IB Mathematics: Arithmetic Sequences – Formulas and Applications | IB数学：等差数列公式与应用

📚 IB Mathematics: Arithmetic Sequences – Formulas and Applications | IB数学：等差数列公式与应用

Arithmetic sequences are one of the fundamental topics in IB Mathematics. They appear in both Analysis and Approaches (AA) and Applications and Interpretation (AI), and they form the basis for many financial and scientific models. This article reviews the key formulas, derivations, and typical applications, with exam-focused tips.

等差数列是IB数学的基础主题之一，出现在分析与方法（AA）和应用与解释（AI）两门课程中，也是许多金融与科学模型的基础。本文将复习核心公式、推导和典型应用，并提供针对考点的提示。

1. What is an Arithmetic Sequence? | 什么是等差数列？

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This constant is called the common difference and is usually denoted by d.

等差数列是指相邻两项之差保持恒定的数列。这个恒定的差值称为公差，通常用 d 表示。

For example, the sequence 2, 5, 8, 11, … is arithmetic because each term is obtained by adding 3 to the previous term, so d = 3.

例如，数列 2, 5, 8, 11, … 是等差数列，因为每一项都是在前一项基础上加 3，因此 d = 3。
Similarly, the sequence 20, 16, 12, 8, … is arithmetic with d = −4, because we subtract 4 each time.

类似地，数列 20, 16, 12, 8, … 是公差为 −4 的等差数列，因为我们每次都减去 4。
The common difference can be zero, positive, or negative. If d > 0, the sequence increases; if d < 0, it decreases; if d = 0, all terms are equal.

公差可以是零、正数或负数。若 d > 0，数列递增；若 d < 0，数列递减；若 d = 0，则所有项相等。

2. The General Term Formula | 通项公式

The n-th term of an arithmetic sequence can be written using the first term u₁ and the common difference d:

等差数列的第 n 项可以用首项 u₁ 和公差 d 表示：

uₙ = u₁ + (n − 1)d

Here, uₙ is the n-th term, u₁ is the first term, n is the position of the term (a positive integer), and d is the common difference.

其中，uₙ 是第 n 项，u₁ 是首项，n 是项的位置（正整数），d 是公差。

To find the 10th term of the sequence 7, 4, 1, −2, …, write u₁ = 7 and d = −3. Then u₁₀ = 7 + (10 − 1)(−3) = 7 − 27 = −20.

要求数列 7, 4, 1, −2, … 的第 10 项，可写出 u₁ = 7，d = −3。因此 u₁₀ = 7 + (10 − 1)(−3) = 7 − 27 = −20。
This formula is on the IB formula booklet, but you must know how to use it quickly in both paper 1 and paper 2.

该公式在IB公式手册中给出，但你必须能够在试卷一和试卷二中快速灵活使用。

3. Finding the First Term and Common Difference | 求首项与公差

Sometimes you are given two terms of an arithmetic sequence and asked to find u₁ and d. This is solved by setting up two equations and solving them simultaneously.

有时题目会给出等差数列中的两项，要求求出 u₁ 和 d。这需要建立两个方程并联立求解。

For example, suppose u₄ = 10 and u₇ = 19. Using the general term formula:

例如，已知 u₄ = 10，u₇ = 19。利用通项公式：

u₁ + 3d = 10

u₁ + 6d = 19

Subtract the first equation from the second: 3d = 9, so d = 3. Substitute back to get u₁ = 1. Thus uₙ = 1 + 3(n − 1) = 3n − 2.

用第二式减去第一式：3d = 9，所以 d = 3。代回可得 u₁ = 1。因此 uₙ = 1 + 3(n − 1) = 3n − 2。

Always check your answer by plugging n = 4 and n = 7 into the formula you found.

一定要将 n = 4 和 n = 7 代入求出的公式进行验证。

4. The Sum of n Terms | 前 n 项和公式

The sum of the first n terms of an arithmetic sequence, denoted Sₙ, can be computed in two equivalent ways.

等差数列的前 n 项和，记作 Sₙ，有两种等价的计算方式。

Sₙ = n/2 × (u₁ + uₙ)

Sₙ = n/2 × [2u₁ + (n − 1)d]

The first formula is useful when you know the first term and the last term uₙ. The second formula is useful when you know the common difference d instead of the last term.

