Arithmetic Sequences and Series: Exercise 1A.2 Walkthrough | 等差数列与级数：习题1A.2精讲

📚 Arithmetic Sequences and Series: Exercise 1A.2 Walkthrough | 等差数列与级数：习题1A.2精讲

Arithmetic sequences form the backbone of many IB Mathematics problems, and Exercise 1A.2 is designed to test your ability to move fluently between the nth term, the sum of n terms, and real-world applications. Whether you are tackling this for Analysis & Approaches SL or HL, a clear, step-by-step approach will help you avoid common errors and build confidence.

等差数列是IB数学中许多问题的基础，练习1A.2旨在检验你在第n项、前n项和以及实际应用之间灵活转换的能力。无论你是在为Analysis & Approaches SL还是HL应对这些题目，一个清晰、循序渐进的解题方法都能帮助你避免常见错误，建立信心。

1. Understanding the Core Definitions | 理解核心定义

An arithmetic sequence is a list of numbers where the difference between consecutive terms, called the common difference d, is constant. The first term is usually denoted by u₁ or a. The general term formula is:

等差数列是一列数字，相邻两项之间的差值（称为公差d）是常数。首项通常记作u₁或a。通项公式为：

uₙ = u₁ + (n – 1)d

For the sum of the first n terms, we use the formula:

对于前n项的和，我们使用公式：

Sₙ = n/2 (2u₁ + (n – 1)d)   or   Sₙ = n/2 (u₁ + uₙ)

Make sure you can derive these formulas; understanding the logic behind averaging the first and last term gives you a powerful tool when the sequence is presented in an unfamiliar way.

请确保你能推导这些公式；理解首项和末项取平均值的逻辑，能在题目以不熟悉的方式呈现时为你提供强有力的工具。

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2. Finding the nth Term from Basic Information | 由基本信息求第n项

Exercise 1A.2 often begins with straightforward questions such as: “Find the 20th term of the sequence 5, 8, 11, 14, …” Here we identify u₁ = 5 and d = 3. Substituting into the formula gives:

练习1A.2通常以直接的问题开头，比如：“求数列5, 8, 11, 14, …的第20项。”这里我们确定u₁ = 5，d = 3。代入公式得：

u₂₀ = 5 + (20 – 1) × 3 = 5 + 57 = 62

Always check your answer mentally by adding the common difference a few times to see if the scale makes sense. Asking “does the 20th term sit around 60?” confirms you haven’t swapped u₁ and d.

总是心算几次加上公差来检查你的答案，看数量级是否合理。问自己“第20项大约在60左右吗？”可以确认你没有混淆u₁和d。

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3. Solving for the Position When Given a Term Value | 已知某项的值求位置

A classic problem type asks: “Which term of the sequence 2, 7, 12, … is equal to 137?” Set uₙ = 137, u₁ = 2, d = 5, and solve for n:

一类经典问题是：“数列2, 7, 12, …中的哪一项等于137？”设uₙ = 137，u₁ = 2，d = 5，求n：

137 = 2 + (n – 1) × 5 → 135 = (n – 1) × 5 → n – 1 = 27 → n = 28

If n turns out to be a non-integer, the value does not belong to the sequence. This is a common exam trick—always state clearly that the given number is not a term of the sequence.

如果n不是整数，则该值不属于这个数列。这是考试中常见的陷阱——始终清楚地说明给定的数字不是数列中的项。

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4. Working with Two Unknown Terms | 处理含有两个未知项的问题

You may be told that the 4th term is 15 and the 10th term is 33. Form two equations:

你可能被告知第4项是15，第10项是33。列出两个方程：

u₄ = u₁ + 3d = 15

u₁₀ = u₁ + 9d = 33

Subtracting the first from the second yields 6d = 18, so d = 3. Substituting back gives u₁ = 15 – 9 = 6. This method of setting up simultaneous equations is used repeatedly, so master it early.

