JEE Main 2014 (Online) 12th AprilQuestion 1
Let $G$ be the geometric mean of two positive numbers $a$ and $b$, and $M$ be the arithmetic mean of $\frac1a$ and $\frac1b$. If $\frac1M:G$ is $4:5$, then $a:b$ can be :
Sequence and Series · Means and Inequalities (AM, GM, HM)
Solution
Formulas used. (1) $\frac1M=\frac{2ab}{a+b}$ (the H.M. of $a,b$), $\ G=\sqrt{ab}$Step 1 - Ratio. $$\frac{1/M}{G}=\frac{2ab}{(a+b)\sqrt{ab}}=\frac{2\sqrt{ab}}{a+b}=\frac45 .$$ Step 2 - Put $t=\sqrt{a/b}$. Divide top and bottom by $b$: $$\frac{2t}{t^2+1}=\frac45\ \Longrightarrow\ 2t^2-5t+2=0\ \Longrightarrow\ t=2\ \text{or}\ \frac12 .$$ Step 3. $\dfrac ab=t^2=4$ or $\dfrac14$, so $a:b$ can be $1:4$. Answer: (A)