📚 Arithmetic Sequences: Finding the nth Term | 等差数列：求第 n 项
In IGCSE Mathematics, sequences appear frequently across both Paper 1 and Paper 2, and arithmetic sequences (also called linear sequences) are one of the most straightforward yet powerful patterns you will learn. Mastering the nth term formula not only helps you calculate any term in the sequence – like the 139th – but also allows you to determine the position of a given number, check whether a value belongs to the sequence, and solve real-world problems involving constant growth or decline. This article walks you through every essential concept, from identifying the common difference to applying the formula with confidence.
在 IGCSE 数学中，数列经常出现在试卷一和试卷二中，而等差数列（也叫线性数列）是你将学习的最简单却又最实用的模式之一。掌握第 n 项公式不仅能帮你求出数列中的任意一项——比如第 139 项——还能让你确定某个数的位置、判断一个值是否属于该数列，以及解决涉及恒定增长或减少的实际问题。本文将带你逐一学习每一个核心概念，从识别公差开始，一直到灵活运用公式。
1. What Is a Sequence? | 什么是数列？
A sequence is simply an ordered list of numbers following a certain rule. Each number in the sequence is called a term. For example, 3, 7, 11, 15, 19, … is a sequence where each term is 4 more than the previous one. The dots at the end indicate that the pattern continues indefinitely. In IGCSE exams, you will often be asked to find the next few terms or to describe the rule in words.
数列就是按照一定规则排列的一列有序数字。数列中的每一个数被称为项。例如 3, 7, 11, 15, 19, … 就是一个数列，其中每一项都比前一项大 4。末尾的省略号表示这个模式会无限延续下去。在 IGCSE 考试中，你经常会被要求写出接下来的几项，或者用文字描述这个规则。
2. Arithmetic Sequences Defined | 等差数列的定义
An arithmetic sequence is a sequence in which the difference between any term and the previous term is constant. This constant difference is called the common difference, often denoted by the letter d. If d is positive, the terms increase; if d is negative, the terms decrease. All linear sequences that produce a straight-line graph when plotting n against term value are arithmetic.
等差数列是指相邻两项的差总保持不变的数列。这个固定的差值被称为公差，通常用字母 d 表示。如果 d 为正数，数列的项就递增；如果 d 为负数，数列的项就递减。所有将 n 与项值对应起来能画出一条直线的线性数列都属于等差数列。
3. Spotting the Common Difference | 找出公差
ind d, subtract any term from the term that follows it. For the sequence 5, 9, 13, 17, 21, … secutive pairs to be certain the difference is constant. This is the foundation for building the nth term formula.
要找出 d，只需用后一项减去前一项。对于数列 5, 9, 13, 17, 21, … 我们计算 9 − 5 = 4，13 − 9 = 4，以此类推。所以 d = 4。一定要检查两到三对相邻的项，确保差值是固定的。这是建立第 n 项公式的基础。
4. The nth Term Formula | 第 n 项公式
For any arithmetic sequence, the nth term (often written as aₙ) can be found using the formula:
aₙ = a₁ + (n − 1)d
₁ = 5 and d = 4, the nth term becomes aₙ = 5 + (n − 1)×4, which simplifies to aₙ = 4n + 1.
对于任何一个等差数列，第 n 项（通常写作 aₙ）都可以用下面的公式求出：
aₙ = a₁ + (n − 1)d
其中 a₁ 是第一项，d 是公差，n 是位置编号。在 Edexcel IGCSE 考试中，你要掌握这个公式，并能在正向和逆向的情形中灵活运用。如果已知 a₁ = 5，d = 4，那么第 n 项就变成 aₙ = 5 + (n − 1)×4，化简后得到 aₙ = 4n + 1。
5. Finding a Specific Term (e.g. the 139th) | 求特定项（例如第 139 项）
term expression, substitute n with the position you need. Using the sequence 5, 9, 13, 17, … where aₙ = 4n + 1, the 139th term is:
a₁₃₉ = 4(139) + 1 = 556 + 1 = 557
hods give the same result. This skill is extremely useful for solving problems where a high position number is given.
