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If $A_1, A_2 ; G_1, G_2$ and $H_1, H_2$ be $A M^{\prime} s, G M^{\prime} s$ and $H M^{\prime} s$ between two quantities, then the value of $\frac{G_1 G_2}{H_1 H_2}$
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Verified Answer
The correct answer is:
$\frac{A_1+A_2}{H_1+H_2}$
The correct option is $\mathbf{1} \frac{A_1+A_2}{H_1+H_2}$
Let the two quantities be $a$ and $b$. Then $a, A_1, A_2, b$ are in A.P.
$\therefore A_1-a=b-A_2 \Rightarrow A_1+A_2=a+b \ldots$ (i)
Again $a, \mathrm{G}_1, \mathrm{G}_2, \mathrm{~b}$ are in G.P.
$\therefore \frac{\mathrm{G}_1}{\mathrm{a}}=\frac{\mathrm{b}}{\mathrm{G}_2} \Rightarrow \mathrm{G}_1 \mathrm{G}_2=\mathrm{ab} \ldots$ (ii)
Also $\mathrm{a}, \mathrm{H}_1, \mathrm{H}_2, \mathrm{~b}$ are in H.P.
$\therefore \frac{1}{\mathrm{H}_1}-\frac{1}{\mathrm{a}}=\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{H}_2} \Rightarrow$ $\frac{1}{\mathrm{H}_1}+\frac{1}{\mathrm{H}_2}=\frac{1}{\mathrm{a}}+\frac{1}{b}$
$\Rightarrow \frac{H_1+H_2}{H_1 H_2}=\frac{a+b}{a b}= \frac{A_1+A_2}{G_1 G_2}$ [By (i) and (ii)]
$\therefore \frac{\mathrm{G}_1 \mathrm{G}_2}{\mathrm{H}_1 \mathrm{H}_2}=\frac{\mathrm{A}_1+\mathrm{A}_2}{\mathrm{H}_1+\mathrm{H}_2}$.
Let the two quantities be $a$ and $b$. Then $a, A_1, A_2, b$ are in A.P.
$\therefore A_1-a=b-A_2 \Rightarrow A_1+A_2=a+b \ldots$ (i)
Again $a, \mathrm{G}_1, \mathrm{G}_2, \mathrm{~b}$ are in G.P.
$\therefore \frac{\mathrm{G}_1}{\mathrm{a}}=\frac{\mathrm{b}}{\mathrm{G}_2} \Rightarrow \mathrm{G}_1 \mathrm{G}_2=\mathrm{ab} \ldots$ (ii)
Also $\mathrm{a}, \mathrm{H}_1, \mathrm{H}_2, \mathrm{~b}$ are in H.P.
$\therefore \frac{1}{\mathrm{H}_1}-\frac{1}{\mathrm{a}}=\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{H}_2} \Rightarrow$ $\frac{1}{\mathrm{H}_1}+\frac{1}{\mathrm{H}_2}=\frac{1}{\mathrm{a}}+\frac{1}{b}$
$\Rightarrow \frac{H_1+H_2}{H_1 H_2}=\frac{a+b}{a b}= \frac{A_1+A_2}{G_1 G_2}$ [By (i) and (ii)]
$\therefore \frac{\mathrm{G}_1 \mathrm{G}_2}{\mathrm{H}_1 \mathrm{H}_2}=\frac{\mathrm{A}_1+\mathrm{A}_2}{\mathrm{H}_1+\mathrm{H}_2}$.
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