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Question

If a, b, c are in G.P. and b-c, c-a, a-b are in H.P. then prove that a+b+c =3(ac)12

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Solution

bc, ca, ab are in HP
1ab+1bc=2ca
Clearing the denominations, we get
a2+c22b2+2ab+2bc4ac=0
This gives us
(a+b+c)2=a2+b2+c2+2ab+2bc+2ac=0+3b2+6ac=9b2
Hence either a+b+c=3b or a+b+c=3b
Now in the first case a+c=2b and ac=b2, Hence we must have a=b=c. This means that ab, ca, bc is not a harmonic progression.
Thus, We must have a+b+c=3b
Now given abc are in GP
b2=ac
b=ac1/2
a+b+c=3b=3(ac)1/2
Hence proved.

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