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If a, b, c are in G.P. and a1/x = b1/y = c1/z, then xyz are in
(a) AP
(b) GP
(c) HP
(d) none of these

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Solution

(a) AP

a, b and c are in G.P. b2 = acTaking log on both the sides:2log b = log a + log c ........iNow, a1x = b1y =c1zTaking log on both the sides:log ax = log by = log cz ........iiNow, comparing i and ii:log ax=log a + log c 2y=log czlog ax=log a + log c 2y and log ax=log czlog a 2y-x=xlog c and log alog c=xzlog alog c=x2y-x and log alog c=xzx2y-x=xz 2y = x + zThus, x, y and z are in A.P.

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