If one A.M., A and two geometric means $G_{1}$ and $G_{2}$ inserted between any two positive numbers, show that $\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}} = 2 A$ .

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Solution

Let the two positive numbers be a and b a, A and b are in A.P.

$∴ 2 A = a + b$ ......... (i)

Also, a, $G_{1}, G_{2}$ and b are in G.P.

$∴ r = {(\frac{b}{a})}^{\frac{1}{3}}$

Also, $G_{1} = a r$ and $G_{2} = a r^{2}$ ........ (ii)

Now, LHS = $\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$

$= \frac{(a r)^{2}}{a r^{2}} + \frac{(a r^{2})^{2}}{a r}$ [Using (ii)]

$= a + a r^{3}$

$= a + a {({(\frac{b}{a})}^{\frac{1}{3}})}^{3}$

$= a + a (\frac{b}{a})$

$= a + b$

$= 2 A$

$= R H S$ [Using (i)]

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