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📊 等差数列

Arithmetic Sequences

等差数列是 AMC 8 中高频出现的数列类型。掌握通项公式和求和公式，是解决数列问题的两大核心武器。

📚 3 章节 💡 5 道例题 ✏️ 8 道练习 🎯 难度：基础~中等 ⏱ 约35分钟
1
数列的基本概念 Basic Concepts of Sequences
基础必考

1.1 项、首项、末项 Terms, First Term, Last Term

数列（sequence）是按照一定规律排列的一组数。数列中的每一个数称为一个项（term）。

A sequence is a list of numbers arranged in a specific order. Each number in the sequence is called a term.

📝 数列的命名 / Naming Terms
a₁ = 第1项（首项，first term）
a₂ = 第2项（second term）
aₙ = 第n项（nth term，通项）
n = 项数（number of terms）
a₁ = first term, a₂ = second term, aₙ = nth term (general term), n = position

举例：数列 3, 7, 11, 15, 19, ...

  • 首项 a₁ = 3
  • a₂ = 7, a₃ = 11, a₄ = 15, a₅ = 19
  • 规律：每一项比前一项多 4

1.2 数列的表示方法 Ways to Represent a Sequence

数列可以用多种方式表示：

Sequences can be represented in several ways:

方式Method示例
列举法Listing2, 5, 8, 11, 14, ...
递推式Recursive formulaa₁=2, aₙ₊₁=aₙ+3
通项公式Explicit formulaaₙ = 3n − 1
💡 邓老师提示：在 AMC 8 中，识别数列的规律（constant difference = 常数差）是第一步。找到规律后，用通项公式或求和公式来解题。
On the AMC 8, the first step is to identify the pattern (constant difference). Then use the formula for the general term or sum to solve.
2
等差数列的定义与通项公式 Arithmetic Sequence: Definition and General Term
核心必考

2.1 公差 d Common Difference

等差数列（arithmetic sequence）是相邻两项之差为常数的数列。这个常数差称为公差（common difference），记为 d。

An arithmetic sequence has a constant difference between consecutive terms. This constant is called the common difference, denoted d.

📝 公差 / Common Difference
d = a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = ...
公差可以是正数（递增）、负数（递减）或零（常数列）
d can be positive (increasing), negative (decreasing), or zero (constant)

举例：

  • 2, 5, 8, 11, ... → d = 5 − 2 = 3（递增）
  • 20, 15, 10, 5, ... → d = 15 − 20 = −5（递减）
  • 7, 7, 7, 7, ... → d = 0（常数列）

2.2 通项公式 General Term Formula

等差数列的通项公式（第 n 项的公式）：

The formula for the nth term of an arithmetic sequence:

📝 通项公式 / General Term
aₙ = a₁ + (n − 1) × d
其中 a₁ 是首项，d 是公差，n 是项数
where a₁ = first term, d = common difference, n = term number

推导过程：

  • a₁ = a₁
  • a₂ = a₁ + d
  • a₃ = a₁ + 2d
  • a₄ = a₁ + 3d
  • ... → aₙ = a₁ + (n−1)d
💡 邓老师提示：如果只知道 aₖ（第 k 项）而不是 a₁，也可以用公式 aₙ = aₖ + (n−k) × d。这在某些题目中更方便。
If you know aₖ instead of a₁, use aₙ = aₖ + (n−k) × d. This can be more convenient for certain problems.
3
等差数列的求和 Sum of an Arithmetic Sequence
核心高频

3.1 求和公式 Sum Formula

等差数列前 n 项的和，有两种常用公式：

There are two common formulas for the sum of the first n terms:

📝 等差数列求和公式 / Arithmetic Sum Formulas
Sn = n × (a₁ + aₙ) ÷ 2   （公式一）
Sn = n × a₁ + n(n−1) × d ÷ 2   （公式二）
Formula 1: S = n(a₁ + aₙ)/2 — use when you know both first and last terms
Formula 2: S = na₁ + n(n−1)d/2 — use when you know first term and common difference
💡 邓老师提示：公式一最常用！S = 项数 × (首项 + 末项) ÷ 2。本质上就是"平均数 × 项数"。
Formula 1 is most commonly used! S = number of terms × (first + last) ÷ 2. It's essentially "average × count".

3.2 中项法 Middle Term Method

当等差数列有奇数项时，中间那一项恰好等于所有项的平均值。

When an arithmetic sequence has an odd number of terms, the middle term equals the average of all terms.

📝 中项法 / Middle Term Method
Sn = n × a中（n 为奇数）
a中 = (a₁ + aₙ) ÷ 2 = 中间项的值
When n is odd, the sum equals n times the middle term.

