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nth Term of A.P

Let A and G b...

Question

Let A and G be arithmetic and geometric means of two positive numbers a and b then A>G

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Solution

If a and b are two unequal positive number then
$A = \frac{a + b}{2} G = \sqrt{a b} A - G = \frac{a + b}{2} - \sqrt{a b} = \frac{a + b - 2 \sqrt{a b}}{2} = \frac{(\sqrt{a} - \sqrt{b})^{2}}{2}$
Since $(\sqrt{a} - \sqrt{b})^{2}$ always $+ v e A - G > 0 A > G$

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