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Let $a_n;\;n> 1$ be a sequence of positive numbers such that $a_1, a_2, a_3$ are in $AP$, $a_2, a_3, a_4$ are in $GP$, $a_3, a_4, a_5$ are in $AP$, $a_4, a_5, a_6$ are in $GP$, and so on. Find an expression for $a_n$ in terms of $a_1$ and $a_2$.

I did
$a_3=2a_2+a_1$ $a_4=a_2+a_1\bigg(\frac{1}{a_2}-4\bigg)$ $a_5=6a_2+a_1\bigg(\frac{2}{a_2}-7\bigg)$ But I can't see how to write $a_n$ in terms of $a_1$ and $a_2$... Can somenone help me? Thanks for attention!

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Hint:

Apply induction to prove: $$ a_n=\frac1{a_2}\times\begin{cases} \left(\frac{n+1}2a_2-\frac{n-1}2a_1\right)\left(\frac{n-1}2a_2-\frac{n-3}2a_1\right),& n\text{ odd},\\ \left(\frac{n}2a_2-\frac{n-2}2a_1\right)^2,& n\text{ even}. \end{cases} $$

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  • $\begingroup$ Very good! But, you know how to do this question without induction? $\endgroup$ – Tas May 28 at 14:42
  • $\begingroup$ Nice job. Let me add, that a couple of conditions needs to be written up for cases that are not true with the above formula: I immediately thought seeing this that if all elements are equal, they satisfy the conditions. But if I plug $a_2 = a_1$ in the above, I get $a_1$ for odd, but $-a_1$ for even. And there might be other solutions like this (or not, I cannot prove...), hence the conditions I mention. $\endgroup$ – Dávid Laczkó May 28 at 15:15
  • $\begingroup$ @DávidLaczkó I don't quite understand how $\frac{a^2}a=-a$ for even $n$. The only condition I see $a_2\ne0$. $\endgroup$ – user May 28 at 21:32
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    $\begingroup$ @DávidLaczkó I just have changed the sign inside the brackets for beauty reasons. $(-1)^2=1$ still... :) $\endgroup$ – user May 29 at 6:03
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    $\begingroup$ Ah, my bad, I missed the squaring. $\endgroup$ – Dávid Laczkó May 29 at 6:06

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