3. Petya sequentially writes down integers, starting from 21, such that each subsequent number is 4 less than the previous one, and Vasya, looking at the current number, calculates the sum of all the numbers written down by this point. Which of the sums found by Vasya will be closest to 55?
Answer: 56.

First solution. Petya sequentially writes down the members of the arithmetic progression $a_{n}=$ $21-4(n-1)$, while Vasya writes down the sum of these members $S_{n}=\frac{2 \cdot 21-4(n-1)}{2} \cdot n=(21-2(n-1)) n=23 n-2 n^{2}$. The equation $23 x-2 x^{2}=55$ has the following solutions: $\frac{23 \pm \sqrt{89}}{4}$. Since $3<\frac{23-\sqrt{89}}{4}<4$ and $8<$ $\frac{23+\sqrt{89}}{4}<9$, it is sufficient to compare $S_{3}=51, S_{4}=60, S_{8}=56$ and $S_{9}=45$. The closest to 55 is the sum $S_{8}=56$.

Second solution. As long as Petya writes down positive numbers, the sums written by Vasya increase, and it is not difficult to calculate that in the increasing sequence of numbers written by Vasya, the closest to 55 will be the number 51. But then Petya will start writing down negative numbers, and the sums calculated by Vasya will begin to decrease. Among the decreasing sequence of sums, the closest to 55 will be the number 56 (see table).

| Petya | 21 | 17 | 13 | 9 | 5 | 1 | -3 | -7 | -11 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Vasya | 21 | 38 | 51 | 60 | 65 | 66 | 63 | 56 | 45 | $\cdots$ |