📚 Arithmetic Sequence | 等差数列
An arithmetic sequence is one of the most fundamental concepts in IB Mathematics. It appears in the Number and Algebra topic and provides the foundation for understanding linear patterns, series, and financial applications. Mastering arithmetic sequences means being comfortable with the common difference, the general term formula, and the sum of a finite number of terms. This article will walk you through every key idea, with worked examples and clear explanations, ensuring you can approach exam questions with confidence.
等差数列是IB数学中最基础的概念之一，属于数与代数主题，为理解线性规律、级数及金融应用奠定基础。掌握等差数列意味着要能熟练运用公差、通项公式以及有限项求和公式。本文将带你逐一梳理每一个关键知识点，配合范例和清晰的说明，帮助你自信应对考试题目。
1. Definition and Common Difference | 定义与公差
An arithmetic sequence (or arithmetic progression) is a list of numbers where the difference between any two consecutive terms is always the same. This fixed difference is called the common difference, usually denoted by the letter d. If we write the first few terms as u₁, u₂, u₃, …, then u₂ − u₁ = u₃ − u₂ = d. The sequence 3, 7, 11, 15, … is arithmetic with d = 4, while 10, 7, 4, 1, … is arithmetic with d = −3.
等差数列（也称算术数列）是这样一列数：任意相邻两项之间的差始终相同。这个固定的差值称为公差，通常用字母 d 表示。若记前几项为 u₁, u₂, u₃, …，则有 u₂ − u₁ = u₃ − u₂ = d。例如数列 3, 7, 11, 15, … 是公差为 4 的等差数列，而 10, 7, 4, 1, … 是公差为 −3 的等差数列。
It is essential to check that the difference is constant across several pairs of terms, especially when working with sequences given by a formula. If d > 0 the sequence is increasing; if d < 0 it is decreasing; and if d = 0 all terms are equal (a constant sequence). In IB questions you may be given the first few terms and asked to find d, or you may need to confirm that a sequence is arithmetic.
关键是要核验多对相邻项的差是否恒定，尤其在处理由公式给出的数列时。如果 d > 0，数列递增；如果 d < 0，数列递减；若 d = 0，则所有项都相等（常数列）。在IB考题中，可能会给出前几项让你求 d，也可能需要你验证一个数列是否为等差数列。
2. The General Term (nth Term) | 通项公式
To describe any term in an arithmetic sequence without listing all previous terms, we use the general term, often called the nth term. If the first term is u₁ and the common difference is d, then the nth term is given by:
为了无需列出之前所有项就能描述等差数列中的任意一项，我们使用通项，也称第 n 项。如果首项为 u₁，公差为 d，那么第 n 项的公式为：
uₙ = u₁ + (n − 1)d
Here n is a positive integer. The formula tells us we start at u₁ and add the common difference (n − 1) times to reach the nth term. For instance, in the sequence 5, 9, 13, 17, … with u₁ = 5 and d = 4, the 10th term is u₁₀ = 5 + (10 − 1)×4 = 5 + 36 = 41.
其中 n 为正整数。公式表明，从首项 u₁ 出发，加上 (n − 1) 倍的公差即可得到第 n 项。例如在数列 5, 9, 13, 17, … 中，u₁ = 5, d = 4，因此第 10 项为 u₁₀ = 5 + (10 − 1)×4 = 5 + 36 = 41。
IB often uses the notation uₙ for sequences, though you may also see Tₙ or aₙ. The logic remains identical. This formula is linear in n, which is why arithmetic sequences model linear growth or decay.
IB 常使用记号 uₙ 表示数列项，不过你也可能见到 Tₙ 或 aₙ，其逻辑完全相同。该公式关于 n 是线性的，这正是等差数列能模拟线性增长或衰减的原因。
3. Working with the General Term | 通项公式的应用
You can use the general term formula in several ways: to find a specific term when n is known, to determine the position n when a term value is given, or to set up equations when two terms are known. For example, if u₁ = 20 and d = −3, find the first term that is negative. Set up uₙ < 0: 20 + (n − 1)(−3) < 0 → 20 − 3n + 3 < 0 → 23 < 3n → n > 7.67, so the 8th term is the first negative term (u₈ = 20 + 7(−3) = −1).
你可以通过多种方式运用通项公式：已知 n 求特定项、已知项的值求其位置 n，或在已知两项时建立方程。例如，若 u₁ = 20，d = −3，求第一个负值项。列出不等式 uₙ < 0：20 + (n − 1)(−3) < 0 → 20 − 3n + 3 < 0 → 23 < 3n → n > 7.67，因此第 8 项是第一个负值项（u₈ = 20 + 7(−3) = −1）。
When a question gives a term and asks for n, simply substitute the known values into uₙ = u₁ + (n − 1)d and solve the linear equation. Be careful with negative d and with terms that might seem counter‑intuitive; the algebra will always guide you to the correct integer n.
