For any real number sequence $A=\left(a_{1}, a_{2}, a_{3}, \cdots\right)$, define $\Delta A$ as the sequence $\left(a_{2}-a_{1}, a_{3}-a_{2}, a_{4}-a_{3}, \cdots\right)$, where its $n$-th term is $a_{n+1}-a_{n}$. Assume that all terms of the sequence $\Delta (\Delta A)$ are 1, and $a_{19}=a_{92}=0$. Try to find $a_{1}$.
(1992 10th American Mathematical Invitational Problem 8)
Let the first term of $\triangle A$ be $d$, then according to the conditions,
$$
\Delta(\Delta A)=(d, d+1, d+2, \cdots),
$$

where the $n$-th term is $d+(n-1)$. Therefore, the sequence $A$ can be written as
$$
\left(a_{1}, a_{1}+d, a_{1}+d+(d+1), a_{1}+d+(d+1)+(d+2), \cdots\right),
$$

where the $n$-th term is $a_{n}=a_{1}+(n-1) d+\frac{1}{2}(n-1)(n-2)$. This shows that $a_{n}$ is a quadratic polynomial in $n$, with the leading coefficient being $\frac{1}{2}$. Since $a_{19}=a_{92}=0$, we have
$$
a_{n}=\frac{1}{2}(n-19)(n-92) \text {. }
$$

Thus,
$$
a_{1}=\frac{1}{2}(1-19)(1-92)=819 .
$$