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
$\frac{G_1G_2}{H_1H_2}=\frac{A_1+A_2}{H_1+H_2}= \frac{(2a+b)(a+2b)}{9ab}$

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_1G_2 = ab $
and $ a,H_1,H_2,b$ are in $HP. $
$\Rightarrow H_1 = \frac{3ab}{2b+a},H_2 = \frac{3ab}{b+2a}$
$\therefore \frac{1}{H_1}+ \frac{1}{H_2}= \frac{1}{a}+\frac{1}{b}$
$\Rightarrow \frac{H_1+ H_2}{H_1 H_2} = \frac{A_1+A_2}{G_1G_2}= \frac{1}{a}+\frac{1}{b} ...(i)$
Now, $ \frac{G_1G_2}{H_1 H_2} = \frac{ab}{\bigg(\frac{3ab}{2b+a}\bigg)+\bigg(\frac{3ab}{b+2a}\bigg)}$
$ = \frac{(2a+b)\, (a+2b)}{9ab} ...(ii)$
From Eqs. (i) and (ii), we get
$ \frac{G_1G_2}{H_1 H_2} = \frac{A_1+A_2}{H_1+ H_2} = \frac{(2a+b)\, (a+2b)}{9ab} $