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Find k such that k + 9, k – 6 and 4 form three consecutive terms of a G.P.

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Best answer

Let a = k + 9; b = k − 6; and c = 4;

We know that a, b and c are in GP, then

b2 = ac {using property of geometric mean}

(k − 6)2 = 4(k + 9)

k2 – 12k + 36 = 4k + 36

k2 – 16k = 0

k = 0 or k = 16

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