3.1.6 ** The terms of an arithmetic sequence are all real numbers, the common difference is 4, and the square of any term plus the sum of the remaining terms does not exceed 100. Find the maximum number of terms in this sequence.
Let the real number sequence $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence satisfying the given conditions, then
i.e., $\square$
$$
\begin{array}{c}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100, \\
a_{1}^{2}+\frac{\left(a_{1}+4\right) \mid\left[a_{1}+(n-1) \cdot 4\right]}{2} \cdot(n-1) \leqslant 100,
\end{array}
$$

Therefore,
$$
a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
$$

Thus, $\Delta=(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0$, i.e., $7 n^{2}-6 n-401 \leqslant 0$. So $n \leqslant$ $\frac{3+\sqrt{2816}}{7}<9$, i.e., $n \leqslant 8$. There exists an 8-term arithmetic sequence: $-4,0,4,8,12,16,20,2$ that satisfies the given conditions, hence the maximum number of terms is 8.