Question

Let $A_1, G_1, H_1$ denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For $n \geq 2$, 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. $G _1> G _2> G _3>\ldots \ldots$
• B. $G _1< G _2< G _3<\ldots .$.
• C. $G _1= G _2= G _3=\ldots .$.
• D. $G _1< G _3< G _5<\ldots \ldots$ and $G _2> G _4> G _6>\ldots \ldots$

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 $H _1< H _2< H _3 \ldots . \Rightarrow$ (B)
(a) $G _1=\sqrt{ ab }, G _2=\sqrt{ A _1 H _1}= G _1 \Rightarrow G _1= G _2= G _3$
[Also $G_3=\sqrt{A_2 H_2}=G_2=G_1]$.