用第二个方程减去第一个方程得到6d = 18，所以d = 3。回代得u₁ = 15 – 9 = 6。这种建立联立方程组的方法会被反复使用，所以尽早掌握。

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5. Calculating the Sum of the First n Terms | 计算前n项的和

Given u₁ = 4 and d = 5, find S₃₀. Use the sum formula directly:

已知u₁ = 4，d = 5，求S₃₀。直接使用求和公式：

S₃₀ = 30/2 × (2 × 4 + (30 – 1) × 5) = 15 × (8 + 145) = 15 × 153 = 2295

If you prefer the Sₙ = n/2 (u₁ + uₙ) version, first calculate u₃₀ = 4 + 29 × 5 = 149, then S₃₀ = 15 × (4 + 149) = 2295. Both routes lead to the same result, but the first avoids computing uₙ separately.

如果你更喜欢Sₙ = n/2 (u₁ + uₙ)版本，先计算u₃₀ = 4 + 29 × 5 = 149，然后S₃₀ = 15 × (4 + 149) = 2295。两条路径结果相同，但第一条避免单独计算uₙ。

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6. Finding n Given a Sum and Other Parameters | 已知求和及其他参数求n

Suppose Sₙ = 570, u₁ = 3, d = 4. Set up the equation:

假设Sₙ = 570，u₁ = 3，d = 4。建立方程：

n/2 (2 × 3 + (n – 1) × 4) = 570 → n/2 (6 + 4n – 4) = 570 → n/2 (4n + 2) = 570 → n(2n + 1) = 570

This simplifies to 2n² + n – 570 = 0. Solve the quadratic (using the formula or factorisation) to get n = 15 (discarding the negative root). Always check that n is a positive integer; if both roots are invalid, the given sum is impossible for that sequence.

这简化为2n² + n – 570 = 0。解二次方程（用求根公式或因式分解）得n = 15（舍去负根）。始终检查n是否为正整数；如果两个根都不成立，则给定和对于该数列是不可能的。

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7. Handling Fractional and Negative Common Differences | 处理分数和负数公差

Sequences with decimal or fractional d, such as u₁ = 100, d = – ½, appear in Exercise 1A.2 to test algebraic confidence. For example, find the first term that becomes negative:

带有小数或分数公差d的数列，比如u₁ = 100，d = -½，会出现在练习1A.2中以检验代数上的自信。例如，求第一个变为负数的项：

100 + (n – 1)(-0.5) < 0 → -0.5(n – 1) < -100 → n – 1 > 200 → n > 201

Thus the 202nd term is the first negative. Pay close attention to inequality direction when multiplying or dividing by negative numbers; this is a common slip.

因此第202项是第一个负数。当乘以或除以负数时，要特别注意不等号的方向；这是一个易犯的错误。

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8. Word Problems: Applied Arithmetic Sequences | 应用题：等差数列的实际应用

A typical applied question: “A theatre has 25 seats in the first row. Each subsequent row has 3 more seats than the previous one. There are 40 rows. Find the total number of seats.”

典型的应用题：“一个剧院第一排有25个座位。后面每一排比前一排多3个座位。共有40排。求座位总数。”

Here u₁ = 25, d = 3, n = 40. Total seats S₄₀ = 40/2 × (2×25 + 39×3) = 20 × (50 + 117) = 20 × 167 = 3340. Always reread the question to check whether it asks for a specific row or the total; many students lose marks by calculating the wrong quantity.

这里u₁ = 25，d = 3，n = 40。座位总数S₄₀ = 40/2 × (2×25 + 39×3) = 20 × (50 + 117) = 20 × 167 = 3340。始终重读题目，检查它是问某一排还是总数；许多学生因为算错了量而失分。

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9. Using Sigma Notation Efficiently | 高效使用西格玛记号

Exercise 1A.2 may present sums in sigma notation, such as Σ (2k + 5) from k=1 to 30. Recognise this as an arithmetic series by writing out the first few terms: when k=1, term=7; k=2, term=9; so u₁=7 and d=2.

