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Question -

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



Answer -

Let a = k + 9; b =k−6; and c = 4;

We know that a, b andc 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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