Q5. Numerical For any sequence of real numbers $A = \{a_1, a_2, a_3, \ldots \}$, a sequence $\Delta A$ is defined such that

$$\Delta A = \{a_2 - a_1, a_3 - a_2, a_4 - a_3, \ldots \}.$$

Suppose that all of the terms of the sequence $\Delta (\Delta A)$ are 1 and that $a_{19} = a_{92} = 0$. Find $a_3$.

Key Concepts & Plan

1. Second Differences Constant ⇒ Quadratic Form
   If $\Delta(\Delta a_n)=1$ for all $n$, the original sequence $a_n$ is quadratic:
   $a_n = An^2 + Bn + C$ with $2A = 1$.
2. Use Given Values
   You know $a_{19}=0$ and $a_{92}=0$. Substitute $n=19$ and $n=92$ to create two equations.
3. Solve for Unknowns $B$ and $C$
4. Evaluate $a_3$

Detailed Walk-Through

1. Recognize Quadratic Nature
   Because each second difference is $1$, we set
   $A = \tfrac{1}{2} \quad\Rightarrow\quad a_n = \tfrac{1}{2}n^2 + Bn + C$ This comes from the discrete analogue of a second derivative.
2. Write Equations from Conditions
   $a_{19}=0 \;\Rightarrow\; \tfrac{1}{2}(19)^2 + 19B + C = 0$ $a_{92}=0 \;\Rightarrow\; \tfrac{1}{2}(92)^2 + 92B + C = 0$
3. Solve for $B$
   Subtract the first equation from the second to eliminate $C$. The algebra gives a single equation in $B$, which is then solved directly.
4. Find $C$
   Put the value of $B$ back into either original equation.
5. Compute $a_3$
   Plug $n=3$ into $a_n$ with the found $A, B, C$.

Each step follows logically: constant second difference ⇒ quadratic, then use boundary conditions, then evaluate the required term.

🧒 ELI5 Version

Imagine you are drawing a line of numbers on the ground.
If you look at how far you step from one number to the next, that is the first list of differences.
If you again look at how far those steps change, that is the second list of differences.

In this puzzle, every time you examine the second jumps, they are always 1. That means your number line must curve like a nice smooth U-shape (a quadratic).
Two of the numbers you land on (the 19th and the 92nd) are zero. Using those two clues you can find the exact curve and then simply count forward (or backward) to see what the 3rd number must be.
After doing the counting neatly, you discover the 3rd number is 712.

Step-by-Step Solution

1. Assume Quadratic Form

Second difference of $a_n=An^2+Bn+C$ equals $2A$. Hence

So

2. Use $a_{19}=0$

That is

3. Use $a_{92}=0$

4. Solve for $B$

Subtract (1) from (2):

5. Find $C$

Insert $B$ into (1):

6. Compute $a_3$

Answer: $\boxed{712}$