Question

If the arithmetic mean of $a$ and $b$ is double of their geometric mean, with $a>b>0$, then a possible value for the ratio $\frac{a}{b}$, to the nearest integer, is

• A. 5
• B. 8
• C. 11
• D. 14

Answer 1

Answer: (D)

14

A.M of a and b$=\frac{a+b}{2}$
G.M. of a and b$=\sqrt{ab}$
A.M=2 G.M
$\frac{a+b}{2}=2\sqrt{ab}$
$\frac{a+b}{2\sqrt{ab}}=\frac{2}{1}$
Apply componendo -dividendo
$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{2+1}{2-1}$
$\frac{{(\sqrt{a}+\sqrt{b})}^2}{{(\sqrt{a}-\sqrt{b})}^2}=\frac{3}{1}$
$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}$
$\sqrt{a}+\sqrt{b}=\sqrt{3a}-\sqrt{3b}$
$(\sqrt{3}-1)\sqrt{a}=(\sqrt{3}+1)\sqrt{b}$
$\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\frac{a}{b}=\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
$\frac{a}{b}=\frac{{(2+\sqrt{3})}^2}{{{\left(2\right)}^2-\left(\sqrt{3}\right)}^2}=\frac{4+3+4\sqrt{3}}{4-3}$
$\frac{a}{b}=\frac{7+4\sqrt{3}}{1}$
$\frac{a}{b}\approx 13.9=14$