If $A$ is the arithmetic mean and $G_{1}, G_{2}$ be two geometric means between any two numbers, then prove that $2 A = \frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$

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Solution

Let the two numbers be $p$ and $q$

$∴$ Arithmetic mean of $p$ and $q$ ,

$A = \frac{p + q}{2}$ ... $(i)$

Since geometric means $G_{1}$ & $G_{2}$ are inserted between $p$ and $q$ .

$p, G_{1}, G_{2}, q$ are in $G P$ .

Let $r$ be the common ratio of the $G P$

$\Rightarrow q = p r^{3}$

$r = {(\frac{q}{p})}^{1 / 3}$

Now, $G_{1} = p {(\frac{q}{p})}^{1 / 3}$

$\Rightarrow G_{1} = q^{1 / 3} p^{2 / 3}$ ... $(i i)$

& $G_{2} = p {(\frac{q}{p})}^{2 / 3}$

$\Rightarrow G_{2} = q^{2 / 3} p^{1 / 3}$ ... $(i i i)$

$R . H . S = \frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$

$\Rightarrow R . H . S = \frac{(q^{1 / 3} p^{2 / 3})^{2}}{q^{2 / 3} p^{1 / 3}} + \frac{(q^{2 / 3} p^{1 / 3})^{2}}{q^{1 / 3} p^{2 / 3}}$

(From equations $(i i)$ & $(i i i)$ )

$\Rightarrow R . H . S = \frac{q^{2 / 3} p^{4 / 3}}{q^{2 / 3} p^{1 / 3}} + \frac{q^{4 / 3} p^{2 / 3}}{q^{1 / 3} p^{2 / 3}}$

$\Rightarrow R . H . S = p^{\frac{4}{3} - \frac{1}{3}} + q^{\frac{4}{3} - \frac{1}{3}}$

$\Rightarrow R . H . S = p + q$

$\Rightarrow R . H . S = 2 \times (\frac{p + q}{2})$

$\Rightarrow R . H . S = 2 A$ (From equaion $(i)$ )

$\Rightarrow R . H . S = L . H . S$

$\Rightarrow 2 A = \frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$

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