9. For any real number sequence $A=\left(a_{1}, a_{2}, a_{3}, \cdots\right)$, define $\Delta A$ as the sequence $\left(a_{2}-a_{1}, a_{3}-a_{2}, a_{4}-\right.$ $\left.a_{3}, \cdots\right)$, where its $n$-th term is $a_{n+1}-a_{n}$. Assume that all terms of $\Delta(\Delta A)$ are 1, and $a_{19}=a_{92}$ $=0$, try to find $a_{1}$.
9. Solution: Let the first term of the sequence $\Delta A$ be $d$, then the sequence $\Delta A$ is $(d, d+1, d+2, \cdots)$, where the $n$-th term is $d+(n-1)$. Therefore, the sequence $A$ can be written as $a_{1}, a_{1}+d, a_{1}+d+(d+1), \cdots$, where the $n$-th term is $a_{n}=a_{1}+(n-1)d+\frac{1}{2}(n-1)(n-2)$.

Thus, $a_{n}$ is a quadratic polynomial in $n$, and the leading coefficient of this polynomial is $\frac{1}{2}$. Since $a_{19}=a_{92}=0$, it must be that $a_{n}=\frac{1}{2}(n-19)(n-92)$, from which we find $a_{1}=\frac{1}{2}(1-19)(1-92)=819$.