Mathematics - Sequence & Series Question with Solution | TestHub

By the given condition, $\frac{a+b}{2}=2\sqrt{ab}\Rightarrow a+b=4\sqrt{ab}$ Now, $(a-b{)}^2=(a+b{)}^2-4ab=16ab-4ab=$ $12ab$ $\therefore a-b=\sqrt{12ab}=2\sqrt{3}\sqrt{ab}$. (Taking + ve sign only as $a>b$) $\therefore \frac{a+b}{a-b}=\frac{4\sqrt{ab}}{2\sqrt{3}\sqrt{ab}}=\frac{2}{\sqrt{3}}$ By componendo and dividendo, $\frac{2a}{2b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ or $\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$