If a, b, c are in A.P. and geometric means of ac and ab, ab and bc, ca and cb are d, e, f respectively then d2, e2, f2 are in

(1) A.P.

(2) G.P.

(3) H.P.

(4) A.G.P

Solution:

Given GM of ac and ab = d

So d2 = ac.ab

d2= a2bc

Given GM of ab and bc = e

e2 = ab2c

Given GM of ca and cb = f

f2 = abc2

Since a, b, c in AP

b = (a+c)/2

d2+f2 = a2bc+ abc2

= b(a2c+ac2)

= ((a+c)/2) ac(a+c)

= ac(a+c)2/2

e2 = ab2c

= ac(a+c)2/4

2e2 = ac(a+c)2/2

So 2e2 = d2+f2

So they are in AP.

Hence option (1) is the answer.

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