第一个公式在已知首项和末项 uₙ 时使用；第二个公式在已知公差 d 而不是末项时使用。

Example: Find the sum of the first 20 terms of the sequence 3, 7, 11, 15, … Here u₁ = 3 and d = 4. Using the second formula: S₂₀ = 20/2 × [2 × 3 + (20 − 1) × 4] = 10 × (6 + 76) = 820.

例：求数列 3, 7, 11, 15, … 前 20 项的和。这里 u₁ = 3，d = 4。使用第二个公式：S₂₀ = 20/2 × [2 × 3 + (20 − 1) × 4] = 10 × (6 + 76) = 820。
The sum formula can be derived by writing the sum forwards and backwards and then adding the two lines; this is known as Gauss’s method.

求和公式可以通过将和式正写与倒写，再把两式相加的方法推导出来，这就是高斯求和法。

5. Equivalence of the Two Sum Formulas | 两个求和公式的等价性

The two sum formulas are equivalent because the last term uₙ is equal to u₁ + (n − 1)d. If you substitute this expression into Sₙ = n/2 × (u₁ + uₙ), you get the second formula.

两个求和公式是等价的，因为末项 uₙ = u₁ + (n − 1)d。将这一表达式代入 Sₙ = n/2 × (u₁ + uₙ)，即可得到第二个公式。

Use Sₙ = n/2 × (u₁ + uₙ) when the last term uₙ is given or easy to find.

当末项 uₙ 已知或容易求出时，使用 Sₙ = n/2 × (u₁ + uₙ)。
Use Sₙ = n/2 × [2u₁ + (n − 1)d] when d is known and the last term is not directly required.

当已知公差 d 且不需要直接求出末项时，使用 Sₙ = n/2 × [2u₁ + (n − 1)d]。
On IB exams, showing clearly which formula you choose and why can earn method marks even if the final arithmetic is incorrect.

在IB考试中，清晰说明你选择了哪个公式以及原因，即使最后计算有误，也可能获得方法分。

6. Solving for n When the Sum Is Given | 已知和求项数 n

In some problems, you are given the sum Sₙ and asked to find n. This often leads to a quadratic equation.

在有些题目中，已知前 n 项和 Sₙ，要求 n。这通常会转化为一元二次方程。

Consider the arithmetic sequence 5, 9, 13, 17, … How many terms must be added to make the sum 945?

考虑等差数列 5, 9, 13, 17, … 求需要加多少项，才能使和为 945？

Here u₁ = 5 and d = 4. The sum formula gives:

这里 u₁ = 5，d = 4。代入求和公式：

Sₙ = n/2 × [2 × 5 + (n − 1) × 4] = n(2n + 3)

Set this equal to 945: n(2n + 3) = 945, so 2n² + 3n − 945 = 0. Factoring or using the quadratic formula gives n = 21 or n = −22.5. Since n must be a positive integer, n = 21.

令其等于 945：n(2n + 3) = 945，即 2n² + 3n − 945 = 0。因式分解或使用求根公式可得 n = 21 或 n = −22.5。由于 n 必须是正整数，因此 n = 21。

Always reject non-integer or negative solutions for n in sequence problems.

在数列问题中，一定要舍去负数或非整数的 n 值。

7. Applications in Finance | 金融中的应用

Arithmetic sequences model situations with constant linear growth, such as simple interest, fixed annual salary increases, or constant monthly rent changes.

等差数列用于建模恒定线性增长的情境，例如单利、固定年度加薪或每月租金恒定变化等。

If you invest $1000 at 5% simple interest per year, the amount after n years is given by Aₙ = 1000 + 50n, which is arithmetic with d = 50.

如果你以每年 5% 的单利投资1000美元，n 年后的金额为 Aₙ = 1000 + 50n，这是一个公差为 50 的等差数列。
In IB Applications and Interpretation, loan repayments with a fixed “capital repayment” component follow an arithmetic pattern, while compound interest follows a geometric pattern. Distinguishing the two is crucial.

