Волейбол арифметическая прогрессия

To solve this problem, we need to determine the number of points the last-place team scored in a volleyball tournament where 7 teams played against each other once. Each win gives 1 point, there are no draws, and the points form an arithmetic progression.

Total Games and Points:

• Total number of games: $\binom{7}{2} = 21$
• Total points awarded: 21 (since each game awards 1 point to the winner)

Arithmetic Progression of Points: Let's denote the points of the teams in decreasing order as an arithmetic progression: $P_1, P_2, P_3, P_4, P_5, P_6, P_7$ with a common difference $-r$ (since points decrease from the top team to the last).

Expressing the Points: $P_n = a - (n-1)r$ where $n$ is the team's position (1 through 7), $a$ is the first term (points of the top team), and $r$ is the common difference.

Calculating Sum of Points: The sum of an arithmetic progression is: $S = \frac{n}{2}(2a - (n-1)r)$ Given that $S = 21$ and $n = 7$: $21 = \frac{7}{2}(2a - 6r)$ Simplify: $21 = 7a - 21r$ $a - 3r = 3$ $a = 3 + 3r$

Finding Possible Values for $r$: Since the maximum points a team can have is 6 (winning all 6 games), $a \leq 6$: $3 + 3r \leq 6$ $r \leq 1$ Similarly, the last-place team's points must be non-negative: $P_7 = a - 6r = 3 + 3r - 6r = 3 - 3r \geq 0$ $r \leq 1$

So, $r$ can be 0 or 1.

Case 1: $r = 0$

• All teams have $a = 3$ points.
• This is acceptable but means all teams are tied, which isn't consistent with the concept of first and last place.

Case 2: $r = 1$

• $a = 3 + 3(1) = 6$
• Points for each team:
  • $P_1 = 6$
  • $P_2 = 5$
  • $P_3 = 4$
  • $P_4 = 3$
  • $P_5 = 2$
  • $P_6 = 1$
  • $P_7 = 0$
• This distribution is valid and matches the total points awarded.

Answer: The last-place team scored 0 points.

Ответ: 0.