Question

If $A_1,A_2;$ $G_1,G_2$ and $H_1,H_2$ are arithmetic mean, geometric mean and harmonic mean between two numbers, then the value of $\frac{G_1{G}_2}{H_1{H}_2}\times \frac{H_1+H_2}{A_1+A_2}$ is

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is A $1$

Let $a$ and $b$ be two numbers.
Sum of $n\text{A.M.'s}=n\times \text{single}\text{A.M.}$
$\Rightarrow A_1+A_2=2\times \left(\frac{a+b}{2}\right)=a+b\cdots \left(1\right)$

Product of $n\text{G.M.'s}=(\text{single G.M.}{)}^n$
$\Rightarrow G_1{G}_2=(\sqrt{ab}{)}^2=ab\cdots (2)$

Dividing equation $\left(1\right)$ by $\left(2\right),$ we get
$\frac{A_1+A_2}{G_1{G}_2}=\frac{a+b}{ab}\cdots \left(3\right)$
Since $a,H_1,H_2,b$ are in $\text{H.P.}$
$\Rightarrow \frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{b}$ are in $\text{A.P.}$
$\Rightarrow \frac{1}{H_1}+\frac{1}{H_2}=2\left(\frac{\frac{1}{a}+\frac{1}{b}}{2}\right)$
$\Rightarrow \frac{1}{H_1}+\frac{1}{H_2}=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$
$\Rightarrow \frac{H_1+H_2}{H_1{H}_2}=\frac{a+b}{ab}\cdots \left(4\right)$

From $\left(3\right)$ and $\left(4\right),$ we get
$\frac{A_1+A_2}{G_1{G}_2}=\frac{H_1+H_2}{H_1{H}_2}$
$\Rightarrow \frac{H_1{H}_2}{H_1+H_2}=\frac{G_1{G}_2}{A_1+A_2}$
$\Rightarrow \frac{G_1{G}_2}{H_1{H}_2}\times \frac{H_1+H_2}{A_1+A_2}=1$