Question

Let $a$ and $b$ be positive real numbers. If $a,A_1,A_2,b$ are in arithmetic progression, $a,G_1,G_2,b$ are in geometric progression and $a,H_1,H_2,B$ are in harmonic progression, then show that
$\frac{G_1{G}_2}{H_1{H}_2}=\frac{A_1+A_2}{H_1+H_2}=\frac{(2a+b)(a+2b)}{9ab}$

Answer 1

Since, $a,A_1,A_2,b$ are in AP.
$\Rightarrow A_1+A_2=a+b$
$a,G_1,G_2,b$ are in $GP\Rightarrow G_1{G}_2=ab$
and $a,H_1,H_2,b$ are in $HP.$
$\Rightarrow H_1=\frac{3ab}{2b+a},H_2=\frac{3ab}{b+2a}$
$\therefore \frac{1}{H_1}+\frac{1}{H_2}=\frac{1}{a}+\frac{1}{b}$
$\Rightarrow \frac{H_1+H_2}{H_1{H}_2}=\frac{A_1+A_2}{G_1{G}_2}=\frac{1}{a}+\frac{1}{b}...(i)$
Now, $\frac{G_1{G}_2}{H_1{H}_2}=\frac{ab}{\left(\frac{3ab}{2b+a}\right)+\left(\frac{3ab}{b+2a}\right)}$
$=\frac{(2a+b)(a+2b)}{9ab}...(ii)$
From Eqs. (i) and (ii), we get
$\frac{G_1{G}_2}{H_1{H}_2}=\frac{A_1+A_2}{H_1+H_2}=\frac{(2a+b)(a+2b)}{9ab}$