Let $A_{1}, G_{1}, H_{1}$ denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For $n > 2$ , let $A_{n - 1}$ and $H_{n - 1}$ have arithmetic, geometric and harmonic means as $A_{n}, G_{n}, H_{n}$ respectively. Then, which one of the following statements is correct?

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$G_{1} < G_{3} < G_{5} \dots$ and $G_{2} > G_{4} > G_{6} > \dots$

$G_{1} > G_{2} > G_{3} > \dots$

$G_{1} = G_{2} = G_{3} = \dots$

$G_{1} < G_{2} < G_{3} < \dots$

answer is C.

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Detailed Solution

Let a and b be two distinct positive numbers. Then,

$A_{1} = \frac{a + b}{2}, G_{1} = \sqrt{a b}$ and $H_{1} = \frac{2 a b}{a + b}$

It is given that for $n \geq 2$

$A_{n} = \frac{A_{n - 1} + H_{n - 1}}{2}, G_{n} = \sqrt{A_{n - 1} H_{n - 1}}$ and $H_{n} = \frac{2 A_{n - 1} H_{n - 1}}{A_{n - 1} + H_{n - 1}}$

$\Rightarrow A_{n} H_{n} = A_{n - 1} H_{n - 1}$

$∴ G_{n + 1} = \sqrt{A_{n} H_{n}} = \sqrt{A_{n - 1} H_{n - 1}} = G_{n}$ for $n \geq 2$

$\Rightarrow G_{n} = G_{n - 1} = G_{n - 2} = \dots = G_{3} = G_{2} = G_{1} = \sqrt{a b} .$

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