Exercise 1B: Arithmetic Sequences and Series | 练习1B：等差数列与求和

📚 Exercise 1B: Arithmetic Sequences and Series | 练习1B：等差数列与求和

Exercise 1B is designed to strengthen your understanding of arithmetic sequences and series, two foundational topics in the IB Mathematics curriculum. Whether you are following the Analysis & Approaches or Applications & Interpretation pathway, the ability to recognise linear patterns, apply the general term formula, and compute finite sums is essential. This article guides you through the core concepts and worked examples that mirror the style of questions found in typical IB practice sets. Each section pairs an English explanation with a Chinese translation, ensuring clarity for bilingual learners.

练习1B旨在加深你对等差数列与求和的理解，这两个主题是IB数学课程的基础。无论你学习的是分析与方法还是应用与解释方向，识别线性模式、应用通项公式以及计算有限项之和都是必备技能。本文通过核心概念和典型例题，模拟IB练习题中常见的题型。每小节均提供英文讲解与中文翻译，帮助双语学习者清晰地掌握知识点。

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1. Understanding the Arithmetic Sequence | 理解等差数列

An arithmetic sequence is a list of numbers where the difference between consecutive terms remains constant. This fixed difference is called the common difference, denoted by d. For example, the sequence 7, 10, 13, 16, … has d = 3 because each term increases by 3. Recognising this constant change is the first step in solving IB problems. Arithmetic sequences appear in contexts ranging from simple savings plans to linear growth models, so a solid grasp of the definition is vital.

等差数列是一组数字，其中连续两项之间的差保持恒定。这个固定的差值称为公差，用 d 表示。例如，数列 7, 10, 13, 16, … 的公差 d = 3，因为每一项增加 3。识别这种恒定变化是解决IB问题的第一步。等差数列出现在从简单储蓄计划到线性增长模型的各种情境中，因此牢固掌握定义至关重要。

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2. The General Term Formula | 通项公式

The nth term of an arithmetic sequence can be found using the formula uₙ = u₁ + (n − 1)d, where u₁ is the first term and d is the common difference. This linear relationship allows you to calculate any term directly without listing all previous ones. In IB exercises, you are often asked to find uₙ given u₁ and d, or to reverse-engineer the sequence from two known terms. Mastery of this formula underpins success in the entire topic.

等差数列的第 n 项可以用公式 uₙ = u₁ + (n − 1)d 求得，其中 u₁ 为首项，d 为公差。这种线性关系使得你可以直接计算任意一项，而无需列出前面所有项。在IB练习题中，经常会要求根据 u₁ 和 d 求 uₙ，或者根据两个已知项反推数列。掌握这一公式是在整个主题中取得成功的基础。

uₙ = u₁ + (n − 1)d

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3. Finding the nth Term | 求第n项

Consider a typical Exercise 1B question: ‘Find the 20th term of the arithmetic sequence whose first term is 12 and common difference is 5.’ Applying the formula, we substitute u₁ = 12, d = 5, and n = 20 to obtain u₂₀ = 12 + (20 − 1) × 5 = 12 + 95 = 107. Always show your substitution clearly – IB examiners value logical working as much as the final answer. When the common difference is negative, the approach remains identical; simply ensure you include the sign.

考虑一道典型的练习1B题：“求首项为12、公差为5的等差数列的第20项。”代入公式，u₁ = 12, d = 5, n = 20，得 u₂₀ = 12 + (20 − 1) × 5 = 12 + 95 = 107。务必清晰地展示代入过程——IB考官既看重最终答案，也看重逻辑推理步骤。当公差为负时，方法完全相同，只需注意带入符号即可。

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4. Determining the Number of Terms | 确定项数

Sometimes you are given the first term, the common difference, and a target value, and must find which term equals that value. For example, an arithmetic sequence starts at 45 and decreases by 3 each time. Which term equals 0? Set uₙ = 0, u₁ = 45, d = −3, and solve 0 = 45 + (n − 1)(−3). This yields (n − 1)(−3) = −45, so n − 1 = 15, giving n = 16. The 16th term is zero. This type of question tests your equation-solving skills within a sequence context.

有时题目会给出首项、公差和一个目标值，要求找出哪一项等于该值。例如，一等差数列从45开始，每次减3，问哪一项等于0？设 uₙ = 0, u₁ = 45, d = −3，解方程 0 = 45 + (n − 1)(−3)。得到 (n − 1)(−3) = −45，因此 n − 1 = 15，n = 16。第16项等于0。这类问题考查在数列背景下解方程的能力。

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5. Arithmetic Series and Sum Formula | 等差数列求和公式

When the terms of an arithmetic sequence are added together, we obtain an arithmetic series. The sum of the first n terms, denoted Sₙ, can be calculated using two equivalent formulas: Sₙ = n/2 [2u₁ + (n − 1)d] or Sₙ = n/2 (u₁ + uₙ). The first is useful when you know the first term and common difference, while the second is handy when the last term uₙ is already known. IB problems frequently require you to choose the most efficient form based on the given information.

当等差数列的各项相加时，就得到等差级数。前 n 项之和记作 Sₙ，可以用两个等价公式计算：Sₙ = n/2 [2u₁ + (n − 1)d] 或 Sₙ = n/2 (u₁ + uₙ)。当已知首项和公差时，第一个公式更方便；当已求得末项 uₙ 时，第二个公式更快捷。IB题目经常要求你根据所给信息选择最有效的公式形式。

Sₙ = n/2 [2u₁ + (n − 1)d] = n/2 (u₁ + uₙ)

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6. Summation of Finite Arithmetic Series | 有限等差数列求和

Let us practise a summation question: ‘Find the sum of the first 30 terms of the arithmetic sequence 5, 9, 13, 17, …’ Here u₁ = 5, d = 4, and n = 30. Using Sₙ = n/2 [2u₁ + (n − 1)d], we get S₃₀ = 30/2 [2×5 + (30 − 1)×4] = 15 × [10 + 116] = 15 × 126 = 1890. Always check that you have used the correct n – a common mistake is to plug in n − 1 instead of n. Writing the values in a small table before substituting can reduce errors.

