Question

If A.M. between positive numbers $p$ and $q(p\ge q)$ is two times the $GM$, then $p$ : $q$ is -

• A. $1:1$
• B. $2:1$
• C. $(2+\sqrt{3}):(2-\sqrt{3})$
• D. $3:1$

Answer 1

Answer: (C)

Given numbers are $p$ and $q$
$\text{ A.M. }=\frac{p+q}{2},GM=\sqrt{pq}$
$=2\text{ (Given) }$
$\therefore \frac{p+q}{2\sqrt{pq}}=2$
$\therefore \frac{p+q+2\sqrt{pq}}{p+q-2\sqrt{pq}}=\frac{3}{1}$
(componendo dividendo method)
$\frac{(\sqrt{p}+\sqrt{q}{)}^2}{(\sqrt{p}-\sqrt{q}{)}^2}=\frac{3}{1}$
$\Rightarrow \frac{\sqrt{p}+\sqrt{q}}{\sqrt{p}-\sqrt{q}}=\sqrt{3}$
$\therefore \frac{2\sqrt{p}}{2\sqrt{q}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\therefore \frac{p}{q}={\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)}^2=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$