Question

If the ratio of the A.M. and G.M. of two positive numbers $a$ and $b$ is $m:n$, then $a:b$ is equal to

• A. $\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)$
• B. $\left(m-\sqrt{m^2-n^2}\right):\left(m+\sqrt{m^2-n^2}\right)$
• C. $\left(m+\sqrt{m^2+n^2}\right):\left(m-\sqrt{m^2+n^2}\right)$
• D. $\left(m-\sqrt{m^2+n^2}\right):\left(m+\sqrt{m^2+n^2}\right)$

Answer 1

Answer: (A)

Let the A.M. of the numbers $a$ and $b$ is $A$ and G.M. of $a$ and $b$ is $G$, then $A=\frac{a+b}{2}$ and $G=\sqrt{ab}$
Given, $A:G=m:n$
i.e., $\frac{A}{G}=\frac{m}{n}\Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac{m}{n}$
Applying componendo and dividendo rule, we get
$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{m+n}{m-n}$
$\Rightarrow \frac{(\sqrt{a}{)}^2+(\sqrt{b}{)}^2+2\sqrt{ab}}{(\sqrt{a}{)}^2+(\sqrt{b}{)}^2-2\sqrt{ab}}=\frac{m+n}{m-n}$
$\Rightarrow \frac{(\sqrt{a}+\sqrt{b}{)}^2}{(\sqrt{a}-\sqrt{b}{)}^2}=\frac{m+n}{m-n}$
$\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}$
Again, applying componendo and dividendo rule, we get
$\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$
$\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$
$\Rightarrow \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$
Now, squaring on both sides, we get
$\frac{a}{b}=\frac{(\sqrt{m+n}+\sqrt{m-n}{)}^2}{(\sqrt{m+n}-\sqrt{m-n}{)}^2}$
$\Rightarrow \frac{a}{b}=\frac{m+n+m-n+2\sqrt{m+n}\sqrt{m-n}}{m+n+m-n-2\sqrt{m+n}\sqrt{m-n}}$
$\left[\because (a+b{)}^2=a^2+b^2+2ab(a-b{)}^2=a^2+b^2-2ab\right]$
$\Rightarrow \frac{a}{b}=\frac{2m+2\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}}\left[\because (a+b)(a-b)=a^2-b^2\right]$
$\Rightarrow \frac{a}{b}-\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$
$\Rightarrow a:b=\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)$