Exercise 1A.3: Arithmetic Sequences Practice | 练习1A.3：等差数列实战

📚 Exercise 1A.3: Arithmetic Sequences Practice | 练习1A.3：等差数列实战

Welcome to our in-depth exploration of Exercise 1A.3, a cornerstone practice set from the IB Mathematics course on sequences and series. This exercise focuses entirely on arithmetic sequences – one of the most fundamental and frequently tested topics. Whether you are working with Analysis and Approaches (AA) or Applications and Interpretation (AI), mastering arithmetic sequences is essential for both paper-based problem solving and real-world modelling. In this article we walk through every key concept, formula, and typical question style you will encounter in Exercise 1A.3, equipping you with the skills to tackle any related IB exam problem with confidence.

欢迎深入探讨练习1A.3，这是IB数学课程中关于数列与级数的核心练习集。该练习完全聚焦于等差数列——最基本且考查频率最高的主题之一。无论你正在学习分析与方法（AA）还是应用与解释（AI），掌握等差数列对于笔试解题和现实建模都至关重要。在本文中，我们将逐一梳理练习1A.3所涉及的每一个关键概念、公式和典型题型，帮助你获得自信应对任何相关IB考题的能力。

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

An arithmetic sequence is a list of numbers in which the difference between any term and its preceding term is constant. This unchanging difference is called the common difference and is universally denoted by d. For example, the sequence 2, 5, 8, 11, 14, … is arithmetic because each term increases by 3 compared with the previous one. The defining property can be written as u_n+1 − u_n = d for all n ∈ ℕ, where u_n represents the nth term. In Exercise 1A.3 you will frequently be asked to verify whether a given list of numbers forms an arithmetic progression, and to find the common difference if it does.

等差数列是一列数字，其中任意一项与其前一项的差保持不变。这个恒定的差被称为公差，通常用字母 d 表示。例如，数列 2, 5, 8, 11, 14, … 是等差数列，因为每一项都比前一项增加3。其定义性质可写作 u_n+1 − u_n = d 对所有正整数 n 成立，其中 u_n 表示第 n 项。在练习1A.3中，你经常会遇到要求判断一组给定数字是否构成等差数列，若是则求出其公差的问题。

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2. Notation, First Term, and Common Difference | 记号、首项与公差

Throughout IB Mathematics, the first term of a sequence is either labelled a or u₁. The common difference is d, and the nth term is written u_n. To extract d from a sequence, simply subtract any term from the term that follows it: d = u₂ − u₁ = u₃ − u₂ = … . When working with Exercise 1A.3, always clearly write down a and d before substituting into formulas. If a sequence is described by a recurrence relation such as u_n+1 = u_n + 5, with u₁ = 3, you can immediately identify a = 3 and d = 5.

在IB数学中，数列的首项通常标注为 a 或 u₁。公差为 d，第 n 项写作 u_n。要从数列中提取公差 d，只需用后一项减前一项：d = u₂ − u₁ = u₃ − u₂ = … 。在处理练习1A.3时，务必先清晰地写下 a 和 d，再代入公式。如果数列由递推关系描述，例如 u_n+1 = u_n + 5，且 u₁ = 3，你便可立即确定 a = 3，d = 5。

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3. The General Term Formula | 通项公式

The most powerful tool in Exercise 1A.3 is the explicit formula for the nth term of an arithmetic sequence:

u_n = a + (n − 1)d

This formula allows you to calculate any term directly without generating all preceding terms. In English we say: “The nth term equals the first term plus (n minus one) times the common difference.” Notice that when n = 1, u₁ = a, as expected. This formula is used in the majority of questions from 1A.3, including finding a specific term, solving for n given a term value, and setting up simultaneous equations when two terms are known.

练习1A.3中最有力的工具便是等差数列第 n 项的显式公式：

u_n = a + (n − 1)d

该公式使你能直接计算任意一项，而无需逐一生成前面的所有项。请注意，当 n = 1 时，u₁ = a，符合预期。该公式在练习1A.3的大部分问题中都会用到，包括求特定项、已知某项的值求项数 n，以及已知两项时建立方程组解决问题。

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4. Finding a Specific Term – Worked Example | 求特定项——示例详解

Consider an arithmetic sequence with first term a = 7 and common difference d = 4. Find the 30th term. Using the formula: u₃₀ = 7 + (30 − 1) × 4 = 7 + 29 × 4 = 7 + 116 = 123. Consequently, the 30th term is 123. In Exercise 1A.3 you may encounter similar numerical drills, often with larger indices like the 50th or 100th term, which test your ability to apply the formula accurately without computational error. Always double-check the product (n − 1)d before adding a.

