If the arithmetic mean of two distinct positive real numbers \(a\) and \(b\) (where \(a>b\) ) is twice their geometri…

From the given condition, we can say that,

\(AM=2\times GM\\ \frac{a+b}{2}=2\sqrt{ab}\\ a+b=4\sqrt{ab}\)

\(\text{we know,}\\ {\left(a-b\right)}^{2}={\left(a+b\right)}^{2}-4ab\\ =16ab-4ab=12ab\\ a-b=2\sqrt{3}\sqrt{ab}\)

\(\text{ (Taking }+ve\text{ sign only as }a>b\text{ ) }\\ ∴\frac{a+b}{a-b}=\frac{4\sqrt{ab}}{2\sqrt{3}\sqrt{ab}}=\frac{2}{\sqrt{3}}\\ \text{ By componendo and dividendo, }\\ \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\)