Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?

保留源文本的换行和格式，所以翻译结果如下：

Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?
Let the arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$, with common difference $d=4$. Then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+\frac{2 a_{1}+4 n}{2}(n-1) \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{array}
$$

It holds for at least one value of $a_{1}$, so the discriminant is non-negative. That is,
$$
\begin{array}{l}
(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0 \\
\Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0 \\
\Leftrightarrow-\frac{\sqrt{2} 816}{7} \leqslant n \leqslant \frac{3+\sqrt{2816}}{7} .
\end{array}
$$

But $n \in \mathbb{N}$, it is easy to see that $n \leqslant 8$.
In fact, the arithmetic sequence with a common difference of 4:
$-4,0,4,8,12,16,20,24$ has 8 terms, and
$$
(-4)^{2}+0+4+8+12+16+20+24=100 \text {. }
$$

Therefore, the maximum value of $n$ is 8. That is, such an arithmetic sequence can have at most 8 terms.