  1. a + ar = 12
    ar^2 = 16

    ar^2/(a(1 + r)) = 16/12
    r^2/(1+r) = 4/3 <==== you had that

    3r^2 = 4 + 4r
    3r^2 - 4r - 4 = 0
    (r-2)(3r+2) = 0
    r = 2 or r = -2/3

    if r = 2,
    ar^2 = 16 , a = 4 , and the sequence is 4, 8, 16, ...
    if r = -2/3,
    a(4/9) = 16, a = 36, and the sequence is 36, -24, 16,

    but it said, term1 > term2, so it must be the 2nd sequence, where
    r = -2/3

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  2. No idea

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