7. For an arithmetic sequence with real number terms, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. The maximum number of terms in such a sequence is $\qquad$.
7. 8 .

Let $a_{n}=a_{1}+(n-1) 4=4 n+a_{1}-4$, according to the condition we get $a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100, a_{1}^{2}+(n-1) a_{1}+2 n(n-1) \leqslant 100$, which is an inequality about $a_{1}$, $a_{1}^{2}+(n-1) a_{1}+2 n^{2}-2 n-100 \leqslant 0$ has a solution, then $\Delta=(n-1)^{2}-8 n^{2}+8 n+400 \geqslant 0$. Therefore, $n<9$, meaning the sequence has at most 8 terms.