Mathematics - Sequence & Series Question with Solution | TestHub

$\mathrm{G}=\sqrt{ab}$ $$\begin{array}{rl}& \mathrm{M}=\frac{\frac{1}{a}+\frac{1}{b}}{2}\\ & \mathrm{M}=\frac{a+b}{2ab}\end{array}$$ Given that $\frac{1}{\mathrm{M}}:\mathrm{G}=4:5$ $$\begin{array}{rl}& \frac{2ab}{(a+b)\sqrt{ab}}=\frac{4}{5}\\ \Rightarrow & \frac{a+b}{2\sqrt{ab}}=\frac{5}{4}\\ \Rightarrow & \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{5+4}{5-4}\end{array}$$ \{Using Componendo \& Dividendo $$\begin{array}{rl}& \Rightarrow \frac{(\sqrt{a}{)}^2+(\sqrt{b}{)}^2+2\sqrt{ab}}{(\sqrt{a}{)}^2+(\sqrt{b}{)}^2-2\sqrt{ab}}=\frac{9}{1}\\ & \Rightarrow {\left(\frac{\sqrt{b}+\sqrt{a}{)}^2}{\sqrt{b}-\sqrt{a}}\right)}^2=\frac{9}{1}\Rightarrow \frac{\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}=\frac{3}{1}\\ & \Rightarrow \frac{\sqrt{b}+\sqrt{a}+\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}-\sqrt{b}+\sqrt{a}}=\frac{3+1}{3-1}\end{array}$$ \{Using Componendo \& Dividendo $$\sqrt{\frac{b}{a}}=\frac{4}{2}=2$$ $$\begin{array}{rl}\frac{b}{a}& =\frac{4}{1}\\ \frac{a}{b}& =\frac{1}{4}\Rightarrow a:b=1:4\end{array}$$