9.11. According to problem $9.10\left(a_{1}+a_{2}+\ldots+a_{7}\right)+\left(a_{2}+\ldots+a_{8}\right)+\ldots$ $\ldots+\left(a_{11}+\ldots+a_{17}\right)=\left(a_{1}+a_{2}+\ldots+a_{11}\right)+\left(a_{2}+\ldots+a_{12}\right)+\ldots+$ $+\left(a_{7}+\ldots+a_{17}\right)$. Therefore, $n<17$.

For $n=16$, such a sequence exists: $5,5,-13,5$, $5,5,-13,5,5,-13,5,5,5,-13,5,5$.
9.11. The numbers $a_{1}, a_{2}, \ldots, a_{n}$ are such that the sum of any seven consecutive numbers is negative, and the sum of any eleven consecutive numbers is positive. For what largest $n$ is this possible?

118 Chapter 9. Computation of Sums and Products

$$
\text { 9.3. Sums } S_{k}(n)=1^{k}+2^{k}+\ldots+n^{k}
$$

The sum $1+2+3+\ldots+n$ can be computed as follows. Add the equations $(k+1)^{2}=k^{2}+2 k+1$ for $k=1,2, \ldots, n$. After simplification, we get $(n+1)^{2}=1+2 S_{1}(n)+n$, where $S_{1}(n)$ is the desired sum. Therefore, $S_{1}(n)=\frac{n(n+1)}{2}$.