Exercise 1A.1: Arithmetic Sequences and Series | 练习 1A.1：等差数列与级数

📚 Exercise 1A.1: Arithmetic Sequences and Series | 练习 1A.1：等差数列与级数

Welcome to Exercise 1A.1, a foundational topic in the IB Mathematics: Analysis and Approaches course. Here we explore arithmetic sequences and series, building the skills to identify patterns, find any term, and calculate sums efficiently. This exercise lays the groundwork for more advanced algebraic reasoning and modelling.

欢迎来到练习 1A.1，这是 IB 数学：分析与方法课程中的基础课题。我们将在此探索等差数列与等差级数，培养识别模式、求出任意项以及高效求和的技能。本练习为更高级的代数推理和建模奠定基础。

1. Understanding Sequences and Series | 理解数列与级数

A sequence is an ordered list of numbers following a specific rule. Each number is called a term. A series is formed when we add the terms of a sequence together. In IB Mathematics, distinguishing between the two is essential: sequences list terms, series sum them.

数列是按照特定规则排列的一列数。每个数称为一项。当我们把数列的各项相加时，就形成了级数。在 IB 数学中，区分两者至关重要：数列列出各项，而级数则对各项求和。

The terms of a sequence are usually denoted by u₁, u₂, u₃, …, uₙ, where the subscript indicates the position. A series is expressed as Sₙ = u₁ + u₂ + … + uₙ for the sum of the first n terms.

数列的各项通常表示为 u₁、u₂、u₃……uₙ，下标表示位置。级数则用 Sₙ = u₁ + u₂ + … + uₙ 表示前 n 项之和。

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2. Definition of an Arithmetic Sequence | 等差数列的定义

An arithmetic sequence is a sequence in which the difference between consecutive terms is constant. This means each term is obtained by adding the same fixed number to the previous term. The constant difference is called the common difference, denoted by d.

等差数列是相邻两项之差为常数的数列。这意味着每一项都是在前一项的基础上加上同一个固定的数而得到的。这个常数差称为公差，用 d 表示。

For example, the sequence 3, 7, 11, 15, … is arithmetic because 7 − 3 = 4, 11 − 7 = 4, and so on. Here d = 4.

例如，数列 3, 7, 11, 15……是等差数列，因为 7 − 3 = 4，11 − 7 = 4，依此类推。此处 d = 4。

A sequence with a negative common difference, such as 20, 15, 10, 5, …, is also arithmetic, with d = −5. The key feature is that the change between terms remains unchanged throughout.

公差为负数的数列，如 20, 15, 10, 5……同样是等差数列，其 d = −5。关键特征在于项与项之间的变化始终保持不变。

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3. The Common Difference | 公差

To find the common difference d, subtract any term from the term that follows it: d = u₍ₙ₊₁₎ − uₙ. This value can be positive, negative, or zero; a sequence with d = 0 is a constant sequence where all terms are equal.

要求出公差 d，用后一项减去前一项：d = u₍ₙ₊₁₎ − uₙ。该值可以是正数、负数或零；当 d = 0 时，就是一个所有项都相等的常数列。

Identifying d early is a powerful problem-solving step. Once you know the first term u₁ and d, the entire sequence is determined. Always check at least two pairs of consecutive terms to confirm the sequence is arithmetic.

尽早识别 d 是解题的有力一步。一旦知道了首项 u₁ 和 d，整个数列就确定了。务必至少检查两对相邻的项，以确认该数列是等差数列。

In IB exam questions, you may be given two non-consecutive terms and asked to find d. Use the relationship between terms: uₘ = uₙ + (m − n)d.

在 IB 考题中，可能会给出两个不相邻的项让你求 d。此时可利用项与项之间的关系：uₘ = uₙ + (m − n)d。

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4. The nth Term Formula | 第 n 项公式

For any arithmetic sequence with first term u₁ and common difference d, the nth term is given by: uₙ = u₁ + (n − 1)d. This formula allows you to jump directly to any term without listing all previous ones.

