Arithmetic Sequences | 等差数列

📚 Arithmetic Sequences | 等差数列

Arithmetic sequences, also known as arithmetic progressions, are one of the most fundamental concepts in IB Mathematics. They form the basis for understanding linear patterns and are widely applied in finance, physics, and everyday problem solving. Mastering arithmetic sequences is essential for success in topics such as series, proof by induction, and modeling real-life situations. Recognizing the constant difference between terms unlocks a wide range of formulas and reasoning skills that are tested across both analysis and applications syllabi.

等差数列（也称为算术数列）是IB数学中最基本的概念之一。它们构成了理解线性模式的基础，并广泛应用于金融、物理和日常问题解决中。掌握等差数列对于成功学习级数、数学归纳法证明以及现实情境建模等主题至关重要。识别项之间的恒定差能够解锁一系列公式和推理技巧，这些在分析与应用两个方向都会考查。

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1. Definition of Arithmetic Sequences | 等差数列的定义

cause each term is obtained by adding 4 to the previous one. The sequence can be finite or infinite, and its terms may be integers, fractions, or algebraic expressions.

等差数列是指任意连续两项之差为常数的数列。这个常数差称为公差，用 d 表示。例如，数列 3, 7, 11, 15, ... 就是公差为 4 的等差数列。因为每一项都是前一项加 4 得到的。数列可以是有限的或无限的，其项可以是整数、分数或代数表达式。

If the difference is not constant throughout, the sequence is not arithmetic. In IB exams, you may be asked to determine whether a given sequence is arithmetic, or to prove that a sequence defined by a recurrence relation is arithmetic.

如果相邻项的差不是恒定的，那么该数列就不是等差 数列。在IB考试中，可能会要求你判断一个给定数列是否为等差数列，或者证明一个由递推关系定义的数列是等差数列。

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2. Common Difference | 公差

the same, yielding a constant sequence. Understanding the sign and magnitude of d immediately tells you the behaviour of the sequence.

公差 d 通过用后一项减去前一项来计算：d = uₙ₊₁ − uₙ。如果 d > 0，数列递增；如果 d < 0，数列递减；如果 d = 0，则每一项都相同，为常数列。理解 d 的符号和大小能让你立即判断数列的行为。

rms decrease by 4 each step. In IB problems, you may need to find d from a word description or from given terms that are not consecutive, requiring the use of the nth term formula.

例如，在数列 17, 13, 9, 5, ... 中，我们得到 d = 13 − 17 = −4，因此每一项递减 4。在IB题目中，你可能需要从文字描述或从非连续的已知项中求出 d，这时需要用到通项公式。

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3. General Term (nth Term) Formula | 通项公式

− 1)d, where u₁ is the first term and n is the term number. This explicit formula lets you jump directly to any term without having to list all previous ones. It is the single most important tool for working with arithmetic sequences.

等差数列的第 n 项由公式 uₙ = u₁ + (n − 1)d 给出，其中 u₁ 为首项，n 为项数。这个显式公式让你可以直接跳到任意一项，而无需列出前面所有项。它是处理等差数列最重要的工具。

uₙ = u₁ + (n − 1)d

sed in the question. The same logic applies if the sequence begins with u₀: uₙ = u₀ + n d.

如果首项 u₁ = 9，d = 3，那么第 15 项 u₁₅ = 9 + 14 × 3 = 51。在IB记法中，数列可能从 u₁ 或 u₀ 开始；一定要检查题目使用的约定。同样的逻辑适用于从 u₀ 开始的情形：uₙ = u₀ + n d。

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4. Finding the First Term and Common Difference | 求首项和公差

Many IB questions give you the values of two specific terms and ask you to determine u₁ and d. This is done by substituting the term numbers into the nth term formula to create a system of two linear equations, which can be solved by elimination or substitution.

许多IB题目会给出某两项的值，要求你求出 u₁ 和 d。方法是将项数代入通项公式，建立两个线性方程组成的方程组，再用消元法或代入法求解。

gives 5d = 15, so d = 3. Substituting back, u₁ = 4. The sequence is thus 4, 7, 10, 13, ...