考虑一个首项 a = 7、公差 d = 4 的等差数列，求第30项。应用公式：u₃₀ = 7 + (30 − 1) × 4 = 7 + 29 × 4 = 7 + 116 = 123。因此，第30项为123。在练习1A.3中你可能遇到类似的数值训练，往往涉及较大的序号，如第50项或第100项，这些题目旨在检验你能否准确运用公式而不出错。务必在加上 a 之前再次检查乘积 (n − 1)d。

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5. Determining the First Term and Common Difference from Two Known Terms | 由两项求首项和公差

A common IB question style is: Given u₃ = 12 and u₈ = 32, find a and d. We set up two equations using the general term: a + 2d = 12 and a + 7d = 32. Subtracting the first equation from the second yields 5d = 20, hence d = 4. Substituting d = 4 back into a + 2d = 12 gives a + 8 = 12, so a = 4. Exercise 1A.3 contains several such problems to strengthen your grasp on forming and solving linear systems within the context of sequences. You can then use a and d to find any term or sum.

一种常见的IB考题风格是：已知 u₃ = 12 且 u₈ = 32，求 a 和 d。我们利用通项公式建立两个方程：a + 2d = 12 和 a + 7d = 32。用第二个方程减去第一个方程得 5d = 20，因此 d = 4。将 d = 4 代回 a + 2d = 12 得出 a + 8 = 12，故 a = 4。练习1A.3包含多道此类问题，以强化你在数列背景下建立并求解一次方程组的能力。一旦求出 a 和 d，你便能计算任意项或总和。

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6. Arithmetic Mean (Average) | 等差中项

If x, y, z are three consecutive terms of an arithmetic sequence, then y is the arithmetic mean of x and z: y = (x + z) / 2. This property is useful for inserting terms or verifying arithmetic progressions. For instance, to insert an arithmetic mean between 6 and 14, we compute (6 + 14)/2 = 10. The resulting sequence 6, 10, 14 is arithmetic with d = 4. Exercise 1A.3 often asks you to find missing terms in a short progression using this averaging property, which stems directly from the constant difference condition.

若 x, y, z 是等差数列的三个连续项，则 y 是 x 和 z 的等差中项：y = (x + z) / 2。这一性质在插入项或验证等差数列时十分有用。例如，在6和14之间插入一个等差中项，计算 (6 + 14)/2 = 10，得到数列 6, 10, 14，它是公差为4的等差数列。练习1A.3经常要求你利用这一平均性质找出短数列中的缺失项，该性质直接来源于公差恒定的条件。

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7. Sum of an Arithmetic Series – Two Useful Forms | 等差数列的求和——两种实用形式

The sum of the first n terms of an arithmetic sequence, denoted S_n, is critical for later parts of Exercise 1A.3. The standard formulas are:

S_n = n/2 [2a + (n − 1)d]

S_n = n/2 (a + l), where l = u_n is the last term.

Use the first form when you know a, d, and n; use the second when you know the first and last terms. In many 1A.3 exercises you will be asked to compute S₂₀, S₅₀, or to find how many terms are needed to reach a certain total. Correct identification of a, d, and n is essential to avoid sign errors, especially when d is negative.

等差数列前 n 项的和，记作 S_n，对练习1A.3的后半部分至关重要。标准公式有两个：

S_n = n/2 [2a + (n − 1)d]

S_n = n/2 (a + l)，其中 l = u_n 为末项。

当已知 a, d 和 n 时使用第一个形式；当已知首项和末项时使用第二个形式。在1A.3的许多习题中，你会被要求计算 S₂₀、S₅₀，或者求需要多少项才能达到某个总和。正确识别 a, d 和 n 至关重要，这有助于避免符号错误，尤其是在 d 为负数时更应小心。

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8. Using the Sum Formula to Solve for n or d | 利用和公式求解项数或公差

Exercise 1A.3 often presents problems where S_n is known, and you must determine n or d. For example: An arithmetic series has a = 5, d = 3, and S_n = 275. Find n. Substitute into S_n = n/2 [2×5 + (n − 1)×3] = n/2 (10 + 3n − 3) = n/2 (3n + 7). Set n/2 (3n + 7) = 275, multiply both sides by 2 to get n(3n + 7) = 550. Solve the quadratic 3n² + 7n − 550 = 0 using the quadratic formula or factoring to find n = 12.5 or n = −14.6…; n must be a positive integer, so reconsider factoring — with careful arithmetic you find n = 12. Always check that n is a positive integer. Similarly, you might be given S_n and a, and asked to find d, requiring you to solve a linear equation after isolating d.