在IB应用与解释课程中，固定本金偿还部分的贷款还款遵循等差模式，而复利遵循等比模式。区分二者至关重要。
For a salary with an annual increase of $2000, the salaries in successive years form an arithmetic sequence with common difference $2000.

对于每年涨薪2000美元的工资，连续年份的工资形成一个公差为2000美元的等差数列。

8. Applications in Physics and Science | 物理与科学中的应用

Arithmetic sequences also appear in kinematics and biology. A classic example is the distance covered in each second during constant acceleration.

等差数列还出现在运动学和生物学中。一个经典例子是匀加速运动中每一秒内经过的距离。

For an object falling from rest under gravity, the distances travelled in the 1st, 2nd, 3rd, … seconds form an arithmetic sequence. With g ≈ 10 m/s², the distances are approximately 5 m, 15 m, 25 m, 35 m, … so d = 10 m.

物体从静止开始下落时，第1秒、第2秒、第3秒……内通过的距离构成等差数列。取 g ≈ 10 m/s²，距离约为 5 m、15 m、25 m、35 m……因此 d = 10 m。
In biology, a population growing by a fixed number of individuals per year can be modelled with an arithmetic sequence, though many real populations grow geometrically.

在生物学中，如果种群每年增加固定个体数，可以用等差数列建模；不过许多真实种群是按等比方式增长的。

9. Common Pitfalls and Exam Tips | 常见易错点与考试提示

Many IB students lose marks by making small definitional errors. Avoid these common mistakes:

许多IB学生因为定义上的小错误而失分。请避免以下常见错误：

Using n instead of n − 1 in the general term. The term u₁ corresponds to n = 1, so the multiplier is n − 1, not n.

在通项公式中把 n 写成 n−1 的正倍数搞混。u₁ 对应 n = 1，因此乘数是 n − 1，而不是 n。
Calculating the common difference as u₁ − u₂. Remember d = u₂ − u₁.

把公差算成 u₁ − u₂。记住 d = u₂ − u₁。
Confusing Sₙ with uₙ. Sₙ is the sum of the first n terms, while uₙ is the n-th term alone.

混淆 Sₙ 与 uₙ。Sₙ 是前 n 项的和，而 uₙ 是第 n 项本身。
When using the formula Sₙ = n/2 × (u₁ + uₙ), remember to find uₙ first if it is not explicitly given.

使用 Sₙ = n/2 × (u₁ + uₙ) 时，如果末项未直接给出，要先求出 uₙ。

10. Problem-Solving Strategies | 解题策略

A systematic approach will help you earn full marks on arithmetic sequence questions.

采用系统化方法能帮助你在等差数列题目中获得满分。

Step 1: Identify which quantities are given — u₁, d, n, uₙ, or Sₙ.

第一步：确认题目给出了哪些量——u₁、d、n、uₙ 或 Sₙ。
Step 2: Write the relevant formula before plugging in numbers. This shows the examiner your method.

第二步：在代入数值前写出相关公式。这向考官展示你的解题思路。
Step 3: If a word problem seems unfamiliar, list the first few terms to confirm it is arithmetic and find d.

第三步：如果应用题看起来不熟悉，列出前几项来确认是否为等差数列并求 d。
Step 4: For final answers, remember the required units and whether n must be an integer.

第四步：对于最终答案，注意单位，以及 n 是否必须为整数。

11. Conclusion | 总结

Arithmetic sequences are a compact but powerful tool in IB Mathematics. Mastering the definitions, the general term formula, and the two sum formulas will let you handle a wide variety of problems in pure mathematics and real-world contexts.

等差数列是IB数学中简洁而强大的工具。掌握定义、通项公式和两个求和公式，将使你能够应对纯数学和现实情境中的各种问题。

Practice identifying arithmetic patterns quickly, and always check your substitutions carefully. With regular revision, arithmetic sequences will become one of the most reliable sources of marks on your IB exam.

练习快速识别等差模式，并始终仔细检查代入过程。通过定期复习，等差数列将成为你IB考试中最稳定的得分点之一。

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IB Mathematics: Arithmetic Sequences – Formulas and Applications | IB数学：等差数列公式与应用