例如：已知 u₃ = 10，u₈ = 25。那么 u₁ + 2d = 10，u₁ + 7d = 25。第二式减第一式得 5d = 15，所以 d = 3。代回得 u₁ = 4。因此数列为 4, 7, 10, 13, ...

This technique is fundamental and appears frequently in Paper 1 and Paper 2. Always define your variables clearly and check your result by generating the required terms.

这一技巧非常基本，在试卷1和试卷2中频繁出现。务必清晰地定义变量，并通过生成所需项来检验结果。

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5. Sum of the First n Terms | 前n项和

The sum of the first n terms of an arithmetic sequence is denoted by Sₙ. Two equivalent formulas are used, both of which you must know:

等差数列的前 n 项和记作 Sₙ。有两个等价公式必须掌握：

Sₙ = n/2 (2u₁ + (n − 1)d)

Sₙ = n/2 (u₁ + uₙ)

The first formula is direct when you know u₁ and d. The second is extremely handy when you already know the last term uₙ. Understanding when to use each saves time in exams.

第一个公式在已知 u₁ 和 d 时直接使用。第二个公式在已知末项 uₙ 时特别方便。明白何时使用哪个公式能在考试中节省时间。

158 = 3160. Alternatively, u₄₀ = 1 + 39×4 = 157, so S₄₀ = 40/2 (1 + 157) = 20×158 = 3160.

例如，求数列 1, 5, 9, ... 前 40 项的和。这里 u₁ = 1，d = 4，n = 40。S₄₀ = 40/2 (2×1 + 39×4) = 20 × (2 + 156) = 20 × 158 = 3160。或者 u₄₀ = 1 + 39×4 = 157，S₄₀ = 40/2 (1 + 157) = 20×158 = 3160。

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6. Arithmetic Mean | 等差中项

(a + b)/2. This idea extends to inserting multiple arithmetic means between two extremes.

两个数 a 和 b 的等差中项是 (a + b)/2。在等差数列中，如果 a, m, b 是连续三项，则 m − a = b − m，由此推出 m = (a + b)/2。这一概念可以扩展到在两数之间插入多个等差中项。

k arithmetic means between p and q, imagine a sequence with k + 2 terms where u₁ = p and u_k+2 = q. Calculate the common difference d = (q − p)/(k + 1), then generate the means.

在 p 和 q 之间插入 k 个等差中项，相当于构造一个有 k + 2 项的数列，其中 u₁ = p，u_k+2 = q。先计算公差 d = (q − p)/(k + 1)，然后生成各个中项。

so u₁ = 6, u₆ = 26. Then u₆ = 6 + 5d = 26 → d = 4. The inserted terms are 10, 14, 18, 22.

例子：在 6 和 26 之间插入 4 个等差中项。总项数为 6，u₁ = 6，u₆ = 26。则 u₆ = 6 + 5d = 26 → d = 4。插入的项为 10, 14, 18, 22。

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7. Solving Word Problems with Arithmetic Sequences | 等差数列解应用题

Arithmetic sequences model real-life situations where a quantity changes by a fixed amount each period. Common examples include simple interest growth, uniform increase in production, stacks of objects, and seating rows in an auditorium. The key is to extract u₁, d, and n from the scenario.

等差数列模拟现实中每个时期按固定量变化的情形。常见例子包括单利增长、产量均匀增加、物体堆叠以及礼堂座位排。关键是从情境中提取出 u₁、d 和 n。

ow, 19 in the second, 23 in the third, and so on. How many seats are there in the 12th row? How many seats in total if there are 12 rows? 15 + 11×4 = 59 seats. Total seats S₁₂ = 12/2 (2×15 + 11×4) = 6 × (30 + 44) = 6 × 74 = 444.

问题：某剧院第一排有 15 个座位，第二排有 19 个，第三排有 23 个，依此类推。第 12 排有多少座位？若共有 12 排，总座位数是多少？这里 u₁ = 15，d = 4。u₁₂ = 15 + 11×4 = 59 个座位。总座位数 S₁₂ = 12/2 (2×15 + 11×4) = 6 × (30 + 44) = 6 × 74 = 444。

Always include units in the final answer and interpret the result in context. Word problems are excellent for checking understanding of both nth term and sum formulas.

