Question

Let $A_1,G_1,H_1$ denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For $n\ge 2$, let $A_{n-1}$ and $H_{n-1}$ have arithmetic, geometric and harmonic means as $A_n$, $G_n,H_n$ respectively.
Which one of the following statements is correct?

• A. $G_1>G_2>G_3>\dots \dots$
• B. $G_1<G_2<G_3<\dots .$.
• C. $G_1=G_2=G_3=\dots .$.
• D. $G_1<G_3<G_5<\dots \dots$ and $G_2>G_4>G_6>\dots \dots$

Answer 1

Answer: (C)

$\therefore G_1=G_2$
$\text{ again }{G}_2^2=A_2{H}_2=G_3^2\text{ from (3) }$
$G_2=G_3$
$\therefore G_1=G_2=G_3\Rightarrow (C)$
$\text{ again }{A}_1=\frac{a+b}{2};A_2=\frac{A_1+H_1}{2}$
$\text{ But }{H}_1<A_1;A_2<\frac{A_1+A_1}{2}$
$\Rightarrow A_2<A_1\Rightarrow A_1>A_2$
$\text{ again }{A}_2=\frac{A_1+H_1}{2};A_3=\frac{A_2+H_2}{2}$
$\text{ But }{H}_2<A_2;A_3<\frac{A_2+H_2}{2}$
$\Rightarrow A_3<A_2\Rightarrow A_2>A_3$
$\therefore A_1>A_2>A_3.\Rightarrow (C)$
$\frac{G_3^2}{H_1}<\frac{G_2^2}{H_2}<\frac{G_1^2}{H_3}\text{ As }{G}_1=G_2=G_3$
$\therefore \frac{1}{H_1}<\frac{1}{H_2}<\frac{1}{H_3}$
Aliter:
hence $H_3>H_2>H_1$ or $H_1<H_2<H_3\dots .\Rightarrow$ (B)
(a) $G_1=\sqrt{ab},G_2=\sqrt{A_1{H}_1}=G_1\Rightarrow G_1=G_2=G_3$
[Also $G_3=\sqrt{A_2{H}_2}=G_2=G_1]$.