Variation of geometric, harmonic and arithmetic means in sequence. A question I got on my test was -

Let ${A}_{1}, {G}_{1}$ and ${H}_{1}$, denote the arithmetic, geometric
  and harmonic means of two distinct positive numbers. For $n\geq 2$,
  Let ${A}_{n-1}$ and ${H}_{n-1}$ have the arithmetic, geometric and
  harmonic means as ${A}_{n}, {G}_{n}$ and ${H}_{n}$ respectively. Then
  which of the following statements are correct?
1) ${G}_{1}>{G}_{2}>{G}_{3}>$ ...
2) ${G}_{1}={G}_{2}={G}_{3}=$ ...
3) ${A}_{1}>{A}_{2}>{A}_{3}>$ ...
4) ${A}_{1}>{A}_{3}>{A}_{5}>$ ... and  ${A}_{2}>{A}_{4}>{A}_{6}>$ ...
5) ${H}_{1}>{H}_{2}>{H}_{3}>$ ...
6) ${H}_{1}<{H}_{2}<{H}_{3}<$ ...
7) ${H}_{1}>{H}_{3}>{H}_{5}>$ ... and  ${H}_{2}<{H}_{4}<{H}_{6}<$ ...
8) ${H}_{1}<{H}_{3}<{H}_{5}<$ ... and  ${H}_{2}>{H}_{4}>{H}_{6}>$ ...

Is there any other way to do this other than taking , for instance 1 and 2 as the first two numbers and calculating AM,GM and HM upto the 4th iteration and then comparing it? 
That's what I ended up doing in the test. :)
 A: Here is perhaps a quick way.  
By the AM-GM-HM inequality,   it must always be the case that $H_n < H_{n+1} < G_{n+1} < A_{n+1} < A_n$, which resolves the sequences of $H_n$ and $A_n$.
It is also easy to see that $A_{n+1} H_{n+1} = A_n H_n$, so we also readily get $G_{n+1} = G_n$.
A: Let $2$ distinct positive no. be $a$ and $b\;,$ Here $A_{1}\;,G_{1}$ and $H_{1}$ are $\bf{A.M}\;,\bf{G.M}$ and $\bf{H.M}$
of these two positive no., So $$\displaystyle A_{1} = \frac{a+b}{2}\;\;,G_{1} = \sqrt{ab}\;\;,H_{1} = \frac{2ab}{a+b}$$.
And given  $A_{n}\;,G_{n}$ and $H_{n}$ are $\bf{A.M}\;,\bf{G.M}$ and $\bf{H.M}$ of no.,s $A_{n-1}$ and $H_{n-1}.$
So $$\displaystyle A_{n} = \frac{A_{n-1}+H_{n-1}}{2}\;\;,G_{n} = \sqrt{A_{n-1}\cdot H_{n-1}}$$ and $$\displaystyle H_{n} = \frac{2A_{n-1}\cdot H_{n-1}}{A_{n-1}+H_{n-1}}$$
So $G_{1} = \sqrt{ab}$ and $\displaystyle G_{2} = \sqrt{A_{1}\cdot H_{1}}=\sqrt{\frac{a+b}{2}\cdot \frac{2ab}{a+b}} = \sqrt{ab}$
Now $$\displaystyle G_{3} = \sqrt{A_{2}\cdot H_{2}} = \sqrt{\frac{A_{1}+H_{1}}{2}\cdot \frac{2A_{1}\cdot H_{1}}{A_{1}+H_{1}}} = \sqrt{ab}$$
So We get $$\boxed{G_{1} = G_{2} = G_{3} = \sqrt{ab}}$$
Given $A_{2}$ is the arithmetic mean of $A_{1},H_{1}$
So $A_{1},A_{2},H_{1}$ are in $A.P$, So $A_{1}>A_{2}>H_{1}$.
Similarly Given Given $A_{3}$ is the arithmetic mean of $A_{2},H_{2}$
So $A_{1},A_{2},H_{1}$ are in $A.P$, So $A_{2}>A_{3}>H_{2}$.
So we get $$A_{1}>A_{2}>A_{3}........$$
Similarly We get $$H_{1}<H_{2}<H_{3}<......$$
