📚 IGCSE Mathematics: Arithmetic Sequences — General Term and Properties | IGCSE数学：等差数列通项与性质
An arithmetic sequence is one of the most fundamental topics in IGCSE Mathematics. In this article, you will learn how to identify an arithmetic sequence, find its general term, use key properties, and avoid common exam mistakes.
等差数列是 IGCSE 数学中最基础的考点之一。本文将帮助你学会识别等差数列、求通项、运用重要性质，并避开常见考试误区。
1. What is an Arithmetic Sequence? | 什么是等差数列？
An arithmetic sequence is a list of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference, usually denoted by d.
等差数列是一列数字，其中任意两个连续项之间的差是固定的。这个固定的差叫做公差，通常用 d 表示。
h term is obtained by adding 4 to the previous term. Here, the common difference is d = 4.
例如数列 3, 7, 11, 15, ... 是等差数列，因为每一项都比前一项大 4。这里公差 d = 4。
The first term of a sequence is usually written as a₁. For the sequence above, a₁ = 3.
数列的第一项通常记作 a₁。在上面的数列中，a₁ = 3。
2. The General Term (nth term) | 通项公式
The nth term of an arithmetic sequence can be written as a linear expression in n. The standard formula is given by the first term plus (n − 1) times the common difference.
等差数列的第 n 项可以写成关于 n 的一次表达式。标准通项公式为首项加上 (n − 1) 倍的公差。
aₙ = a₁ + (n − 1)d
Here, aₙ is the nth term, a₁ is the first term, n is the term number, and d is the common difference.
其中 aₙ 是第 n 项，a₁ 是第一项，n 是项数，d 是公差。
, 8, 11, 14, ..., we have a₁ = 5 and d = 3. The nth term is aₙ = 5 + (n − 1) × 3 = 3n + 2.
例如，对于数列 5, 8, 11, 14, ...，有 a₁ = 5，d = 3。通项为 aₙ = 5 + (n − 1) × 3 = 3n + 2。
To find the 10th term, substitute n = 10: a₁₀ = 3 × 10 + 2 = 32.
求第 10 项时，代入 n = 10：a₁₀ = 3 × 10 + 2 = 32。
3. Finding the First Term and Common Difference | 求首项和公差
q always represents an arithmetic sequence with first term p + q and common difference p.
从等差数列的表达式中，你可以直接读出首项和公差。一般形式 aₙ = pn + q 一定表示等差数列，其首项为 p + q，公差为 p。
s aₙ = 7 − 2n, then the first term is a₁ = 7 − 2 = 5, and the common difference is d = −2.
例如，若通项为 aₙ = 7 − 2n，则首项 a₁ = 7 − 2 = 5，公差 d = −2。
ents incorrectly think that aₙ = 7 − 2n starts with 7. Always check by substituting n = 1.
注意：n 的系数是公差，不是首项。很多同学误认为 aₙ = 7 − 2n 的首项是 7。一定要代入 n = 1 来检验。
4. Using Two Terms to Find the Sequence | 利用两项确定数列
If you are given two terms of an arithmetic sequence, you can set up equations to find both a₁ and d. Suppose you know that the 3rd term is 12 and the 7th term is 28.
如果已知等差数列的两项，可以通过列方程求出 a₁ 和 d。例如，已知第 3 项为 12，第 7 项为 28。
Use the general term formula to write two equations:
利用通项公式列出两个方程：
a₃ = a₁ + 2d = 12
a₇ = a₁ + 6d = 28
Then substitute back to get a₁ = 4. Therefore, the nth term is aₙ = 4 + (n − 1) × 4 = 4n.
用第二个方程减去第一个方程：4d = 16，所以 d = 4。代回可得 a₁ = 4。因此通项为 aₙ = 4 + (n − 1) × 4 = 4n。
This method of simultaneous equations is essential for IGCSE problems that give two separate terms of a sequence.