当题目给出某一项的值并要求找出 n 时，只需将已知值代入 uₙ = u₁ + (n − 1)d，解出一次方程即可。注意处理负公差以及那些乍看之下反常的项；代数运算总能引导你得到正确的整数 n。
4. Finding the Common Difference | 求公差
If you are given two terms of an arithmetic sequence, you can always find d. Suppose you know the p‑th term uₚ and the q‑th term u_q, with p ≠ q. Using the general term formula twice gives:
如果已知等差数列中的两项，你总能求出 d。假设已知第 p 项 uₚ 和第 q 项 u_q，且 p ≠ q。两次运用通项公式可得：
d = (u_q − uₚ) / (q − p)
This works because u_q − uₚ = [u₁ + (q − 1)d] − [u₁ + (p − 1)d] = (q − p)d. For instance, if the 3rd term is 8 and the 7th term is 20, then d = (20 − 8) / (7 − 3) = 12 / 4 = 3. Once d is known, you can back‑substitute to find u₁.
这是因为 u_q − uₚ = [u₁ + (q − 1)d] − [u₁ + (p − 1)d] = (q − p)d。例如，若第 3 项为 8，第 7 项为 20，则 d = (20 − 8) / (7 − 3) = 12 / 4 = 3。求出 d 之后，再回代即可求得 u₁。
This method is extremely reliable and is often tested in IB Paper 1 and Paper 2. Always double‑check that the computed d is consistent with the given context, especially if the sequence is defined in a word problem.
这种方法非常可靠，在 IB 试卷一和试卷二中经常考查。务必复核计算出的 d 是否与题目情境一致，尤其当数列以应用题形式给出时。
5. Sum of an Arithmetic Sequence | 等差数列的求和
The sum of the first n terms of an arithmetic sequence is called an arithmetic series. The sum, often denoted Sₙ, can be expressed in two equivalent forms:
等差数列前 n 项的和称为等差级数。该和通常记作 Sₙ，有两种等价表达式：
Sₙ = n/2 (u₁ + uₙ)
Sₙ = n/2 [2u₁ + (n − 1)d]
The first formula is handy when you know the first and the nth term. The second is useful when you only know u₁ and d. For example, to sum the first 20 terms of 3 + 7 + 11 + … , u₁ = 3, d = 4. Using the second formula: S₂₀ = 20/2 [2×3 + (20 − 1)×4] = 10 [6 + 76] = 820.
第一个公式在已知首项和第 n 项时非常方便。第二个公式则适用于仅知道 u₁ 和 d 的情形。例如，求 3 + 7 + 11 + … 的前 20 项之和：u₁ = 3，d = 4。用第二个公式：S₂₀ = 20/2 [2×3 + (20 − 1)×4] = 10 [6 + 76] = 820。
When using these formulas, be meticulous with the order of operations. It is often helpful to compute the bracket first, then multiply by n/2. In IB exams, forgetting the factor ½ is a common mistake.
使用这些公式时，要仔细遵循运算顺序。通常先计算括号内的值，再乘以 n/2。在 IB 考试中，遗漏因子 ½ 是常见失误。
6. Derivation of the Sum Formula | 求和公式的推导
IB assessments often value understanding the derivation of Sₙ, and questions may ask you to prove or explain the formula. The classic approach is to write the sum forwards and backwards and add them. Let Sₙ = u₁ + u₂ + … + uₙ. Since the sequence is arithmetic, the terms equally spaced from the ends have the same sum: u₁ + uₙ = u₂ + uₙ₋₁ = u₃ + uₙ₋₂ = … . Writing the sequence in reverse gives Sₙ = uₙ + uₙ₋₁ + … + u₁. Adding the two representations:
IB 评估常重视对 Sₙ 推导过程的理解，题目可能会要求你证明或解释该公式。经典方法是正向和反向写出和式再相加。令 Sₙ = u₁ + u₂ + … + uₙ。由于数列是等差的，首尾等距离的两项之和相等：u₁ + uₙ = u₂ + uₙ₋₁ = u₃ + uₙ₋₂ = …。反向写出数列得到 Sₙ = uₙ + uₙ₋₁ + … + u₁。将两种表示相加：
2Sₙ = (u₁ + uₙ) + (u₂ + uₙ₋₁) + … + (uₙ + u₁) = n(u₁ + uₙ)
Dividing by 2 yields Sₙ = n/2 (u₁ + uₙ). This proof highlights why the average of the first and last term, multiplied by the number of terms, gives the total sum. It also reinforces the idea that arithmetic sequences are symmetric around their centre.
除以 2 即得 Sₙ = n/2 (u₁ + uₙ)。这一证明凸显了为何首项与末项的平均值乘以项数就能得出总和，也强化了等差数列关于中心对称的直观认识。
7. Solving Problems Involving Sum | 涉及求和的问题
Many IB problems combine the sum formula with the general term formula. For example, you might be given Sₙ and u₁, and asked to find n or d. Suppose the sum of the first n terms of an arithmetic sequence is Sₙ = 3n² − n. To check whether the sequence is arithmetic, find uₙ by using uₙ = Sₙ − Sₙ₋₁. For n ≥ 2, uₙ = (3n² − n) − (3(n−1)² − (n−1)) = 6n − 4. Since uₙ is linear in n, the sequence is arithmetic with d = 6 and u₁ = 2.
