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16 August, 21:40

If the sum of three consecutive terms of GP are k-4, k, k+5. Find k

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  1. 16 August, 23:37
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    k = 20

    Step-by-step explanation:

    t1 = a = k-4

    a = k - 4

    t2 = ar = k

    (k-4) r = k - 4

    r = k / (k-4)

    t3 = ar^2 = k + 5

    (k - 4) (k/k - 4) ^2 = k+5

    k^2/k-4 = k+5

    k^2 = (k+5) (k-4)

    k^2 = k^2 - 4k + 5k - 20

    0 = k - 20

    k = 20
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