Question

If the harmonic mean and the geometric mean of two numbers, $a$ and $b$ are $4$ and $3\sqrt{2}$ respectively then the interval $\left[a,b\right]=$

• A. $\left[3,6\right]$
• B. $\left[2,7\right]$
• C. $\left[4,5\right]$
• D. $\left[1,8\right]$

Answer 1

The correct option is A

$\left[3,6\right]$

Explanation for the correct option:

The given numbers are $a$ and $b$.

The harmonic mean can be given by, $HM=\frac{2}{{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}}$

$$\begin{gathered} \Rightarrow HM=\frac{2}{{\displaystyle \frac{a+b}{ab}}} \\ \Rightarrow HM=\frac{2ab}{{\displaystyle a+b}} \end{gathered}$$

The given harmonic mean is $4$.

$$\begin{gathered} \Rightarrow \frac{2ab}{{\displaystyle a+b}}=4 \\ \Rightarrow 4a+4b=2ab \\ \Rightarrow 2b-ab=-2a \\ \Rightarrow b\left(2-a\right)=-2a \\ \Rightarrow b=-\frac{2a}{2-a} \\ \Rightarrow b=\frac{2a}{a-2} \end{gathered}$$

The geometric mean can be given by, $GM=\sqrt{ab}$.

The given geometric mean is $3\sqrt{2}$.

$$\begin{gathered} \Rightarrow \sqrt{ab}=3\sqrt{2} \\ \Rightarrow {\left(\sqrt{ab}\right)}^2={\left(3\sqrt{2}\right)}^2 \\ \Rightarrow ab=18 \\ \Rightarrow a\times \frac{2a}{a-2}=18\mathbf{[}\mathbf{\because }\mathbf{b}\mathbf{=}\frac{\mathbf{2}\mathbf{a}}{\mathbf{a}\mathbf{-}\mathbf{2}}\mathbf{]} \\ \Rightarrow 2a^2=18a-36 \\ \Rightarrow a^2-9a+18=0 \\ \Rightarrow a^2-6a-3a+18=0 \\ \Rightarrow a\left(a-6\right)-3\left(a-6\right)=0 \\ \Rightarrow \left(a-6\right)\left(a-3\right)=0 \end{gathered}$$

Thus, either $a=6$ or $a=3$.

Hence, $b=\frac{18}{a}=\frac{18}{6}=3$.

Or, $b=\frac{18}{a}=\frac{18}{3}=6$.

For the interval $\left[a,b\right]$, $a<b$.

Thus, $\left[a,b\right]=\left[3,6\right]$.

Hence, option (A) $\left[3,6\right]$ is the correct answer.