我们练习一道求和题：“求等差数列 5, 9, 13, 17, … 的前30项之和。”这里 u₁ = 5, d = 4, n = 30。使用 Sₙ = n/2 [2u₁ + (n − 1)d]，得到 S₃₀ = 30/2 [2×5 + (30 − 1)×4] = 15 × [10 + 116] = 15 × 126 = 1890。务必检查使用的是正确的 n——一个常见错误是代入 n − 1 而不是 n。在代入前将数值列成小表格可以减少失误。

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7. Using the Sum to Find a Term | 利用和求项

In some IB-style exercises, you are given Sₙ and need to determine the value of a specific term. For instance, if Sₙ = 3n² + n, the nth term uₙ can be found using the relationship uₙ = Sₙ − Sₙ₋₁. This technique is powerful because it works even when the sequence is not explicitly given. Calculate Sₙ₋₁ by replacing n with (n − 1): Sₙ₋₁ = 3(n − 1)² + (n − 1) = 3n² − 6n + 3 + n − 1 = 3n² − 5n + 2. Then uₙ = (3n² + n) − (3n² − 5n + 2) = 6n − 2. The sequence is arithmetic because uₙ is linear in n.

在一些IB风格的练习题中，题目给出 Sₙ 并要求你确定某一项的值。例如，若 Sₙ = 3n² + n，则第 n 项 uₙ 可以利用关系式 uₙ = Sₙ − Sₙ₋₁ 求得。这种方法非常有效，即便数列没有明确给出也可使用。将 n 替换为 (n − 1) 计算 Sₙ₋₁：Sₙ₋₁ = 3(n − 1)² + (n − 1) = 3n² − 6n + 3 + n − 1 = 3n² − 5n + 2。然后 uₙ = (3n² + n) − (3n² − 5n + 2) = 6n − 2。该数列是等差数列，因为 uₙ 关于 n 是线性的。

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8. Solving for First Term and Common Difference | 求首项和公差

Questions that provide two non-consecutive terms, such as ‘the 4th term is 17 and the 10th term is 41’, require setting up simultaneous equations. Write u₄ = u₁ + 3d = 17 and u₁₀ = u₁ + 9d = 41. Subtract the first equation from the second to eliminate u₁: (u₁ + 9d) − (u₁ + 3d) = 41 − 17 → 6d = 24 → d = 4. Substitute back to find u₁ = 17 − 3×4 = 5. Always verify your solution by checking that the terms indeed produce the correct values – this habit saves time in exams.

题目给出两个非连续项，如“第4项是17，第10项是41”，需要建立联立方程组。写出 u₄ = u₁ + 3d = 17 和 u₁₀ = u₁ + 9d = 41。将第二式减去第一式以消去 u₁：(u₁ + 9d) − (u₁ + 3d) = 41 − 17 → 6d = 24 → d = 4。代回求出 u₁ = 17 − 3×4 = 5。务必验证所求出的项确实给出正确数值——这个习惯能在考试中节省时间。

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9. Application: Seating Capacity Problem | 应用题：座位容量问题

Application problems often appear at the end of Exercise 1B. Consider a concert hall where the first row has 24 seats, and each subsequent row has 3 additional seats. There are 20 rows in total. To find the total number of seats, recognise that the row capacities form an arithmetic sequence with u₁ = 24, d = 3, and n = 20. First find the last term: u₂₀ = 24 + 19×3 = 24 + 57 = 81. Then use S₂₀ = 20/2 (24 + 81) = 10 × 105 = 1050. The hall contains 1050 seats. Always define your variables clearly in context – this is an IB criterion.

应用题常出现在练习1B的末尾。设想一个音乐厅，第一排有24个座位，之后每排增加3个座位，总共20排。求总座位数。识别出每排座位数构成等差数列，u₁ = 24, d = 3, n = 20。先求末项：u₂₀ = 24 + 19×3 = 24 + 57 = 81。然后用 S₂₀ = 20/2 (24 + 81) = 10 × 105 = 1050。该厅共有1050个座位。务必在情境中清晰定义变量——这是IB的评分标准之一。

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10. The Sum Formula Given in Terms of n | 已知Sₙ求通项

Another common Exercise 1B problem provides the sum of the first n terms as a quadratic expression and asks whether the sequence is arithmetic. Since an arithmetic series has Sₙ = n/2 [2u₁ + (n − 1)d], expanding gives a quadratic of the form (d/2)n² + (u₁ − d/2)n. If the given expression matches this pattern, the sequence is arithmetic. For example, if Sₙ = 5n² + 3n, compare coefficients: d/2 = 5 → d = 10, and u₁ − d/2 = 3 → u₁ = 3 + 5 = 8. The original sequence is 8, 18, 28, … with d = 10. This approach links algebraic manipulation with sequence properties.

练习1B中还常见一类问题：给出前 n 项之和为一个二次表达式，问数列是否为等差数列。由于等差级数的 Sₙ = n/2 [2u₁ + (n − 1)d]，展开后得到形如 (d/2)n² + (u₁ − d/2)n 的二次式。若给定表达式符合此形式，该数列即为等差数列。例如，若 Sₙ = 5n² + 3n，比较系数：d/2 = 5 → d = 10，u₁ − d/2 = 3 → u₁ = 3 + 5 = 8。原数列为 8, 18, 28, …，公差 d = 10。这种方法将代数操作与数列性质联系起来。

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