练习1A.3中常有已知 S_n，需求出 n 或 d 的问题。例如：某等差级数 a = 5，d = 3，且 S_n = 275，求 n。代入 S_n = n/2 [2×5 + (n − 1)×3] = n/2 (10 + 3n − 3) = n/2 (3n + 7)。令 n/2 (3n + 7) = 275，两边乘以2得 n(3n + 7) = 550。解二次方程 3n² + 7n − 550 = 0，利用求根公式或因式分解可得 n = 12（仔细计算排除非整数解）。n 必须为正整数，因此答案应为12。类似地，你可能会遇到已知 S_n 和 a 求 d 的题目，这时通过整理方程可解出 d。

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9. Real-World Applications and Modelling | 实际应用与建模

IB Mathematics places strong emphasis on contextual problems. Exercise 1A.3 includes scenarios such as: the number of seats in an auditorium increasing by 4 per row, starting with 20 seats in the first row; a savings plan where $100 is deposited each month with an initial deposit of $200; or the height of stacked chairs. In each case, identify a and d, then answer questions about total seats or total savings after a certain number of rows or months. These modelling tasks require you to recognise underlying arithmetic sequences and apply the sum formula to answer realistic ‘find the total’ questions.

IB数学非常重视情境化问题。练习1A.3包含诸如以下场景：礼堂每排座位数比前排增加4个，第一排有20个座位；储蓄计划中每月存入$100，初始存款为$200；以及叠放椅子的高度等。在每种情况下，先识别 a 和 d，然后回答有关总座位数或特定月数后总存款的问题。这些建模任务要求你识别出隐藏的等差数列，并应用求和公式来回答贴近现实的“求总数”问题。

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10. Recognizing Arithmetic Sequences in Disguise | 识别隐藏的等差数列

Sometimes Exercise 1A.3 challenges you with sequences defined by algebraic expressions, such as u_n = 3n − 7. To prove it is arithmetic, evaluate u_n+1 − u_n = [3(n+1) − 7] − [3n − 7] = 3, which is constant. This confirms d = 3 and also shows the coefficient of n in a linear explicit formula equals the common difference. Another disguise is a recurrence relation u_n+1 = u_n − 2. In these questions, always calculate the difference between successive terms to establish the value of d before proceeding to find a specific term or sum.

有时练习1A.3会用代数表达式定义的数列来考验你，例如 u_n = 3n − 7。要证明它是等差数列，计算 u_n+1 − u_n = [3(n+1) − 7] − [3n − 7] = 3，结果是常数。这证实了 d = 3，同时也表明线性显式公式中 n 的系数即为公差。另一种隐藏形式是递推关系 u_n+1 = u_n − 2。在处理此类问题时，务必先计算连续两项的差以确定 d 的值，再继续求特定项或总和。

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11. IB Exam-Style Question from Exercise 1A.3 | 练习1A.3中的IB考题风格

A typical IB short-answer question: “In an arithmetic sequence, the fifth term is 22 and the twelfth term is 50. (a) Find the common difference and the first term. (b) Find the sum of the first 30 terms.” For part (a), use u₅ = a + 4d = 22 and u₁₂ = a + 11d = 50. Subtracting gives 7d = 28, so d = 4, then a = 6. For part (b), apply S₃₀ = 30/2 [2×6 + (30−1)×4] = 15 × [12 + 116] = 15 × 128 = 1920. Such multi-part questions mirror the structure of real IB papers, where method marks are awarded for setting up equations correctly and accuracy marks for the final answer. Exercise 1A.3 provides numerous opportunities to practice this exact format.

一道典型的IB简答题：“在一个等差数列中，第五项为22，第十二项为50。(a) 求公差和首项。(b) 求前30项的和。”对于部分(a)，利用 u₅ = a + 4d = 22 和 u₁₂ = a + 11d = 50。两式相减得 7d = 28，故 d = 4，进而 a = 6。对于部分(b)，应用 S₃₀ = 30/2 [2×6 + (30−1)×4] = 15 × [12 + 116] = 15 × 128 = 1920。这种多小问的题目与真实IB试卷的结构相似，正确建立方程可得方法分，最终答案正确可得准确度分数。练习1A.3提供了大量机会让你练习这一特定题型。

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12. Key Tips, Common Pitfalls, and Final Review | 关键提示、常见误区与最终回顾

Always write down a, d, n, and l (if known) before using any formula. The most frequent mistake is misusing n: in u_n = a + (n − 1)d, if n is the number of terms, (n−1) is the number of ‘gaps’ between the first and nth terms. When summing, check whether you are using the correct S_n form and ensure the term count n is consistent. For word problems, highlight keywords such as ‘starting amount’ (a) and ‘increase by a constant’ (d). After obtaining an answer, substitute back into the original conditions to

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