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If the AM and GM between two number are in the ratio $m: n$, then what is the ratio between the two numbers?
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Verified Answer
The correct answer is:
$\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
Let 'a' and 'b' two numbers.
A.M $=\frac{a+b}{2}$ and $G \cdot M=\sqrt{a b}$
According to the question, $A: G=m: n$
$\Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n} \Rightarrow \frac{(a+b)^{2}}{4 a b}=\frac{m^{2}}{n^{2}}$ ...(i)
and $\frac{(a+b)^{2}-4 a b}{4 a b}=\frac{m^{2}-n^{2}}{n^{2}}$
$\Rightarrow \frac{(a-b)^{2}}{4 a b}=\frac{m^{2}-n^{2}}{n^{2}}$ ...(ii)
Since, on dividing Equation (i) and (ii), we get
$\frac{(a+b)^{2}}{(a-b)^{2}}=\frac{m^{2}}{m^{2}-n^{2}} \Rightarrow \frac{(a+b)}{(a-b)}=\frac{m}{\sqrt{m^{2}-n^{2}}}$
$\Rightarrow \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
(Using componendo dividendo rule)
$\Rightarrow \quad \frac{2 a}{2 b}=\frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
A.M $=\frac{a+b}{2}$ and $G \cdot M=\sqrt{a b}$
According to the question, $A: G=m: n$
$\Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n} \Rightarrow \frac{(a+b)^{2}}{4 a b}=\frac{m^{2}}{n^{2}}$ ...(i)
and $\frac{(a+b)^{2}-4 a b}{4 a b}=\frac{m^{2}-n^{2}}{n^{2}}$
$\Rightarrow \frac{(a-b)^{2}}{4 a b}=\frac{m^{2}-n^{2}}{n^{2}}$ ...(ii)
Since, on dividing Equation (i) and (ii), we get
$\frac{(a+b)^{2}}{(a-b)^{2}}=\frac{m^{2}}{m^{2}-n^{2}} \Rightarrow \frac{(a+b)}{(a-b)}=\frac{m}{\sqrt{m^{2}-n^{2}}}$
$\Rightarrow \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
(Using componendo dividendo rule)
$\Rightarrow \quad \frac{2 a}{2 b}=\frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
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