📚 Arithmetic Sequences | 等差数列
An arithmetic sequence is one of the most fundamental concepts in IGCSE Mathematics. It is a list of numbers where the difference between consecutive terms remains constant. Understanding arithmetic sequences will help you solve problems involving patterns, predictions, and series sums.
等差数列是IGCSE数学中最基本的概念之一。它是一个数列，其中相邻两项的差始终保持不变。理解等差数列能帮助你解决涉及规律、预测和级数求和的问题。
1. What is an Arithmetic Sequence? | 什么是等差数列？
An arithmetic sequence is a sequence of numbers where each term after the first is found by adding a fixed number, called the common difference, to the previous term. For example, 3, 7, 11, 15, ... is an arithmetic sequence with a common difference of 4.
等差数列是指从第二项起，每一项都加上一个固定的数（称为公差）而得到的数列。例如，3, 7, 11, 15, ... 就是一个公差为4的等差数列。
2. Common Difference | 公差
ny term from the term that follows it: d = a₂ − a₁. It can be positive, negative, or zero.
公差 d 可以通过后一项减去前一项得到：d = a₂ − a₁。它可以是正数、负数或零。
he sequence is increasing; if d < 0, it is decreasing; if d = 0, the sequence is constant.
若 d > 0，数列递增；若 d < 0，数列递减；若 d = 0，则数列为常数列。
3. The nth Term Formula | 第 n 项公式
The nth term (aₙ) of an arithmetic sequence is given by the formula:
aₙ = a₁ + (n − 1)d
where a₁ is the first term, d is the common difference, and n is the term number.
等差数列的第 n 项 aₙ 由公式给出：aₙ = a₁ + (n − 1)d，其中 a₁ 是首项，d 是公差，n 是项数。
This formula arises because to reach the nth term, you start at a₁ and add d repeatedly (n − 1) times.
这个公式的来源是：要得到第 n 项，从首项 a₁ 开始，累加 d (n − 1) 次。
4. Finding a Specific Term | 求特定项
5, 9, 13, 17, ... (a₁ = 5, d = 4), the 10th term is a₁₀ = 5 + (10 − 1) × 4 = 5 + 36 = 41.
要求特定项，将 n 代入公式。对于数列 5, 9, 13, 17, ... (a₁ = 5, d = 4)，第 10 项为 a₁₀ = 5 + (10 − 1) × 4 = 5 + 36 = 41。
ifference, and you need to find the first term. Rearrange the formula: a₁ = aₙ − (n − 1)d.
有时会给出某一项和公差，要求首项，这时可将公式变形：a₁ = aₙ − (n − 1)d。
5. Using the nth Term to Find the Term Number | 用第 n 项公式求项数
If you know a term' olving aₙ = a₁ + (n − 1)d for n. For example, is 100 a term in the sequence 7, 12, 17, ... ? 5 → n − 1 = 18.6 → n = 19.6. Since n is not an integer, 100 is not a term of the sequence.
如果已知某一项的值，可以通过解方程 aₙ = a₁ + (n − 1)d 求出它的位置 n。例如，100 是数列 7, 12, 17, ... 中的一项吗？设 100 = 7 + (n − 1) × 5 → 93 = (n − 1) × 5 → n − 1 = 18.6 → n = 19.6，不是整数，因此 100 不是该数列的项。
6. Introduction to Arithmetic Series | 等差数列求和简介
The sum of the terms in an arithmetic sequence is called an arithmetic series. For example, the series 2 + 5 + 8 + 11 is the sum of the first four terms of the sequence 2, 5, 8, 11, ...
等差数列各项的和称为等差级数（等差序列的和）。例如，2 + 5 + 8 + 11 是数列 2, 5, 8, 11, ... 前四项的和。
7. Sum of an Arithmetic Series | 等差数列求和
The sum of the first n terms, Sₙ, is given by:
Sₙ = (n/2)(a₁ + aₙ)
or equivalently
Sₙ = (n/2)[2a₁ + (n − 1)d]
等差数列前 n 项的和 Sₙ 由公式给出：Sₙ = (n/2)(a₁ + aₙ) 或 Sₙ = (n/2)[2a₁ + (n − 1)d]。
₁ = 3, d = 4. First find a₂₀ = 3 + 19 × 4 = 79. Then S₂₀ = (20/2)(3 + 79) = 10 × 82 = 820.
对于数列 3, 7, 11, 15, ... 求前 20 项的和。这里 a₁ = 3, d = 4。先求 a₂₀ = 3 + 19 × 4 = 79，然后 S₂₀ = (20/2)(3 + 79) = 10 × 82 = 820。
8. Using the Sum Formula | 求和公式应用
n a quadratic. For example, how many terms of the sequence 5, 8, 11, ... give a sum of 345? → 3n² + 7n − 690 = 0. Solving the quadratic gives n = 15 (the positive integer solution).
如果已知总和 Sₙ，可以通过列方程（常为二次方程）求出项数 n。例如，数列 5, 8, 11, ... 的前多少项和为 345？这里 a₁ = 5, d = 3。使用 Sₙ = (n/2)[2a₁ + (n − 1)d]：345 = (n/2)[10 + (n − 1)×3] = (n/2)(3n + 7)。两边乘 2：n(3n + 7) = 690 → 3n² + 7n − 690 = 0。解这个二次方程得 n = 15（取正整数解）。
9. Word Problems | 应用题
such as saving money, stacking objects, or seating arrangements. For instance, if you save £5 in the first week, £8 in the second, £ c sequence with a₁ = 5, d = 3. The total saved after n weeks is the arithmetic series sum.
等差数列出现在许多实际情境中，如存钱、堆叠物品、座位安排等。例如，如果第一周存 5 英镑，第二周存 8 英镑，第三周存 11 英镑，以此类推，每周存款构成一个首项为 5、公差为 3 的等差数列。n 周后的总存款就是等差级数的和。
10. Linear Sequences vs Arithmetic Sequences | 线性序列与等差数列的关系
ar function of n. This means that plotting term number n against term value aₙ gives a straight line.
等差数列也被称为线性数列，因为第 n 项公式 aₙ = a₁ + (n − 1)d 可以改写为 aₙ = dn + (a₁ − d)，这是 n 的线性函数。这意味着将项数 n 与项值 aₙ 作图会得到一条直线。
11. Common Mistakes | 常见错误
A common mistake is confusing n − 1 with n when using the nth term formula. Another is forgetting to divide by 2 in the sum formula, or using aₙ instead of a₁ in the first sum formula. Always double-check your substitutions and calculations.
常见错误包括在使用第 n 项公式时将 n − 1 误用为 n，以及在求和公式中忘记除以 2，或者在第一个求和公式中误用 aₙ 代替 a₁。请始终仔细检查代入和计算。
Also, when finding n from a term value, remember that n must be a positive integer. If your solution gives a decimal or negative number, that value is not a term in the sequence.
此外，当通过项值求 n 时，记住 n 必须是正整数。如果得出的解是小数或负数，那么该值就不是数列中的项。
12. Summary | 总结
a₁ + (n − 1)d]. Practice identifying a₁ and d from a sequence, and apply the formulas confidently in both routine and problem-solving contexts. Recognising patterns and linking the linear form will strengthen your understanding for IGCSE exams.
要掌握等差数列，记住关键公式：aₙ = a₁ + (n − 1)d 以及 Sₙ = (n/2)(a₁ + aₙ) 或 Sₙ = (n/2)[2a₁ + (n − 1)d]。练习从数列中识别出首项和公差，并在常规和问题解决情境中自信地应用这些公式。识别规律并联想到线性形式，将强化你对 IGCSE 考试的理解。
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