The sum of the absolute values of the terms of a finite arithmetic progression is 100. If all its terms are increased by 1 or all its terms are increased by 2, then in both cases the sum of the absolute values of the terms of the resulting progression will also be equal to 100. What values can the quantity $n^{2} d$ take under these conditions, where $d$ is the common difference of the progression, and $n$ is the number of its terms?
Let's denote the sum of the absolute values of the terms of an arithmetic progression by $S$. We will show that the quantity $S /\left(n^{2} d\right)$ is constant for progressions satisfying the condition of the problem and equals $1 / 4$ if the given progression $a_{1}, a_{2}, \ldots, a_{n}$ is increasing (for a decreasing progression, this quantity equals -1/4). From the condition of the problem, it follows that the function

$S(x)=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\ldots+\left|x-a_{n}\right|$

takes the same value at three different points. Since

$$
S(x)=(2 i-n) x+\mathrm{const}, \quad \text { for } \quad \begin{cases}x \leqslant a_{i}, & i=1 \\ a_{i} \leqslant x \leqslant a_{i+1}, & i=1, \ldots, n-1 \\ x \geqslant a_{i}, & i=n\end{cases}
$$

then for $x \leq a_{i+1}$ and $i > n / 2$ it increases. Therefore, the condition of the problem can only be satisfied when the number $n=2 k$ is even and

$$
S=S\left(a_{k}+\frac{d}{2}\right)=2 \cdot \frac{d / 2+d / 2+(k-1) d}{2} \cdot k=k^{2} d=\frac{n^{2} d}{4}
$$

## Answer

The problem