Question

If the $A.M.$ between two numbers exceeds their $G.M.$ by $2$ and the $G.M.$ exceeds their $H.M.$ by $8/5$, find the smaller of the two numbers.

Answer 1

If the numbers be a and b, then
$A-G=2\therefore (a+b)/2-\sqrt{ab}=2$
or ${(\sqrt{a}-\sqrt{b})}^2=4$ ...(1)
$G-H=\frac{8}{5}\therefore \sqrt{ab}-\frac{2ab}{a+b}=\frac{8}{5}$
or $\sqrt{ab}\frac{[a+b-2\sqrt{ab}]}{a+b}=\frac{8}{5}$
or $5\sqrt{ab}{(\sqrt{a}-\sqrt{b})}^2=8(a+b)$
or $5\sqrt{ab}.4=8(a+b),$ by (1)
$\therefore 2a-5\sqrt{ab}+2b+0$
or $(2\sqrt{a}-\sqrt{b})(\sqrt{a}-2\sqrt{b})=0$
$\therefore \sqrt{a}=2\sqrt{b}or\left(\sqrt{b}/2\right)\therefore a=4bor\left(b/4\right)$
Form (1), ${(2\sqrt{b}-\sqrt{b})}^2=4orb=4$
$\therefore \sqrt{a}=2\sqrt{b}=4ora=16$
Hence the numbers are $16$ and $4$. The other factor will give numbers as $4$ and $16$.