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
Open in App
Solution
Now a, H1,H2 ,b are in H.P. ∴1a,1H1,1H2,1b are in A.P. ∴1b=1a+3D∴D=13(1b−1a) ∴1H1=1a+D=1a+13(1b−1a)=2b+a3ab ∴1H2=1a+2D=1a+23(1b−1a)=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.