Question

If the A.M. and G.M. between two numbers are in ratio of $m:n,$ then the ratio of numbers can be

• A. $\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$
• B. $\frac{m+\sqrt{m^2-n^2}}{n-\sqrt{m^2-n^2}}$
• C. $\frac{m+\sqrt{m^2+n^2}}{m-\sqrt{m^2+n^2}}$
• D. $\frac{n+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$

Answer 1

The correct option is A $\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$

Let the two numbers be $a$ and $b$, so
$A=\frac{a+b}{2},G=\sqrt{ab}\Rightarrow a+b=2A,ab=G^2\cdots \left(1\right)$

The equation whose roots are $a$ and $b$ is
$x^2-(a+b)x+ab=0\Rightarrow x^2-2Ax+G^2=0\Rightarrow x=\frac{2A\pm \sqrt{4A^2-4G^2}}{2}\Rightarrow x=A\pm \sqrt{A^2-G^2}$

So, the two numbers are $a=A+\sqrt{A^2-G^2}$ and
$b=A-\sqrt{A^2-G^2}$
It is given that
$A:G=m:n\Rightarrow A=km,G=kn$
Substituting the values of $A$ and $G$, we get
$\frac{a}{b}=\frac{km+\sqrt{k^2{m}^2-k^2{n}^2}}{km-\sqrt{k^2{m}^2-k^2{n}^2}}\therefore \frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$