Question

If $A_1,A_2,G_1,G_2$ and $H_1,H_2$ be two AM's, GM's and HM's between two numbers respectively, then the value of $\frac{\left(G_1{G}_2\right)}{\left(H_1{H}_2\right)}\times \frac{(H_1+H_2)}{(A_1+A_2)}$

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is A

$1$

Step 1. Find the relations between $A_1,A_2,G_1,G_2$ and $H_1,H_2$:

Let the two numbers are $a$ and $b$

So, $a,A_1,A_2,b$ are in AP.

Then$(a+b)/2=(A_1+A_2)/2$

$\Rightarrow (a+b)=(A_1+A_2)..........\left(i\right)$

and $a,G_1,G_2,b$ are in GP.

$$\begin{gathered} \Rightarrow \sqrt{ab}=\sqrt{G_1{G}_2} \\ \Rightarrow ab=G_1{G}_2.........\left(ii\right) \end{gathered}$$

also $a,H_1,H_2,b$ are in HP.

So ,$\frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{b}$ are in AP.

$$\begin{gathered} \Rightarrow \frac{\left[\right(1/a)+(1/b\left)\right]}{2}=\frac{\left[\right(1/H_1)+(1/H_2\left)\right]}{2} \\ \Rightarrow \frac{(a+b)}{ab}=\frac{(H_1+H_2)}{H_1{H}_2}...........\left(iii\right) \end{gathered}$$

Step 2. Find the value of $\frac{\left(G_1{G}_2\right)}{\left(H_1{H}_2\right)}\times \frac{(H_1+H_2)}{(A_1+A_2)}$

Now from the equations (i), (ii) and (iii), we get

$$\begin{gathered} \Rightarrow \frac{(A_1+A_2)}{G_1{G}_2}=\frac{(H_1+H_2)}{H_1{H}_2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathbf{}\mathbf{(}\mathit{i}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathit{i}\mathit{i}\mathbf{)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{(}\mathit{i}\mathit{i}\mathit{i}\mathbf{\left)}\mathbf{\right)} \\ \Rightarrow \frac{G_1{G}_2}{H_1{H}_2}\times \frac{(H_1+H_2)}{(A_1+A_2)}=1 \end{gathered}$$

Hence the correct option is A.