11. In an arithmetic sequence with all terms being real numbers, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most $\qquad$ terms.
11. 8 Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then $a_{n}=a_{1}+4(n-1)$. From the given information,
$$
\begin{aligned}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 & \Leftrightarrow a_{1}^{2}+\frac{\left(a_{2}+a_{n}\right)(n-1)}{2} \leqslant 100 \\
& \Leftrightarrow a_{1}^{2}+\frac{\left[a_{1}+4+a_{1}+4(n-1)\right](n-1)}{2} \leqslant 100 \\
& \Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0,
\end{aligned}
$$

which means the quadratic inequality in terms of $a_{1}$ has a solution, so it should satisfy
$$
\begin{aligned}
\Delta=(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0 & \Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0 \\
& \Leftrightarrow \frac{3-\sqrt{2816}}{7} \leqslant n \leqslant \frac{3+\sqrt{2816}}{7}, \text{ so the maximum value of } n \text{ is 8, meaning the sequence can have at most 8 terms.}
\end{aligned}
$$