8. For any real number sequence $A=\left(a_{1}, a_{2}, \cdots\right)$, define
$$
\Delta A=\left(a_{2}-a_{1}, a_{3}-a_{2}, \cdots\right),
$$

where the $n$-th term is $a_{n+1}-a_{n}$. Suppose that all terms of the sequence $\Delta(\Delta A)$ are 1, and $a_{19}=a_{92}=0$. Then the value of $a_{1}$ is $\qquad$
8. 819.

Let the first term of $\Delta A$ be $d$. Then according to the condition, we have
$$
\Delta A=(d, d+1, \cdots) \text {, }
$$

the $n$-th term of which is $d+(n-1)$.
Therefore, the $n$-th term of sequence $A$ is
$$
a_{n}=a_{1}+(n-1) d+\frac{1}{2}(n-1)(n-2) \text {. }
$$

Since $a_{19}=a_{92}=0$, we have
$$
\left\{\begin{array}{l}
a_{1}+18 d+9 \times 17=0, \\
a_{1}+91 d+45 \times 91=0
\end{array} \Rightarrow a_{1}=819\right.
$$