对于首项为 u₁、公差为 d 的任意等差数列，第 n 项为：uₙ = u₁ + (n − 1)d。利用此公式，你可以直接找出任何一项，而无需逐一列出前面的所有项。

Here n represents the term number, and n must be a positive integer. Note that the multiplier (n − 1) is used because the first term has zero additions of d. If u₁ = 2 and d = 3, then u₄ = 2 + (4 − 1)×3 = 11.

此处 n 代表项数，且 n 必须为正整数。注意乘数是 (n − 1)，因为首项加零次 d。若 u₁ = 2 且 d = 3，则 u₄ = 2 + (4 − 1)×3 = 11。

This linear form explains why arithmetic sequences grow steadily. The formula can also be rearranged to find n if a term value is known, making it a versatile tool for problem solving.

这一线性形式解释了为什么等差数列会稳定增长。如果已知某项的值，还可将公式变形求出 n，使其成为解题中的多面手。

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5. Finding the nth Term – Worked Examples | 求第 n 项——示例

Example 1: Given an arithmetic sequence with u₁ = 5 and d = 2, find the 20th term. Solution: u₂₀ = 5 + (20 − 1)×2 = 5 + 38 = 43. Always substitute carefully and perform the multiplication before addition.

示例1：已知等差数列 u₁ = 5，d = 2，求第 20 项。解：u₂₀ = 5 + (20 − 1)×2 = 5 + 38 = 43。务必仔细代入，先乘后加。

Example 2: The 4th term of an arithmetic sequence is 11 and the 10th term is 29. Find u₁ and d. Set up equations: u₄ = u₁ + 3d = 11 and u₁₀ = u₁ + 9d = 29. Subtracting gives 6d = 18, so d = 3. Then u₁ = 11 − 9 = 2. The sequence is 2, 5, 8, …

示例2：等差数列的第 4 项为 11，第 10 项为 29。求 u₁ 和 d。建立方程组：u₄ = u₁ + 3d = 11，u₁₀ = u₁ + 9d = 29。两式相减得 6d = 18，故 d = 3。然后 u₁ = 11 − 9 = 2。该数列为 2, 5, 8……

These examples show how to move between specific terms and the general formula. Practice with numeric and algebraic data builds the fluency needed for IB Paper 1 and Paper 2.

这些例子展示了如何在具体项与通项公式之间转换。通过数值和代数数据的练习，可以培养 IB 试卷一和试卷二所需的熟练度。

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6. Arithmetic Series – Sum of Terms | 等差级数——各项之和

An arithmetic series is the sum of the terms of an arithmetic sequence. The sum of the first n terms is denoted Sₙ. For example, if the sequence is 3, 5, 7, 9, then S₄ = 3 + 5 + 7 + 9 = 24.

等差级数是等差数列各项之和。前 n 项之和记为 Sₙ。例如，若数列为 3, 5, 7, 9，则 S₄ = 3 + 5 + 7 + 9 = 24。

There are two commonly used formulas for Sₙ: Sₙ = n/2 (u₁ + uₙ) and Sₙ = n/2 [2u₁ + (n − 1)d]. The first requires the first and last terms; the second uses the first term and common difference.

Sₙ 有两个常用公式：Sₙ = n/2 (u₁ + uₙ) 和 Sₙ = n/2 [2u₁ + (n − 1)d]。第一个公式需要首项和末项；第二个公式则使用首项和公差。

Choosing the right form depends on the given information. If you are given or can quickly find uₙ, the first is simpler. If you only have u₁ and d, the second formula is more direct.

选择哪种形式取决于所给的信息。如果已知或能快速求出 uₙ，用第一个公式更简单。如果只有 u₁ 和 d，那么第二个公式更直接。

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7. Sum Formula Derivation | 求和公式推导

The sum formula arises from a clever pairing trick. Write Sₙ forwards and backwards: Sₙ = u₁ + u₂ + … + uₙ, and Sₙ = uₙ + u₍ₙ₋₁₎ + … + u₁. Adding these two equations pairs terms that each sum to u₁ + uₙ.

