Question

If $H_1$ and $H_2$ are two harmonic means between two positive numbers $a$ and $b\left(a\ne b\right)$. $A$ and $G$ are the arithmetic and geometric means between $a$ and $b$, then $\frac{H_2+H_1}{H_2{H}_1}$ is

• A. $\frac{A}{G}$
• B. $\frac{2A}{G}$
• C. $\frac{A}{G^2}$
• D. $\frac{2A}{G^2}$

Answer 1

The correct option is D

$\frac{2A}{G^2}$

Explanation for the correct option:

Step-1: Formation of equation:

Let, $a,H_1,H_2,b$ are in harmonic progression.

Then, $\frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{b}$ are in arithmetic progression.

Let, $d$ be the common difference, then,

$$\begin{gathered} \frac{1}{H_1}=\frac{1}{a}+d\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \frac{1}{b}=\frac{1}{a}+3d\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

We know that Arithmetic mean $A$ and Geometric mean $G$ are:

$$\begin{gathered} 2A=a+b\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ {G}^2=ab\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Now, subtracting $\left(1\right)$ from $\left(2\right)$.

$d=\frac{1}{2}\left[\frac{1}{b}-\frac{1}{H_1}\right]$

Step- 2: Find the value $H_1$.

Substitute value of $d$ in $\left(1\right)$.

$$\begin{gathered} \frac{1}{H_1}=\frac{1}{a}+\frac{1}{2}\left[\frac{1}{b}-\frac{1}{H_1}\right] \\ \frac{1}{H_1}+\frac{1}{2H_1}=\frac{1}{a}+\frac{1}{2b} \\ \frac{3}{2H_1}=\frac{2b+a}{2ab} \\ {H}_1=\frac{3ab}{2b+a} \end{gathered}$$

Step-3: Find the value $H_2$:

$$\begin{gathered} \frac{1}{H_2}=\frac{1}{a}+2d \\ \frac{1}{H_2}=\frac{1}{a}+2\left[\frac{1}{2}\left[\frac{1}{b}-\frac{1}{H_1}\right]\right] \\ \frac{1}{H_2}=\frac{1}{a}+\frac{1}{b}-\frac{\left(2b+a\right)}{3ab} \\ \frac{1}{H_2}=\frac{\left(2a+b\right)}{3ab} \\ {H}_2=\frac{3ab}{2a+b} \end{gathered}$$

Step- 4: Find the value of $\frac{H_2+H_1}{H_2{H}_1}$.

$$\begin{gathered} =\frac{1}{H_1}+\frac{1}{H_2} \\ =\frac{1}{{\displaystyle \frac{3ab}{a+2b}}}+\frac{1}{{\displaystyle \frac{3ab}{2a+b}}} \\ =\frac{3a+3b}{3ab} \\ =\frac{2A}{G^2}\left[\because From\left(3\right)and\left(4\right)\right] \end{gathered}$$

Hence, the correct option is D.