一旦你得到了第 n 项表达式，就可以把 n 代入你所需的位置。对于数列 5, 9, 13, 17, …，其中 aₙ = 4n + 1，第 139 项就是：
a₁₃₉ = 4(139) + 1 = 556 + 1 = 557
也可以直接套用公式：a₁₃₉ = 5 + (139 − 1)×4 = 5 + 138×4 = 5 + 552 = 557。可以看出两种方法答案一致。这种技能在解决给出较大位置编号的问题时非常实用。
6. Using the nth Term to Find the Position | 利用第 n 项公式求项的位置
l to that value and solve for n. For instance, is 139 a term in the sequence 3, 7, 11, 15, …? Here a₁ = 3, d = 4, so aₙ = 3 + (n − 1)×4 = 4n − 1. Solve 4n − 1 = 139:
4n − 1 = 139 → 4n = 140 → n = 35
is indeed the 35th term. If n turns out to be a fraction or negative, the number does not belong to the sequence. This process is often tested in IGCSE questions where you need to decide whether a given number appears in the list.
如果你知道某一项的值，想知道它的位置，可以把第 n 项表达式设成这个值，然后解出 n。例如，139 是数列 3, 7, 11, 15, … 中的项吗？这里 a₁ = 3，d = 4，因此 aₙ = 3 + (n − 1)×4 = 4n − 1。解方程 4n − 1 = 139：
4n − 1 = 139 → 4n = 140 → n = 35
因为 n = 35 是一个正整数，所以 139 正是第 35 项。如果 n 算出来是分数或负数，那就说明这个数不属于该数列。IGCSE 考题经常要求学生判断某个给出的数是否出现在数列中，这正是需要掌握的解题过程。
7. Writing a Sequence from the nth Term | 由第 n 项公式写出数列
asks you to generate the first few terms. For aₙ = 4n + 1, simply substitute n = 1, 2, 3, … :
- n = 1: a₁ = 4(1) + 1 = 5
- n = 2: a₂ = 4(2) + 1 = 9
- n = 3: a₃ = 13
- n = 4: a₄ = 17
This confirms the sequence 5, 9, 13, 17, … You can also use the formula to jump to any term without listing all previous terms, which is a key advantage of using algebra over simple pattern spotting.
有时考题会给出第 n 项表达式，让你写出前几项。对于 aₙ = 4n + 1，只需要依次代入 n = 1, 2, 3, … ：
- n = 1：a₁ = 4(1) + 1 = 5
- n = 2：a₂ = 4(2) + 1 = 9
- n = 3：a₃ = 13
- n = 4：a₄ = 17
这验证了数列 5, 9, 13, 17, … 你也可以利用公式直接跳到任意一项，而不需要把前面的项都列出来，这正是使用代数方法比单靠观察规律更胜一筹的关键所在。
8. Finding the First Term and Common Difference from Two Terms | 由两项求首项和公差
A classic IGCSE problem gives you two non-consecutive terms and asks for the formula. Suppose the 5th term is 23 and the 10th term is 43. Write two equations:
a₁ + 4d = 23
a₁ + 9d = 43
method uses simultaneous equations and reinforces the link between algebra and sequences.
IGCSE 中有一类经典题目，是给出两个不相邻的项，让你求出公式。假设第 5 项是 23，第 10 项是 43，那么可以列出两个方程：
a₁ + 4d = 23
a₁ + 9d = 43
用第二个方程减去第一个：5d = 20 → d = 4。再代回去：a₁ + 4(4) = 23 → a₁ = 7。因此第 n 项公式为 aₙ = 7 + (n − 1)×4 = 4n + 3。这个方法用到了联立方程，同时巩固了代数与数列之间的联系。
9. Real-Life Applications | 实际应用
of seats in a theatre row, or reducing a drug dosage by a constant amount. If Emily saves £5 in week 1, £9 in week 2, £ s 4n + 1. You can predict savings in week 139, or find out in which week she saves exactly £139: solve 4n + 1 = 139 → n = 34.5, which is not an integer, so she never saves exactly £ se contextual problems make the algebra meaningful and are common in Edexcel IGCSE papers.