求和公式源于一种巧妙的配对技巧。将 Sₙ 正写和倒写：Sₙ = u₁ + u₂ + … + uₙ，以及 Sₙ = uₙ + u₍ₙ₋₁₎ + … + u₁。两式相加后，每对项的和均为 u₁ + uₙ。

There are n such pairs, so 2Sₙ = n(u₁ + uₙ). Dividing by 2 gives Sₙ = n/2 (u₁ + uₙ). This elegant derivation highlights why the formula works and is worth understanding deeply.

共有 n 对这样的组合，因此 2Sₙ = n(u₁ + uₙ)。两边同除以 2 即得 Sₙ = n/2 (u₁ + uₙ)。这一优雅的推导揭示了公式为何成立，值得深入理解。

To obtain the alternative form, substitute uₙ = u₁ + (n − 1)d into the expression: Sₙ = n/2 [u₁ + u₁ + (n − 1)d] = n/2 [2u₁ + (n − 1)d]. This connects the sum directly to u₁ and d.

要得到另一种形式，将 uₙ = u₁ + (n − 1)d 代入表达式：Sₙ = n/2 [u₁ + u₁ + (n − 1)d] = n/2 [2u₁ + (n − 1)d]。这就将和与 u₁ 和 d 直接联系起来了。

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8. Using the Sum Formula | 求和公式的应用

To compute Sₙ, identify u₁, n, and either uₙ or d. For instance, find the sum of the first 30 terms of the sequence 7, 10, 13, … Here u₁ = 7, d = 3. Using Sₙ = n/2 [2u₁ + (n − 1)d]: S₃₀ = 30/2 [2×7 + (30 − 1)×3] = 15 [14 + 87] = 15 × 101 = 1515.

要计算 Sₙ，需确定 u₁、n 以及 uₙ 或 d。例如，求数列 7, 10, 13……前 30 项的和。这里 u₁ = 7，d = 3。使用 Sₙ = n/2 [2u₁ + (n − 1)d]：S₃₀ = 30/2 [2×7 + (30 − 1)×3] = 15 [14 + 87] = 15 × 101 = 1515。

Sometimes you are asked to find n given Sₙ. Set up the equation and solve the resulting quadratic. For example, if u₁ = 4, d = 2, and Sₙ = 450, then 450 = n/2 [8 + (n − 1)×2] simplifies to n² + 3n − 450 = 0, giving n = 18 (positive root).

有时题目会要求根据 Sₙ 求 n。建立方程并求解所得的一元二次方程。例如，若 u₁ = 4，d = 2，且 Sₙ = 450，则 450 = n/2 [8 + (n − 1)×2] 可化简为 n² + 3n − 450 = 0，解得 n = 18（正根）。

Always check that your answer is a positive integer. In context, a non-integer n would be rejected because the number of terms must be a whole number.

务必检查答案是否为正整数。在语境中，非整数的 n 将被舍去，因为项数必须是整数。

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9. Applications of Arithmetic Sequences | 等差数列的应用

Arithmetic sequences model many real-world situations. Common examples include simple interest accumulation, the depreciation of assets by a fixed amount, or the number of seats in rows of an auditorium increasing by a constant number.

等差数列可以模拟许多现实世界中的情形。常见例子包括单利累积、资产按固定金额折旧，或者礼堂中每排座位数按常数递增。

In a word problem, extract u₁, d, and n from the description. For instance, if a person saves $50 the first month and increases savings by $10 each month, then u₁ = 50, d = 10. The amount saved in the 12th month is u₁₂ = 50 + 11×10 = 160 dollars.

在应用题中，应从描述中提取 u₁、d 和 n。例如，若某人第一个月存 50 美元，之后每月增加储蓄 10 美元，则 u₁ = 50，d = 10。第 12 个月的储蓄额为 u₁₂ = 50 + 11×10 = 160 美元。

The total savings after 12 months is the sum: S₁₂ = 12/2 (50 + 160) = 6 × 210 = 1260 dollars. Recognising these patterns helps turn a word problem into manageable arithmetic.

