Maximal length of a finite real sequence

Question

Suppose that $(a_n)_{1 \le n \le N}$ is a finite sequence of reals such that the sum of any 7 consecutive terms is (strictly) negative and the sum of any 11 consecutive terms is (strictly) positive.

What is the maximal length of this finite sequence of reals?

I tried creating finite sequences having the given property. I succeeded in creating a sequence of 16 terms. Namely $5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5$... but wasn't able to do more.

Answer 1

Assume there is a sequence $(a_1,\ldots,a_{17})$ satisfying the restrictions. Then take all the inequalities you can get: $$\{a_i+\ldots+a_{i+6} < 0\}_{i \in [11]}$$ $$\{0 < a_i+\ldots+a_{i+10}\}_{i \in [7]}$$ (here $[n]$ stays for $\{1,\ldots,n\}$).

Add all these inequalities together, you will get $0 < 0$.

So a sequence of 17 terms is impossible (and any sequence with more elements is impossible as well).

Answer 2

$x_1+\dots+x_7<0$ and $x_8+\dots+x_{14}<0$, so $x_1+\dots+x_{14}<0$. But $x_4+\dots+x_{14}>0$, so $x_1+x_2+x_3<0$.

$x_5+\dots+x_{11}<0$ and $x_1+\dots+x_{11}>0$, so $x_4>0$. From there is it easy to get a contradiction if you have 17 terms.