Question

If $A$ be the A.M. and $H$ the H.M. between two numbers $a$ and $b$, then show
$\frac{a-A}{a-H}\times \frac{b-A}{b-H}=\frac{A}{H}.$

Answer 1

$A$ is $A.M$ of $a,b$ then $A=\frac{a+b}{2}$
$H$ is $H.M$ of $a,b$ then $H=\frac{2ab}{a+b}$
Now subettuting this in
$\frac{a-A}{a-H}\times \frac{b-A}{b-H}=\frac{a-\frac{a+b}{2}}{a-\frac{2ab}{a+b}}\times \frac{b-\frac{a+b}{2}}{b-\frac{2ab}{a+b}}$
$=\frac{a-b}{2(a^2-ab)}\times \frac{(b-a)}{2(b^2-ab)}=\frac{1}{2a}\times \frac{1}{2b}$
$=\frac{1}{4ab}=\frac{\left(\frac{a+b}{2}\right)}{\frac{2ab}{a+b}}=\frac{A}{H}$