12 个月后的总储蓄额为和：S₁₂ = 12/2 (50 + 160) = 6 × 210 = 1260 美元。识别这些模式有助于将文字题转化为易于计算的算术问题。

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10. Problem-Solving Strategies | 解题策略

When tackling an arithmetic sequence problem, start by listing all given information and converting it into u₁, d, n, uₙ, or Sₙ. Decide which variable you need to find. Then select the appropriate formula.

在解决等差数列问题时，首先列出所有已知信息，并将其转化为 u₁、d、n、uₙ 或 Sₙ。确定需要求哪个变量，然后选择适当的公式。

If two quantities are unknown, use two pieces of information to form simultaneous equations. This is a frequent scenario in IB questions where you are given two terms or a term and a sum. Solve the system using substitution or elimination.

如果有两个未知量，则利用两条信息建立联立方程组。这在 IB 考题中很常见，你会得到两项，或者一项与一个和。使用代入法或消元法解方程组。

After finding a solution, verify it by plugging back into the original conditions. A quick sanity check can catch sign errors or misread values. For series sums, estimate the ballpark figure mentally to confirm the order of magnitude.

求出解后，代回原始条件进行验证。快速的合理性检查能发现符号错误或看错数值。计算级数和时，可以在脑中估算大致数值，以确认数量级是否正确。

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11. Common Mistakes to Avoid | 常见错误

A common error is confusing (n − 1) and n in the nth term formula. Remember that u₁ corresponds to n = 1, so the factor should be (n − 1) to avoid overcounting the common difference.

一个常见错误是混淆第 n 项公式中的 (n − 1) 和 n。请记住，u₁ 对应 n = 1，因此乘数应为 (n − 1)，以避免多算公差。

Another mistake is using the wrong sum formula when uₙ is not the last term given. Always ensure the term you plug into Sₙ = n/2 (u₁ + uₙ) is indeed the nth term. If only u₁ and d are known, stick to the second sum formula to be safe.

另一个错误是当 uₙ 不是所给的末项时，误用了求和公式。务必确保代入 Sₙ = n/2 (u₁ + uₙ) 的项确实是第 n 项。如果只知道 u₁ 和 d，为稳妥起见，应坚持使用第二个求和公式。

Watch out for units and context in applied problems. Forgetting to convert units or misinterpreting the starting value can lead to a completely wrong model. Always read the problem statement twice.

留意应用题中的单位和语境。忘记换算单位或误解初始值，都可能导致完全错误的模型。一定要将题目读两遍。

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12. Practice Questions and Summary | 练习题与总结

To master Exercise 1A.1, practice a variety of problems: finding the nth term, computing sums, solving for n or d, and tackling word problems. Try these: (1) An arithmetic sequence has u₅ = 22 and u₉ = 42. Find u₁ and d. (2) Find the sum of the first 40 terms of the sequence −5, −2, 1, 4, … (3) A loan is repaid with monthly payments starting at $200 and increasing by $15 each month. What is the total paid over 24 months?

要掌握练习 1A.1，需要练习各种类型的问题：求第 n 项、计算和、解出 n 或 d，以及解决应用题。不妨尝试以下题目：(1) 等差数列满足 u₅ = 22，u₉ = 42，求 u₁ 和 d。(2) 求数列 −5, −2, 1, 4……前 40 项的和。(3) 一笔贷款每月偿还额从 200 美元开始，每月增加 15 美元。24 个月的总还款额是多少？

Summary: An arithmetic sequence has a constant common difference d. Its nth term is uₙ = u₁ + (n − 1)d. The sum of the first n terms is Sₙ = n/2 (u₁ + uₙ) or Sₙ = n/2 [2u₁ + (n − 1)d]. These tools are fundamental for progress in IB mathematics and beyond.

总结：等差数列具有恒定的公差 d。其第 n 项为 uₙ = u₁ + (n − 1)d。前 n 项和为 Sₙ = n/2 (u₁ + uₙ) 或 Sₙ = n/2 [2u₁ + (n − 1)d]。这些工具是 IB 数学及更远学习中打基础的关键。

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