For any sequence of real numbers $A=\left\{a_{1}, a_{2}, \ldots\right\}$, we define $\Delta A$ as the sequence $\left\{a_{2}-a_{1}, a_{3}-a_{2}, \ldots\right\}$.

We assume that all terms of the sequence $\Delta(\Delta A)$ are 1 and that $a_{19}=a_{92}=0$.

Determine $a_{1}$.
We denote $d$ as the first term of $\Delta A$. Therefore, we have: $\Delta A=\{d, d+1, d+2, \ldots\}$ where the $n^{\text{th}}$ term is written as $d+n-1$.

It follows that $A=\left\{a_{1}, a_{1}+d, a_{1}+d+(d+1), a_{1}+d+(d+1)+(d+2), \ldots\right\}$ where the $n^{\text{th}}$ term is written as $a_{n}=a_{1}+(n-1)d+\frac{(n-1)(n-2)}{2}$.

$a_{n}$ is therefore a polynomial of the second degree in $n$ with a leading coefficient of $\frac{1}{2}$.

Since $a_{19}=a_{92}=0$, we necessarily have:

$$
a_{n}=\frac{1}{2}(n-19)(n-92)
$$

Thus, $a_{1}=819$.

## Intermediate Exercises