最终答案务必带上单位，并在情境中解释结果。应用题是检验对通项公式和求和公式理解的好方式。

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8. Relationship with Linear Functions | 与线性函数的关系

when n = 0. This connection means that arithmetic sequences are discrete linear functions.

通项 uₙ = u₁ + (n − 1)d 可改写为 uₙ = d n + (u₁ − d)，这是一个 n 的线性函数。公差 d 扮演斜率的角色，(u₁ − d) 是 n = 0 时的截距。这一关联意味着等差数列是离散的线性函数。

ch tells us that arithmetic series sum to a quadratic expression in n. These relationships are particularly useful when moving between sequences, graphs, and algebra in the IB course.

将 uₙ 对 n 作图得到共线的点。和 Sₙ 是二次函数：Sₙ = (d/2)n² + (u₁ − d/2)n，这表明等差级数总和是 n 的二次式。在IB课程中，当你在数列、图像和代数之间切换时，这些关系尤其有用。

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9. Summation Notation and Arithmetic Series | 求和符号与等差级数

frequently test the ability to interpret and evaluate expressions given in sigma form, and to express a sum in sigma notation.

等差级数是等差数列各项之和。用西格玛记号可写作 Σ (u₁ + (k − 1)d)，k 从 1 到 n。IB考试经常考查解释和计算西格玛形式给出的表达式的能力，以及将和式表达为西格玛记号。

rm u₁ = 3×1 − 2 = 1, d = 3, n = 20. The sum S₂₀ = 20/2 (2×1 + 19×3) = 10 × (2 + 57) = 590.

例子：计算 Σ (3k − 2)，k 从 1 到 20。这表示首项 u₁ = 3×1 − 2 = 1，公差 d = 3，项数 n = 20 的等差级数。和 S₂₀ = 20/2 (2×1 + 19×3) = 10 × (2 + 57) = 590。

Be comfortable changing between explicit sum formulas and sigma notation, as well as finding n given a sum. This skill is essential for later topics like limits and definite integrals with series.

要能自如地在显式和与西格玛记号之间切换，以及给定和求 n。这项技能对后续学习极限和用级数求定积分等主题至关重要。

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10. Proof of Sum Formula | 求和公式推导

quals u₁ + uₙ. There are n such pairs, so 2Sₙ = n(u₁ + uₙ), and therefore Sₙ = n/2 (u₁ + uₙ).

IB试卷1常要求给出等差求和公式的经典证明。写出 Sₙ = u₁ + u₂ + ... + uₙ。将顺序颠倒得 Sₙ = uₙ + uₙ₋₁ + ... + u₁。把这两式逐项相加得 2Sₙ = (u₁ + uₙ) + (u₂ + uₙ₋₁) + ... + (uₙ + u₁)。由于是等差数列，括号中的每一对都等于 u₁ + uₙ。共有 n 对，所以 2Sₙ = n(u₁ + uₙ)，因此 Sₙ = n/2 (u₁ + uₙ)。

This elegant proof uses only the definition of an arithmetic progression. It not only justifies the formula but also deepens your understanding of why the average of the first and last term is multiplied by n. Knowing this proof is a syllabus requirement.

这个优雅的证明只用到了等差数列的定义。它不仅证明了公式的合理性，还加深了你对为何用首末项平均值乘以 n 的理解。掌握这个证明是课程大纲的要求。

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11. Applications in Real Life | 实际应用

Arithmetic sequences model any phenomenon with a constant rate of change. Examples include simple interest (interest calculated always on the original principal), straight-line depreciation of assets, stacking pipes or logs, salary increments by a fixed amount, and even the flight of stairs if each step has the same height and depth. In IB, these contexts appear in internal assessments and exam questions.

等差数列可对任何具有恒定变化率的现象建模。例子包括单利（利息始终按原始本金计算）、资产的直线折旧、堆叠管子或圆木、按固定金额增长的

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