Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of $\frac{1}{a}$  and  $\frac{1}{b}$. If  $\frac{1}{M}:G$  is  $4:5$, then $a:b$ can be : 

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