Question

If the H.M. of $a$ and $c$ is $b$, G.M. of $b$ and $d$ is $c$ and A.M. of $c$ and $e$ is $d$, then the G.M. of $a$ and $e$ is

• A. $b$
• B. $c$
• C. $d$
• D. $ae$

Answer 1

The correct option is B

$c$

Explanation for the correct option:

It is given that the H.M. of $a$ and $c$ is $b$.

Thus, $b=\frac{2}{{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{c}}}$

$$\begin{gathered} \Rightarrow b=\frac{2}{{\displaystyle \frac{a+c}{ac}}} \\ \Rightarrow b=\frac{2ac}{a+c}\_\left(1\right) \end{gathered}$$

It is given that the G.M. of $b$ and $d$ is $c$.

Thus, $c=\sqrt{bd}$.

$\Rightarrow c^2=bd\_\left(2\right)$

It is given that the A.M. of $c$ and $e$ is $d$.

Thus, $d=\frac{c+e}{2}\_\left(3\right)$

Put, $b=\frac{2ac}{a+c}$ and $d=\frac{c+e}{2}$ in the equation $\left(2\right)$.

$$\begin{gathered} c^2=\frac{2ac}{a+c}\times \frac{c+e}{2} \\ \Rightarrow c^2=\frac{ac\left(c+e\right)}{a+c} \\ \Rightarrow c=\frac{a\left(c+e\right)}{a+c} \\ \Rightarrow c\left(a+c\right)=a\left(c+e\right) \\ \Rightarrow ac+c^2=ac+ae \\ \Rightarrow c^2=ae \\ \Rightarrow c=\sqrt{ae} \end{gathered}$$

Therefore, the G.M. of $a$ and $e$ can be given by $\sqrt{ae}=c$.

Hence, option (B) is the correct answer.