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Question

Let a, b be positive real numbers. If a, G1, G2, b are in geometric progression and a, H1, H2, b are in harmonic progression, show that
G1G2H1H2==(2a+b)(a+2b)9ab

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Solution

Now a, H1,H2 ,b are in H.P.
1a,1H1,1H2,1b are in A.P.
1b=1a+3D D=13(1b1a)
1H1=1a+D=1a+13(1b1a)=2b+a3ab
1H2=1a+2D=1a+23(1b1a)=2a+b3ab
G1G2H1H2=ab(2b+a)(2a+b)9a2b2=(2b+a)(2a+b)9ab
Note : The value of H2 can be obtained from the value of H1 if we interchange a and b.

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