举例：求数列 1, 3, 5, 7, 9 的和。

  • 方法一：S = 5 × (1 + 9) ÷ 2 = 5 × 5 = 25
  • 方法二：中间项是 5，S = 5 × 5 = 25
⚠️ 注意：中项法只适用于奇数项的等差数列！如果是偶数项，需要用标准公式。
The middle term method only works for an odd number of terms! For even terms, use the standard formula.
4
例题精讲 Worked Examples
5 题含历年真题
📌 例题 1 通项公式

等差数列的首项为 3，公差为 4。第 20 项是多少？An arithmetic sequence has first term 3 and common difference 4. What is the 20th term?

解题思路
aₙ = a₁ + (n−1)d = 3 + (20−1) × 4 = 3 + 19 × 4 = 3 + 76 = 79 aₙ = 3 + (20−1)×4 = 3 + 76 = 79.
📌 例题 2 等差数列求和

求前 50 个正偶数之和：2 + 4 + 6 + 8 + ... + 100。Find the sum of the first 50 positive even numbers: 2 + 4 + 6 + 8 + ... + 100.

解题思路
首项 a₁ = 2，末项 a₅₀ = 100，项数 n = 50
S = n(a₁ + aₙ) ÷ 2 = 50 × (2 + 100) ÷ 2 = 50 × 51 = 2550 a₁=2, a₅₀=100, n=50. S = 50×(2+100)÷2 = 50×51 = 2550.
📌 例题 3 AMC 8 真题改编

一个等差数列的第 5 项是 17，第 15 项是 47。这个数列的首项是多少？In an arithmetic sequence, the 5th term is 17 and the 15th term is 47. What is the first term?

解题思路
a₅ = a₁ + 4d = 17 ... ①
a₁₅ = a₁ + 14d = 47 ... ②
②−①：10d = 30，d = 3
代入①：a₁ + 12 = 17，a₁ = 5 a₅=a₁+4d=17, a₁₅=a₁+14d=47. Subtracting: 10d=30, d=3. So a₁+12=17, a₁=5.
📌 例题 4 求项数

等差数列 5, 8, 11, 14, ... 中，有多少项不超过 200？In the arithmetic sequence 5, 8, 11, 14, ..., how many terms are ≤ 200?

解题思路
a₁ = 5，d = 3
aₙ = 5 + (n−1) × 3 ≤ 200
3(n−1) ≤ 195 → n−1 ≤ 65 → n ≤ 66
所以有 66 项不超过 200 aₙ = 5+(n−1)×3 ≤ 200. So 3(n−1) ≤ 195, n−1 ≤ 65, n ≤ 66. There are 66 terms ≤ 200.
📌 例题 5 中项法

求 1 + 3 + 5 + 7 + ... + 99 的值（所有两位数以内的正奇数之和）。Find the value of 1 + 3 + 5 + 7 + ... + 99 (sum of all positive odd numbers up to 99).

解题思路
首项 a₁ = 1，末项 aₙ = 99，公差 d = 2
项数 n = (99−1)÷2 + 1 = 50
方法一：S = 50 × (1+99) ÷ 2 = 50 × 50 = 2500
方法二：中间两项是 49 和 51，中位数 = (49+51)÷2 = 50。S = 50 × 50 = 2500 a₁=1, aₙ=99, d=2, n=50. S = 50×(1+99)÷2 = 50×50 = 2500.
5
巩固练习 Practice Problems
8 题提交即判

第1题 等差数列的首项为 10，公差为 −3。第 15 项是多少？An arithmetic sequence has first term 10 and common difference −3. What is the 15th term?

第2题 等差数列 4, 7, 10, 13, ... 的第 100 项是多少？What is the 100th term of the arithmetic sequence 4, 7, 10, 13, ...?

第3题 求等差数列 1, 4, 7, 10, ..., 100 的所有项之和。Find the sum of the arithmetic sequence 1, 4, 7, 10, ..., 100.

第4题 一个等差数列的第 3 项是 11，第 8 项是 26。公差 d 是多少？The 3rd term of an arithmetic sequence is 11, and the 8th term is 26. What is d?

第5题 等差数列 100, 93, 86, 79, ... 中，第一个负数项是第几项？In the arithmetic sequence 100, 93, 86, 79, ..., which term is the first negative number?

第6题 将 1 到 100 所有能被 3 整除的数相加，和是多少？What is the sum of all integers from 1 to 100 that are divisible by 3?

第7题 等差数列的和为 S = n² × 3（其中 n 为项数）。若首项 a₁ = 3，则公差 d 是多少？An arithmetic sequence has sum S = 3n² (where n is the number of terms). If a₁ = 3, what is d?

第8题 三个数成等差数列，它们的和是 36，积是 960。这三个数中最小的数是多少？Three numbers form an arithmetic sequence. Their sum is 36 and their product is 960. What is the smallest number?

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