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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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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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