3. Gea thought of a number. If she were to form a sequence of 2021 terms such that the first term is the thought number, and each subsequent term is 20.21 greater than the previous one, the sum of all the numbers in this sequence would be 11.1 times greater than $2021 \cdot 2021$. What number did Gea think of?
First method:

Let $x$ be the number Gea thought of.

The sequence of 2021 terms that arises in the described manner is a sequence of numbers:

$$
x, x+1 \cdot 20.21, x+2 \cdot 20.21, \ldots, x+2020 \cdot 20.21
$$

Then we have:

$x+x+1 \cdot 20.21+x+2 \cdot 20.21+\ldots+x+2020 \cdot 20.21=11.1 \cdot 2021 \cdot 2021$

$2021 \cdot x+20.21 \cdot(1+2+\ldots+2020)=11.1 \cdot 2021 \cdot 2021$

$2021 \cdot x+20.21 \cdot(2021 \cdot 2020): 2=11.1 \cdot 2021 \cdot 2021$

$2021 \cdot x+20.21 \cdot 1010 \cdot 2021=11.1 \cdot 2021 \cdot 2021 /: 2021$

$x+20.21 \cdot 1010=11.1 \cdot 2021$

$x+2021 \cdot 10.1=11.1 \cdot 2021$

$x=11.1 \cdot 2021-2021 \cdot 10.1$

$x=2021 \cdot(11.1-10.1)$

$x=2021 \cdot 1$

$x=2021$

Gea thought of the number 2021.