许多 IB 问题会将求和公式与通项公式结合。例如，给出 Sₙ 和 u₁，要求你求 n 或 d。假设某等差数列前 n 项和为 Sₙ = 3n² − n。要检验该数列是否为等差数列，可利用 uₙ = Sₙ − Sₙ₋₁ 求得通项。当 n ≥ 2 时，uₙ = (3n² − n) − (3(n−1)² − (n−1)) = 6n − 4。由于 uₙ 关于 n 是线性的，该数列是等差数列，公差 d = 6，首项 u₁ = 2。
Another common task is finding n when Sₙ is given, which often leads to a quadratic equation. Always reject non‑integer or negative solutions for n, since n must be a positive integer. Check your final answer in the context of the problem.
另一种常见任务是已知 Sₙ 求 n，这往往会导出一个二次方程。必须舍去非整数或负数的解，因为 n 必须是正整数。应结合题目背景检验最终答案。
8. Arithmetic Mean | 等差中项
The arithmetic mean of two numbers a and c is the middle term b such that a, b, c form an arithmetic sequence. It follows that b − a = c − b, so 2b = a + c and therefore b = (a + c) / 2. This concept extends naturally: any term in an arithmetic sequence (except the first and last) is the arithmetic mean of its two neighbours.
两数 a 与 c 的等差中项是指一个中间项 b，使得 a, b, c 构成等差数列。由此可得 b − a = c − b，于是 2b = a + c，即 b = (a + c) / 2。这一概念可自然推广：等差数列中除首末项外的任意一项都是其左右相邻两项的等差中项。
In IB questions you might be asked to insert a certain number of arithmetic means between two given numbers. For instance, insert four arithmetic means between 5 and 25. This creates an arithmetic sequence with u₁ = 5 and u₆ = 25, so d = (25 − 5)/(6 − 1) = 4. The sequence is 5, 9, 13, 17, 21, 25, where the four means are 9, 13, 17, and 21.
在 IB 题目中，你可能会被要求在两个给定数之间插入若干个等差中项。例如，在 5 和 25 之间插入四个等差中项，这将构成一个等差数列，其中 u₁ = 5，u₆ = 25，于是 d = (25 − 5)/(6 − 1) = 4。所得数列为 5, 9, 13, 17, 21, 25，四个等差中项依次为 9, 13, 17, 21。
9. Applications in Real Life and Sequences | 实际应用与数列
Arithmetic sequences model any situation where a quantity increases or decreases by a fixed amount each step. Typical IB examples include the total number of seats in a theatre where each row has the same increment of seats, the accumulation of simple interest, or the height of a stack of books. Interpreting the problem involves identifying u₁ and d, then applying the sum or term formula as needed.
等差数列可用于模拟任何每一步增加或减少固定数量的情形。IB 中常见的例子包括：剧院中每排座位数等量递增的情况、单利的累积、以及书本叠放的高度等。解读题意需要识别出 u₁ 和 d，然后根据需要应用求和公式或通项公式。
For example, a student saves £5 in the first week, £8 in the second week, £11 in the third week, and so on. The amounts form an arithmetic sequence with u₁ = 5, d = 3. To find the total saved after 10 weeks, calculate S₁₀ = 10/2 [2×5 + (10 − 1)×3] = 5 [10 + 27] = £185. Always state the units clearly in your answer.
例如，某学生第一周存 5 英镑，第二周存 8 英镑，第三周存 11 英镑，以此类推。存款金额构成公差为 3 的等差数列，u₁ = 5。若要求 10 周后的存款总额，计算 S₁₀ = 10/2 [2×5 + (10 − 1)×3] = 5 [10 + 27] = 185 英镑。作答时务必清晰标明单位。
10. Summary and Key Points | 总结与要点
Arithmetic sequences are a core part of the IB Mathematics syllabus, and fluency with the formulas can earn you easy marks. Keep the following summary in mind as you revise:
等差数列是IB数学课程的核心内容，熟练运用公式能轻松得分。复习时请牢记以下总结：
| Common difference | d = uₙ₊₁ − uₙ | 公差 |
| nth term | uₙ = u₁ + (n − 1)d | 第 n 项 |
| d from two terms | d = (u_q − uₚ)/(q − p) | 由两项求公差 |
| Sum (first + last) | Sₙ = n/2 (u₁ + uₙ) | 求和（首项+末项） |
| Sum (first + d) | Sₙ = n/2 [2u₁ + (n − 1)d] | 求和（首项+公差） |
| Arithmetic mean | b = (a + c)/2 when a, b, c are in AP | 等差中项 |
Always define your variables, check that n is a positive integer, and verify the context of the problem. With these principles, you can confidently tackle arithmetic sequence questions across IB papers.
始终明确定义变量，检查 n 是否为正整数，并结合题意进行验证。掌握这些原理，你就能自信地应对 IB 试卷中各类等差数列问题。
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