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. $None\, of\, these$

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 $\frac{a}{b} = \frac{2+\sqrt{3}}{ 2-\sqrt{3}}$