Arithmetic Sequences | 等差数列

📚 Arithmetic Sequences | 等差数列

An arithmetic sequence is one of the most fundamental topics in the Edexcel IGCSE Mathematics syllabus. It appears in both Paper 1 and Paper 2, often in the form of formula-based problems, word problems, or combined questions with algebra. Mastering this topic requires a clear understanding of the two key formulas: the nth term formula and the sum formula.

等差数列是 Edexcel IGCSE 数学大纲中最基础的知识点之一。它在 Paper 1 和 Paper 2 中都会出现，通常以公式计算题、应用题或与代数结合的综合题形式考查。掌握这一主题需要清晰理解两个关键公式：通项公式与求和公式。

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

An arithmetic sequence (also called an arithmetic progression) is a sequence of numbers in which the difference between any two consecutive terms is always constant. This constant difference is known as the common difference. For example, 2, 5, 8, 11, 14, … is an arithmetic sequence because each term is obtained by adding 3 to the previous term. Similarly, 20, 16, 12, 8 is an arithmetic sequence because each term is obtained by subtracting 4 from the previous term.

等差数列（也称等差级数）是一列数字，其中任意两个相邻项之间的差值始终保持不变。这个恒定的差值称为公差。例如，2, 5, 8, 11, 14, … 是一个等差数列，因为每一项都是通过在前一项上加 3 得到的。同样，20, 16, 12, 8 也是等差数列，因为每一项都是通过从前一项中减去 4 得到的。

The terms of a sequence are written as a₁, a₂, a₃, a₄, … where the subscript indicates the position of the term. The first term is denoted by a₁ (or sometimes just a), and the nth term is denoted by aₙ.

数列的项写作 a₁, a₂, a₃, a₄, …，其中下标表示项的位置。首项记作 a₁（有时简写为 a），第 n 项记作 aₙ。

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2. The Common Difference | 公差

The common difference, denoted by d, is the fixed amount added to (or subtracted from) each term to obtain the next term. It can be positive, negative, or zero. If d > 0, the sequence is increasing; if d < 0, the sequence is decreasing; if d = 0, all terms are identical, forming a constant sequence.

公差用 d 表示，是从一个项得到下一项时固定加上（或减去）的量。它可以是正数、负数或零。如果 d > 0，数列递增；如果 d < 0，数列递减；如果 d = 0，则所有项都相同，形成常数数列。

To find the common difference, we use the formula:

d = aₙ₊₁ – aₙ

In practice, we simply subtract any term from the term that immediately follows it. For example, in the sequence 10, 14, 18, 22, we have d = 14 – 10 = 18 – 14 = 22 – 18 = 4. This single value of d applies to every pair of consecutive terms.

在实际计算中，只需用后一项减去前一项即可。例如，在数列 10, 14, 18, 22 中，d = 14 – 10 = 18 – 14 = 22 – 18 = 4。这个唯一的 d 值适用于每一对相邻项。

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3. The nth Term Formula | 通项公式

The nth term (also called the general term) of an arithmetic sequence can be found using the following formula:

aₙ = a₁ + (n – 1)d

Here, a₁ is the first term, d is the common difference, and n is the position of the required term. The formula works because to reach the nth term, we start from a₁ and add the common difference (n – 1) times.

其中 a₁ 是首项，d 是公差，n 是所需项的位置序号。这个公式的原理是：要从首项到达第 n 项，我们从 a₁ 出发，需要加上 (n – 1) 次公差。

Let us look at a worked example. For the sequence 3, 7, 11, 15, …, find the 10th term and the 50th term.

我们先看一个例题。对于数列 3, 7, 11, 15, …，求第 10 项和第 50 项。

Here a₁ = 3 and d = 7 – 3 = 4. Using the formula:

这里 a₁ = 3，d = 7 – 3 = 4。代入公式：

a₁₀ = 3 + (10 – 1) × 4 = 3 + 36 = 39

a₅₀ = 3 + (50 – 1) × 4 = 3 + 196 = 199

Notice that we do not need to write out all the terms; the formula gives the answer directly. The term “nth term” is often referred to by the expression for aₙ in terms of n, for example aₙ = 4n – 1 for this sequence.