这种列二元一次方程组的方法是 IGCSE 考试中已知两项求数列问题的核心方法。
5. Key Properties of Arithmetic Sequences | 等差数列的重要性质
Arithmetic sequences have several important properties that are frequently tested in IGCSE. Understanding these properties can save time and help you verify answers.
等差数列有几个常考的重要性质。理解这些性质可以节省时间，并帮助验证答案。
The difference between any two terms is a multiple of d. Specifically, aₘ − aₙ = (m − n)d.
性质1：任意两项之差是 d 的倍数。具体地，aₘ − aₙ = (m − n)d。
an arithmetic sequence, then y = (x + z) / 2. This is called the arithmetic mean property.
性质2：若三个数 x, y, z 是等差数列的连续三项，则 y = (x + z) / 2。这叫做等差中项性质。
Property 3: Adding a constant to every term, or multiplying every term by a non-zero constant, produces another arithmetic sequence.
性质3：若将每一项都加上同一个常数，或都乘以同一个非零常数，得到的仍是等差数列。
nth term aₙ = pn + q is always arithmetic, and its graph is a straight line with slope p.
性质4：通项为 aₙ = pn + q 的数列一定是等差数列，其图像是一条斜率为 p 的直线。
6. Arithmetic Mean | 等差中项
For any three consecutive terms a, b, c of an arithmetic sequence, the middle term b is the arithmetic mean of a and c. This means that b is exactly halfway between a and c on the number line.
对于等差数列中任意连续三项 a, b, c，中间项 b 是 a 和 c 的等差中项。也就是说，b 在数轴上正好位于 a 和 c 的正中间。
b = (a + c) / 2
sequence, then 10 = (x + 16) / 2. Multiplying both sides by 2 gives 20 = x + 16, so x = 4.
例如，若 x, 10, 16 是等差数列的连续三项，则 10 = (x + 16) / 2。两边乘以 2 得 20 = x + 16，所以 x = 4。
This property is extremely useful when a problem only gives you the middle term and one other term, or when you need to insert terms between two given numbers.
这个性质在题目只给出中间项和另一个项，或者需要你在两个数之间插入若干项时特别有用。
7. The Sum of the First n Terms | 前 n 项和
Although this topic is often linked to arithmetic sequences, the sum formula is also a key part of the IGCSE syllabus. For an arithmetic sequence with first term a₁ and common difference d, the sum of the first n terms is:
虽然本章主要讲数列通项，但求和公式同样是 IGCSE 大纲的重要部分。对于首项为 a₁、公差为 d 的等差数列，前 n 项和为：
Sₙ = n / 2 × [2a₁ + (n − 1)d]
Equivalently, if you know the first term a₁ and the last term aₙ, the sum is:
等价地，如果已知首项 a₁ 和末项 aₙ，则和为：
Sₙ = n / 2 × (a₁ + aₙ)
of the first 20 terms of the arithmetic sequence 2, 5, 8, ... Here a₁ = 2, d = 3, n = 20.
例如，求等差数列 2, 5, 8, ... 的前 20 项和。这里 a₁ = 2，d = 3，n = 20。
S₂₀ = 20 / 2 × [2 × 2 + (20 − 1) × 3] = 10 × (4 + 57) = 610
Notice that the sum formula uses the number of terms n, not the last term number directly. Always check that you are using the correct n.
注意求和公式中使用的是项数 n，而不是末项本身。一定要确认你代入的 n 正确。
8. Problem-Solving Strategies | 解题策略
When solving arithmetic sequence problems, follow a clear strategy: identify the known values, choose the correct formula, and then solve step by step. Here is a structured approach.
解决等差数列问题时，应遵循清晰的策略：确认已知量、选择正确公式，然后逐步求解。下面是一个结构化方法。
Step 1: Write down a₁, d, n, aₙ, or Sₙ as given in the problem.
第1步：写下题目给出的 a₁、d、n、aₙ 或 Sₙ。
Step 2: Determine which quantity you are trying to find.
第2步：确定你要求的是哪一个量。
Step 3: Select the general term formula, the arithmetic mean property, or the sum formula.