等差数列可以描述许多日常情景：每周存下相同金额的钱、在剧院每一排增加固定数量的座位，或者每次减少固定剂量的药物。如果 Emily 第一周存下 5 英镑，第二周 9 英镑，第三周 13 英镑，以此类推，那么第 n 周的金额就是 4n + 1。你可以预测第 139 周的储蓄额，或者计算她在哪一周刚好存下 139 英镑：解方程 4n + 1 = 139 得 n = 34.5，这不是整数，所以她在这种规律下永远不会刚好存下 139 英镑。这些情境题让代数变得更有意义，在 Edexcel IGCSE 试卷中也很常见。
10. Common Mistakes to Avoid | 常见错误
of (n − 1) would incorrectly add one extra d. Another pitfall is misidentifying the common difference – r n correctly and interpret non-integer n as meaning the number is not a term. Finally, don't confuse the position n with the value of the term.
许多学生忘记公式中与 d 相乘的是 (n − 1) 而不是 n。对于第一项，n = 1 得到 a₁ + 0×d，结果正确；如果错误地用 n 代替 (n − 1)，就会多加上一个 d。另一个容易出错的地方是用错公差——如果数列是递减的，d 就是负数，一旦忘记负号，整个公式都会出错。此外，还要注意正确解出 n，并理解非整数的 n 意味着这个数不是数列中的项。最后，不要把位置 n 与项的值混淆了。
11. Using a Table to Build Understanding | 用表格辅助理解
value, especially when first learning the topic. Consider the sequence with a₁ = 7, d = 3:
| Position n | Calculation | Term value |
|---|---|---|
| 1 | 7 + 0×3 | 7 |
| 2 | 7 + 1×3 | 10 |
| 3 | 7 + 2×3 | 13 |
| … | … | … |
| 139 | 7 + 138×3 | 7 + 414 = 421 |
This visual approach shows why we multiply d by (n − 1) and makes it easy to spot patterns. In an exam you could draw a quick table to check your formula.
用表格整理信息可以让位置与项值之间的联系一目了然，在初学阶段尤其有帮助。来看一个首项 a₁ = 7、公差 d = 3 的数列：
| 位置 n | 计算过程 | 项值 |
|---|---|---|
| 1 | 7 + 0×3 | 7 |
| 2 | 7 + 1×3 | 10 |
| 3 | 7 + 2×3 | 13 |
| … | … | … |
| 139 | 7 + 138×3 | 7 + 414 = 421 |
这种直观的方法很清楚地说明了为什么 d 要乘以 (n − 1)，并且能让你轻松发现规律。在考试中你也可以通过画一个速查表格来验证自己的公式。
12. Examination Tips for Edexcel IGCSE | Edexcel IGCSE 考试技巧
mula aₙ = a₁ + (n − 1)d immediately. If the question involves a large number like 139, don't panic – hmetic slip occurs. For reverse problems, always test your answer by plugging several n values back into your formula to ensure it generates the original sequence. Finally, pay attention to units in contextual questions and state whether a term exists based on n being a positive integer.
在遇到等差数列问题时，首先要列出已知的项，并找出 a₁ 和 d。立即写下第 n 项公式 aₙ = a₁ + (n − 1)d。如果题目中出现了像 139 这样比较大的数字，不要慌张——直接把它代入你的表达式即可。一定要清晰展示代入过程，因为 Edexcel IGCSE 会按步骤给分，即使出现计算失误也能获得方法分。对于逆向问题，记得把几个 n 值代回你的公式，检验是否能生成原来的数列。最后，要留意情境题中的单位，并根据 n 是否为正整数来判断某一项是否存在。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图