注意，我们不需要逐项写出来；公式直接给出答案。”通项”也常指用 n 表示 aₙ 的代数式，例如这个数列的通项为 aₙ = 4n – 1。

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4. Finding the First Term and Common Difference | 求首项和公差

In exam problems, you will often be given two different terms of an arithmetic sequence and asked to find a₁ and d. This requires setting up two equations and solving them simultaneously.

在考试中，通常会给出等差数列的两个不同项，要求求出 a₁ 和 d。这需要建立两个方程并联立求解。

Example: The 4th term of an arithmetic sequence is 14 and the 9th term is 34. Find the first term and the common difference.

例题：已知等差数列的第 4 项是 14，第 9 项是 34。求首项和公差。

Using the nth term formula aₙ = a₁ + (n – 1)d, we write:

利用通项公式 aₙ = a₁ + (n – 1)d，我们写出：

a₁ + 3d = 14   … (1)

a₁ + 8d = 34   … (2)

Subtracting equation (1) from equation (2) eliminates a₁:

用方程 (2) 减去方程 (1) 可以消去 a₁：

(a₁ + 8d) – (a₁ + 3d) = 34 – 14

5d = 20

d = 4

Substituting d = 4 back into equation (1):

将 d = 4 代回方程 (1)：

a₁ + 3 × 4 = 14

a₁ = 2

Therefore, the first term is 2 and the common difference is 4. The general nth term is aₙ = 2 + 4(n – 1) = 4n – 2.

因此，首项为 2，公差为 4。通项为 aₙ = 2 + 4(n – 1) = 4n – 2。

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5. Sum of the First n Terms | 前 n 项和

The sum of the first n terms of an arithmetic sequence is denoted by Sₙ. There are two equivalent formulas:

等差数列前 n 项的和记作 Sₙ。有两个等价公式：

Sₙ = n/2 × [2a₁ + (n – 1)d]

Sₙ = n/2 × (a₁ + aₙ)

The second formula is useful when the last term aₙ is already known. The first formula is more common when a₁, d and n are given.

当末项 aₙ 已知时，第二个公式更为方便。当已知 a₁、d 和 n 时，通常使用第一个公式。

Example: Find the sum of the first 20 terms of the arithmetic sequence 5, 8, 11, 14, …

例题：求等差数列 5, 8, 11, 14, … 前 20 项的和。

Here a₁ = 5, d = 3, n = 20. Using the first formula:

这里 a₁ = 5，d = 3，n = 20。代入第一个公式：

S₂₀ = 20/2 × [2 × 5 + (20 – 1) × 3]

S₂₀ = 10 × [10 + 57] = 10 × 67 = 670

The sum of the first 20 terms is 670. We can check this by noting that the 20th term is a₂₀ = 5 + 19 × 3 = 62, and using the second formula: S₂₀ = 20/2 × (5 + 62) = 10 × 67 = 670. Both formulas agree.

前 20 项和为 670。我们可以验证：第 20 项为 a₂₀ = 5 + 19 × 3 = 62，用第二个公式：S₂₀ = 20/2 × (5 + 62) = 10 × 67 = 670。两个公式结果一致。

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6. Working Backwards from the Sum | 由和反推项数

Some questions give you the sum Sₙ and ask you to find n, the number of terms. This typically leads to a quadratic equation in n, which you then solve by factorisation or the quadratic formula.

有些题目会给出和 Sₙ，要求求出项数 n。这通常会得到一个关于 n 的二次方程，需要用因式分解法或求根公式来解。

Example: How many terms of the sequence 3, 7, 11, 15, … are

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