练习1A.2可能会用西格玛记号给出求和，例如从k=1到30的Σ (2k + 5)。通过写出前几项来判断这是一个等差级数：当k=1时，项为7；k=2时，项为9；所以u₁=7，d=2。

Then S₃₀ = 30/2 × (2×7 + 29×2) = 15 × (14+58) = 15×72 = 1080. Alternatively, splitting the sum as 2Σk + Σ5 gives the same result. Use whichever method you find more transparent.

然后S₃₀ = 30/2 × (2×7 + 29×2) = 15 × (14+58) = 15×72 = 1080。或者，将求和拆分为2Σk + Σ5会得到相同结果。使用你认为更直观的方法。

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10. Recognising When a Sequence Is Not Arithmetic | 识别数列何时不是等差数列

If the difference between consecutive terms is not constant, the sequence is not arithmetic. For instance, 1, 4, 9, 16, … has differences 3, 5, 7, so it is quadratic. Do not force arithmetic formulas onto such sequences. The IB exam often mixes arithmetic conditions with other types to see if you check d first.

如果相邻两项的差不是常数，该数列就不是等差数列。例如，1, 4, 9, 16, …的差依次为3, 5, 7，所以它是二次的。不要将等差公式强加于此类数列。IB考试常常混合等差条件与其他类型，以考察你是否首先检验了d。

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11. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法

Many mistakes stem from confusing the term index with the number of terms, or using n/d incorrectly. A classic error: using Sₙ formula with uₙ from the wrong index. Always write down u₁, d, n, and the target variable before substituting. If a problem involves “the sum from the 5th to the 15th term,” calculate S₁₅ – S₄, not S₁₅ – S₅.

许多错误源于混淆项的序号与项数，或错误使用n/d。经典错误：使用Sₙ公式时代入了错误序号的uₙ。在代入前总是先写下u₁、d、n和目标变量。如果问题涉及“从第5项到第15项的和”，应计算S₁₅ – S₄，而不是S₁₅ – S₅。

Another pitfall is forgetting to use parentheses when substituting negative d. For example, u₁ = 10, d = -3, n = 5: u₅ = 10 + (5-1)(-3) = 10 – 12 = -2, not 10 + 4 – 3. Little slips like these can derail an otherwise perfect solution.

另一个陷阱是代入负数d时忘记使用括号。例如u₁ = 10，d = -3，n = 5：u₅ = 10 + (5-1)(-3) = 10 – 12 = -2，而不是10 + 4 – 3。这样的小失误会破坏原本完美的解答。

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12. Linking to IB Exam-Style Questions | 与IB考试风格题目的衔接

Exercise 1A.2 builds the fundamental skills needed for longer IB questions that combine arithmetic sequences with compound interest, geometry, or piecewise functions. When you see a problem asking for the year in which a salary exceeds a certain amount, think arithmetic sequence: u₁ = initial salary, d = annual increment. Document each step clearly—exam marks are awarded for method, not just the final number.

练习1A.2为更长的IB题目打下基础，这些题目将等差数列与复利、几何或分段函数结合起来。当你看到要求计算工资超过某数额的年份的问题时，就要想到等差数列：u₁ = 起薪，d = 年增长额。每一步都清晰记录——考试分数不仅给最终答案，也会给解题步骤。

Practice with a timer, gradually reducing the time you need for the mechanical part of calculations so your working memory is free for interpretation. Over time, the formulas will become second nature, and Exercise 1A.2 will feel like a series of routine yet satisfying puzzles.

用计时器练习，逐渐减少机械计算所需的时间，这样你的工作记忆就能解放出来进行解读。随着时间的推移，公式将成为你的第二天性，练习1A.2将感觉像一系列常规但令人满足的谜题。

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