第3步：选择通项公式、等差中项性质或求和公式。
Step 4: Substitute the known values and solve the equation. If two equations are needed, solve them simultaneously.
第4步：代入已知值并解方程。如果需要两个方程，就联立求解。
Step 5: Check whether your answer is reasonable by substituting back into the original information.
第5步：将答案代回原题条件，检查是否合理。
9. Common Mistakes | 常见错误
Many IGCSE students lose marks on arithmetic sequence questions because of careless errors. The most common mistakes are listed below with explanations of how to avoid them.
许多 IGCSE 学生在等差数列题目中因粗心失分。下面列出最常见的错误及避免方法。
| Common Mistake | 常见错误 | Correct Approach | 正确做法 |
|---|---|
| Using aₙ = a₁ + nd instead of aₙ = a₁ + (n − 1)d | first term corresponds to n = 1, so there are only n − 1 differences before the nth term. |
| Mistaking the coefficient of n for the first term | Substitute n = 1 into the formula to find the first term directly. |
| Using the sum formula with the last term instead of n | Check carefully: Sₙ = n/2 × (a₁ + aₙ) always uses n, the number of terms. |
| Forgetting that d can be negative | A decreasing sequence has a negative common difference. Treat d as a signed number. |
In Chinese:
中文对照：
错误：使用 aₙ = a₁ + nd，而忘记了正确公式是 aₙ = a₁ + (n − 1)d。正确做法：首项对应 n = 1，因此第 n 项之前只有 n − 1 个公差。
错误：把 n 的系数当成首项。正确做法：将 n = 1 代入公式直接求首项。
错误：求和公式中把末项当成 n 使用。正确做法：Sₙ = n/2 × (a₁ + aₙ) 中的 n 永远是项数。
错误：忘记 d 可以为负数。正确做法：递减数列的公差是负数，d 要带符号计算。
10. Practice Questions | 练习
Try the following questions to test your understanding. These questions are designed to match the style of Edexcel IGCSE examinations.
尝试完成下面的练习，测试你的理解程度。这些题目按照 Edexcel IGCSE 考试风格设计。
Question 1: The first term of an arithmetic sequence is 6 and the common difference is 5. Find the 12th term.
第1题：等差数列首项为 6，公差为 5。求第 12 项。
. State whether the sequence is arithmetic, and find its first term and common difference.
第2题：某数列通项为 aₙ = 15 − 3n。判断它是否为等差数列，并求首项和公差。
Question 3: The 5th term of an arithmetic sequence is 17 and the 10th term is 32. Find the first term and the common difference.
第3题：等差数列第 5 项为 17，第 10 项为 32。求首项和公差。
Question 4: Insert three numbers between 2 and 22 so that all five numbers form an arithmetic sequence.
第4题：在 2 和 22 之间插入三个数，使这五个数构成等差数列。
then a₁ = 5. (4) Let d be the common difference; 2 + 4d = 22, so d = 5, giving 7, 12, 17.
答案：(1) a₁₂ = 6 + 11 × 5 = 61。(2) 是，a₁ = 12，d = −3。(3) 两式相减：5d = 15，所以 d = 3；再得 a₁ = 5。(4) 设公差为 d，则 2 + 4d = 22，所以 d = 5，插入的数为 7、12、17。
11. Summary | 总结
+ c) / 2 and the linear form aₙ = pn + q are two powerful tools for quick problem solving.
等差数列由固定的公差定义。通项公式为 aₙ = a₁ + (n − 1)d，前 n 项和为 Sₙ = n/2 × [2a₁ + (n − 1)d]。等差中项性质 b = (a + c) / 2 以及线性形式 aₙ = pn + q 是快速解题的两个强大工具。
Always check the value of n carefully, remember that d may be negative, and verify your answer by substituting back into the original conditions. With regular practice, arithmetic sequence problems become straightforward and highly rewarding in exams.
解题时务必仔细检查 n 的取值，牢记 d 可以为负，并将答案代回原题验证。只要坚持练习，等差数列题目就会变得简单，并且成为考试中的得分亮点。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图