Question

Let $G$ be the geometric mean of two positive numbers $a$ and $b$, and $M$ be the arithmetic means of $\frac{1}{a}$ and $\frac{1}{b}$. If $\frac{1}{M}:G$ is $4:5$, then $a:b$ can be :

• A. 1 : 4
• B. 1 : 2
• C. 2 : 3
• D. 3 : 4

Answer 1

The correct option is A 1 : 4

$G=\sqrt{ab},M=\frac{a+b}{2ab}$

$MG\times G=\frac{a+b}{2}$

$\frac{1}{MG}=\frac{4}{5}$

$\sqrt{ab}=\frac{2}{5}(a+b)$

$ab=\frac{4}{25}(a^2+b^2+2ab)$

$4\frac{a^2}{b^2}+4-17\frac{a}{b}\Rightarrow 0$

Let 'x' $=\frac{a}{b}$

$4x^2+4-17x=0$

$x=4$ or $\frac{1}{4}$

$\frac{a}{b}=4$ or $\frac{a}{b}=\frac{1}{4}$.