A strictly increasing sequence of positive integers $a_1, a_2, a_3,...$ has the property that for every positive integer $k$, the subsequence $a_{2k-1}, a_{2k}, a_{2k+1}$ is geometric and the subsequence $a_{2k}, a_{2k+1}, a_{2k+2}$ arithmetic. If $a_{13}=539$. Then how can I find $a_5.$
Any help will be appreciated.
Thank you.
Just work backwards. You know that $539 a_{11} = a_{12}^2$, but $539=7^2\cdot 11$, hence $a_{11}=11 n^2$ and $a_{12}=77 n$. The sequence has to be strictly increasing, hence $n\leq 6$. What happens if we take $n=6$? It follows that $a_{11}=11\cdot 6^2 = 396$ and $a_{12}=462$, so $a_{10}=2\cdot a_{11}-a_{12}=330$ and $a_9=a_{10}^2/a_{11}=275$. It follows that $a_8=2a_9-a_{10}=220$ and $a_7=a_8^2/a_9 = 176$. Then $a_6=2a_7-a_8=132$ and
$$ a_5 = \frac{a_6^2}{a_7} = \frac{132^2}{176} = \color{red}{99}.$$
The whole sequence then is: $11,22,44,66,99,132,176,220,275,330,396,462,539,616,704,792,891,990,1100,1210,\ldots $
It is left to check that smaller values of $n$ do not really work.
Hint: Let $a_1=a,a_2=ka$. Then it is not hard to show that $a_{2n+1}=a(nk-n+1)^2$. We have $539=11\cdot7^2$, so evidently we take $k=2,a=11$. That gives $a_5=11(2\cdot 2-1)^2=99$.
