### Instruction: For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$, define $\Delta A$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$, whose $n^{\text{th}}$ term is $a_{n+1}-a_n$. Suppose that all of the terms of the sequence $\Delta(\Delta A)$ are $1$, and that $a_{19}=a_{92}=0$. Find $a_1$.### Response: I notice that the sequence $\Delta(\Delta A)$ is constant, which means that the sequence $\Delta A$ is arithmetic, and that the sequence $A$ is quadratic.
This means I can write $a_n$ as a quadratic function of $n$, say $a_n=pn^2+qn+r$ for some constants $p,q,r$.
To find these constants, I can use the given information that $a_{19}=a_{92}=0$.
This gives me two equations: $p(19)^2+q(19)+r=0$ and $p(92)^2+q(92)+r=0$.
Subtracting these equations, I get $p(92^2-19^2)+q(92-19)=0$.
Factoring, I get $p(73)(111)+73q=0$.
Dividing by $73$, I get $p(111)+q=0$.
This means that $q=-p(111)$.
Plugging this back into the first equation, I get $p(19)^2-p(111)(19)+r=0$.
Simplifying, I get $p(361-2091)+r=0$.
This means that $r=p(1730)$.
Now I have $a_n=pn^2-p(111)n+p(1730)$.
To find $p$, I can use the fact that $\Delta(\Delta A)$ is $1$.
This means that the second difference of $a_n$ is constant and equal to $1$.
The second difference of $a_n$ is the difference of the difference of consecutive terms, or $a_{n+2}-2a_{n+1}+a_n$.
If I plug in $a_n=pn^2-p(111)n+p(1730)$, I get $a_{n+2}-2a_{n+1}+a_n=p((n+2)^2-2(n+1)^2+n^2)-p(111)((n+2)-2(n+1)+n)+p(1730)(1-2+1)$.
Simplifying, I get $a_{n+2}-2a_{n+1}+a_n=2p$.
Since this is constant and equal to $1$, I can conclude that $p=\frac{1}{2}$.
Now I have $a_n=\frac{1}{2}n^2-\frac{1}{2}(111)n+\frac{1}{2}(1730)$.
To find $a_1$, I just plug in $n=1$ and get $a_1=\frac{1}{2}(1)^2-\frac{1}{2}(111)(1)+\frac{1}{2}(1730)$.
Simplifying, I get $a_1=\frac{1}{2}(1-111+1730)$.
This is $a_1=810$.