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

$\mathrm{G}=\sqrt{a b}$G=ab​ $$\begin{aligned} &\mathrm{M}=\frac{\frac{1}{a}+\frac{1}{b}}{2} \\ &\mathrm{M}=\frac{a+b}{2 a b} \end{aligned}$$​M=2a1​+b1​​M=2aba+b​​ Given that $\frac{1}{\mathrm{M}}: \mathrm{G}=4: 5$M1​:G=4:5 $$\begin{aligned} & \frac{2 a b}{(a+b) \sqrt{a b}}=\frac{4}{5} \\ \Rightarrow & \frac{a+b}{2 \sqrt{a b}}=\frac{5}{4} \\ \Rightarrow & \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{5+4}{5-4} \end{aligned}$$⇒⇒​(a+b)ab​2ab​=54​2ab​a+b​=45​a+b−2ab​a+b+2ab​​=5−45+4​​ {Using Componendo & Dividendo $$\begin{aligned} &\Rightarrow \frac{(\sqrt{a})^2+(\sqrt{b})^2+2 \sqrt{a b}}{(\sqrt{a})^2+(\sqrt{b})^2-2 \sqrt{a b}}=\frac{9}{1} \\ &\Rightarrow\left(\frac{\sqrt{b}+\sqrt{a})^2}{\sqrt{b}-\sqrt{a}}\right)^2=\frac{9}{1} \Rightarrow \frac{\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}=\frac{3}{1} \\ &\Rightarrow \frac{\sqrt{b}+\sqrt{a}+\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}-\sqrt{b}+\sqrt{a}}=\frac{3+1}{3-1} \end{aligned}$$​⇒(a​)2+(b​)2−2ab​(a​)2+(b​)2+2ab​​=19​⇒(b​−a​b​+a​)2​)2=19​⇒b​−a​b​+a​​=13​⇒b​+a​−b​+a​b​+a​+b​−a​​=3−13+1​​ {Using Componendo & Dividendo $$\sqrt{\frac{b}{a}}=\frac{4}{2}=2$$ab​​=24​=2 $$\begin{aligned} \frac{b}{a} &=\frac{4}{1} \\ \frac{a}{b} &=\frac{1}{4} \Rightarrow a: b=1: 4 \end{aligned}$$ab​ba​​=14​=41​⇒a:b=1:4​