Question

If arithmetic mean of two distinct positive real numbers $a$ and $b(a>b)$ be twice their geometric mean, then $a:b=$

• A. $\left(2+\sqrt{3}\right):\left(2-\sqrt{3}\right)$
• B. $\left(2+\sqrt{5}\right):\left(2-\sqrt{5}\right)$
• C. $\left(2+\sqrt{2}\right):\left(2-\sqrt{2}\right)$
• D. $Noneofthese$

Answer 1

Answer: (A)

By the given condition,
$\frac{a+b}{2}=2\sqrt{ab}$
$\Rightarrow a+b=4\sqrt{ab}$
Now, ${\left(a-b\right)}^2={\left(a+b\right)}^2-4ab$
$=16ab-4ab$
$=12ab$
$\therefore a-b=\sqrt{12ab}=2\sqrt{3}\sqrt{ab}$.a
(Taking $+ve$ sign only as $a>b$)
$\therefore \frac{a+b}{a-b}=\frac{4\sqrt{ab}}{2\sqrt{3}\sqrt{ab}}=\frac{2}{\sqrt{3}}$
By componendo and dividendo,
$\frac{2a}{2b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ or $\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$