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Q.
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
Sequences and Series
Solution:
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 $G=\sqrt{a b}$
Given, $A: G=m: n$
i.e., $ \frac{A}{G}=\frac{m}{n} \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}$
Applying componendo and dividendo rule, we get
$ \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}} =\frac{m+n}{m-n} $
$\Rightarrow \frac{(\sqrt{a})^2+(\sqrt{b})^2+2 \sqrt{a b}}{(\sqrt{a})^2+(\sqrt{b})^2-2 \sqrt{a b}} =\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
$\frac{a}{b}=\frac{(\sqrt{m+n}+\sqrt{m-n})^2}{(\sqrt{m+n}-\sqrt{m-n})^2}$
$\Rightarrow \frac{a}{b}=\frac{m+n+m-n+2 \sqrt{m+n} \sqrt{m-n}}{m+n+m-n-2 \sqrt{m+n} \sqrt{m-n}}$
$\left[\because(a+b)^2=a^2+b^2+2 a b (a-b)^2=a^2+b^2-2 a b\right]$
$\Rightarrow \frac{a}{b}=\frac{2 m+2 \sqrt{m^2-n^2}}{2 m-2 \sqrt{m^2-n^2}} \left[\because(a+b)(a-b)=a^2-b^2\right]$
$\Rightarrow \frac{a}{b}-\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$
$\Rightarrow a: b=\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)$