Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{G}$ be the geometric mean of two positive numbers $\mathrm{a}$ and $\mathrm{b}$, and $\mathrm{M}$ be the arithmetic mean of $\frac{1}{\mathrm{a}}$ and $\frac{1}{\mathrm{~b}}$. If $\frac{1}{\mathrm{M}}: \mathrm{G}$ is $4: 5$, then $\mathrm{a}: \mathrm{b}$ can be:
Options:
Solution:
1963 Upvotes
Verified Answer
The correct answer is:
$1: 4$
$1: 4$
$\mathrm{G}=\sqrt{a b}$
$$
\begin{aligned}
&\mathrm{M}=\frac{\frac{1}{a}+\frac{1}{b}}{2} \\
&\mathrm{M}=\frac{a+b}{2 a b}
\end{aligned}
$$
Given that $\frac{1}{\mathrm{M}}: \mathrm{G}=4: 5$
$$
\begin{aligned}
& \frac{2 a b}{(a+b) \sqrt{a b}}=\frac{4}{5} \\
\Rightarrow & \frac{a+b}{2 \sqrt{a b}}=\frac{5}{4} \\
\Rightarrow & \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{5+4}{5-4}
\end{aligned}
$$
\{Using Componendo \& Dividendo
$$
\begin{aligned}
&\Rightarrow \frac{(\sqrt{a})^2+(\sqrt{b})^2+2 \sqrt{a b}}{(\sqrt{a})^2+(\sqrt{b})^2-2 \sqrt{a b}}=\frac{9}{1} \\
&\Rightarrow\left(\frac{\sqrt{b}+\sqrt{a})^2}{\sqrt{b}-\sqrt{a}}\right)^2=\frac{9}{1} \Rightarrow \frac{\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}=\frac{3}{1} \\
&\Rightarrow \frac{\sqrt{b}+\sqrt{a}+\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}-\sqrt{b}+\sqrt{a}}=\frac{3+1}{3-1}
\end{aligned}
$$
\{Using Componendo \& Dividendo
$$
\sqrt{\frac{b}{a}}=\frac{4}{2}=2
$$
$$
\begin{aligned}
\frac{b}{a} &=\frac{4}{1} \\
\frac{a}{b} &=\frac{1}{4} \Rightarrow a: b=1: 4
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{M}=\frac{\frac{1}{a}+\frac{1}{b}}{2} \\
&\mathrm{M}=\frac{a+b}{2 a b}
\end{aligned}
$$
Given that $\frac{1}{\mathrm{M}}: \mathrm{G}=4: 5$
$$
\begin{aligned}
& \frac{2 a b}{(a+b) \sqrt{a b}}=\frac{4}{5} \\
\Rightarrow & \frac{a+b}{2 \sqrt{a b}}=\frac{5}{4} \\
\Rightarrow & \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{5+4}{5-4}
\end{aligned}
$$
\{Using Componendo \& Dividendo
$$
\begin{aligned}
&\Rightarrow \frac{(\sqrt{a})^2+(\sqrt{b})^2+2 \sqrt{a b}}{(\sqrt{a})^2+(\sqrt{b})^2-2 \sqrt{a b}}=\frac{9}{1} \\
&\Rightarrow\left(\frac{\sqrt{b}+\sqrt{a})^2}{\sqrt{b}-\sqrt{a}}\right)^2=\frac{9}{1} \Rightarrow \frac{\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}=\frac{3}{1} \\
&\Rightarrow \frac{\sqrt{b}+\sqrt{a}+\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}-\sqrt{b}+\sqrt{a}}=\frac{3+1}{3-1}
\end{aligned}
$$
\{Using Componendo \& Dividendo
$$
\sqrt{\frac{b}{a}}=\frac{4}{2}=2
$$
$$
\begin{aligned}
\frac{b}{a} &=\frac{4}{1} \\
\frac{a}{b} &=\frac{1}{4} \Rightarrow a: b=1: 4
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.