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

Let a = k + 9; b = k−6;


c = 4


Since, a, b and c are in GP, then


b2 = ac {using idea 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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