[ $\quad$ Arithmetic progression $\quad]$ [Integer and fractional parts. Archimedes' principle ]

For the numbers $1000^{2}, 1001^{2}, 1002^{2}, \ldots$, the last two digits are discarded. How many of the first terms of the resulting sequence form an arithmetic progression?
The general term of the original sequence is: $a_{n}=\left(10^{3}+n\right)^{2}=10^{6}+2 n \cdot 10^{3}+n^{2}\left(a_{0}-\right.$ the first term $)$.

Let the general term of the resulting sequence be denoted by $b_{n}$, then

$b_{n}=\left[\frac{a_{n}}{100}\right]=\left[\frac{10^{6}+2 \cdot n \cdot 10^{3}+n^{2}}{100}\right]=10^{4}+20 n+\left[\frac{n^{2}}{100}\right]$

Therefore, the difference between consecutive terms of the new sequence will be 20 as long as $\left[\frac{n^{2}}{100}\right]=0$, which is if $n^{2}<100$. Considering that $n$ is a natural number or zero, we get: $0 \leq n \leq 9$. Thus, the first 10 terms of the new sequence form an arithmetic progression.

## Answer

10.