Let a and b are positive real numbers. If a, A_1, A_2, b are in arithmetic progression a, G_1, G_2, b are in G.P. and a, H_1, H_2, b are in H.P. show that (G_1 G_2 / H_1 H_2)=(A_1+A_2 / H_1+H_2)=((2 a+b)(a+2 b) / 9 a b)

Since a, A_1, A_2, b are in A.P. => A_1+A_2=a+b a, G_1, G_2, b are in G.P.\ => G_1 G_2=a b a, H_1, H_2, b are in H.P.\ => H_1=(3 a b / 2 b+a), H_2=(3 a b / b+2 a) \ => (H_1+H_2 / H_1 H_2)=(a+b / a b)=(A_1+A_2 / G_1 G_2) (From (1) and (2))\ => (G_1 G_2 / H_1 H_2)=(A_1+A_2 / H_1+H_2) We also know a, H_1, H_2, b, are in H.P. \ (1 / H_1)=(1 / a)+(1 / 3)((1 / b)-(1 / a)) We get H_1=(3 a b / 2 b+a) and similarly for H_2=(3 a b / 2 a+b) Substituting (3), we get desired result (A_1+A_2 / H_1+H_2)\ =((2 b+a)(2 a+b) / 9 a b) .Hence proved.

Let a and b are positive real numbers. If a, A_1, A_2, ... | Sequence & Series (Mathematics)

Since $a, A_1, A_2, b$ are in A.P. $\Rightarrow A_1+A_2=a+b\\$ $a, G_1, G_2, b$ are in G.P. $\\ \Rightarrow G_1 G_2=a b\\$ $a, H_1, H_2, b$ are in H.P. $\\ \Rightarrow H_1=\frac{3 a b}{2 b+a}, H_2=\frac{3 a b}{b+2 a}$ $\\ \Rightarrow \frac{H_1+H_2}{H_1 H_2}=\frac{a+b}{a b}=\frac{A_1+A_2}{G_1 G_2}\\$ (From (1) and (2)) $\\ \Rightarrow \frac{G_1 G_2}{H_1 H_2}=\frac{A_1+A_2}{H_1+H_2}\\$ We also know $a, H_1, H_2, b$ , are in H.P. $\\ \frac{1}{H_1}=\frac{1}{a}+\frac{1}{3}\left(\frac{1}{b}-\frac{1}{a}\right)\\$ We get $H_1=\frac{3 a b}{2 b+a}$ and similarly for $H_2=\frac{3 a b}{2 a+b}\\$ Substituting (3), we get desired result $\frac{A_1+A_2}{H_1+H_2}\\ =\frac{(2 b+a)(2 a+b)}{9 a b}$ .Hence proved.

Explanation

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