Question

If the ratio of AM between two positive real numbers a and b to their HM is m:n then a:b isequal to

Answer 1

$$\begin{gathered} A.M.oftwopositiverealnumbersaandb=\frac{a+b}{2} \\ And,H.M.oftwopositiverealnumbersaandb=\frac{2ab}{a+b} \\ Now, \\ \frac{A.M.}{H.M.}=\frac{\frac{a+b}{2}}{\frac{2ab}{a+b}} \\ \Rightarrow \frac{m}{n}=\frac{{\left(a+b\right)}^2}{4ab} \\ \Rightarrow \frac{2m}{n}=\frac{{\left(a+b\right)}^2}{2ab} \\ \Rightarrow \frac{2m}{n}=\frac{a^2+b^2+2ab}{2ab} \\ \Rightarrow \frac{2m}{n}-1=\frac{a^2+b^2}{2ab} \\ \Rightarrow \frac{2m-n}{n}=\frac{a^2+b^2}{2ab} \\ Applycomponendoanddividendo, \\ \Rightarrow \frac{2m-n+n}{2m-n-n}=\frac{a^2+b^2+2ab}{a^2+b^2-2ab} \\ \Rightarrow \frac{2m}{2\left(m-n\right)}=\frac{{\left(a+b\right)}^2}{{\left(a-b\right)}^2} \\ \Rightarrow \frac{\sqrt{m}}{\sqrt{\left(m-n\right)}}=\frac{\left(a+b\right)}{\left(a-b\right)} \\ Applycomponendoanddividendo, \\ \Rightarrow \frac{\sqrt{m}+\sqrt{\left(m-n\right)}}{\sqrt{m}-\sqrt{\left(m-n\right)}}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(a+b\right)-\left(a-b\right)} \\ \Rightarrow \frac{\sqrt{m}+\sqrt{\left(m-n\right)}}{\sqrt{m}-\sqrt{\left(m-n\right)}}=\frac{a}{b} \\ \Rightarrow \frac{\mathbf{a}}{\mathbf{b}}\mathbf{=}\frac{\sqrt{\mathbf{m}}\mathbf{+}\sqrt{\left(\mathbf{m}\mathbf{-}\mathbf{n}\right)}}{\sqrt{\mathbf{m}}\mathbf{-}\sqrt{\left(\mathbf{m}\mathbf{-}\mathbf{n}\right)}} \end{gathered}$$