Arithmetic Progressions: Sequences and Sum Formulas | 算术级数（等差数列）及其求和公式

📚 Arithmetic Progressions: Sequences and Sum Formulas | 算术级数（等差数列）及其求和公式

An arithmetic progression (AP) is one of the most fundamental topics in algebra and appears frequently in exam questions. Understanding its definition, nth term, and sum formulas is essential for solving a wide range of problems quickly and confidently.

算术级数（等差数列）是代数中最基础的内容之一，也是考试中的高频考点。掌握它的定义、通项公式以及求和公式，能帮助你快速而准确地解决多种题型。

1. What Is an Arithmetic Progression? | 什么是算术级数（等差数列）？

An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is always the same. This fixed difference is called the common difference and is usually denoted by d. For example, 5, 8, 11, 14, … is an AP because each term is 3 more than the previous one, so d = 3.

算术级数是指相邻两项之差恒为常数的数列。这个常数称为公差，通常用 d 表示。例如，5, 8, 11, 14, … 就是一个等差数列，因为每一项比前一项多 3，所以 d = 3。

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2. The nth Term Formula | 第n项公式（通项公式）

For an AP with first term a₁ and common difference d, the nth term aₙ is formed by starting from a₁ and adding (n-1) copies of d. Thus the formula for the nth term is:

对于首项为 a₁、公差为 d 的等差数列，第 n 项 aₙ 等于首项加上 (n-1) 个公差。因此，通项公式为：

aₙ = a₁ + (n-1)d

This formula allows you to find any term directly without listing the whole sequence. For instance, if a₁ = 2 and d = 5, then a₁₀ = 2 + 9 × 5 = 47.

这个公式能让你不必逐个写出数列就能直接求出任意一项。例如，若 a₁ = 2，d = 5，则 a₁₀ = 2 + 9 × 5 = 47。

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3. Finding the Common Difference and Specific Terms | 求公差与具体项

To determine whether a sequence is arithmetic, check if the difference between successive terms is constant. The common difference can be found by subtracting any term from the term that follows it: d = aₖ₊₁ – aₖ. If you know two given terms, you can substitute them into the nth-term formula and solve for d and n.

要判断一个数列是否为等差数列，只需检查相邻两项之差是否相等。公差可以用后一项减前一项来求：d = aₖ₊₁ – aₖ。如果已知某两项，可将它们代入通项公式，联立求出 d 和该项的序号 n。

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4. Sum of the First n Terms | 前n项求和公式

The sum of the first n terms of an AP is written as Sₙ. It can be calculated by multiplying the average of the first and last terms by the number of terms:

等差数列前 n 项和记为 Sₙ，可以通过首项与末项的平均值乘以项数来计算：

Sₙ = n/2 (a₁ + aₙ)

Because aₙ = a₁ + (n-1)d, we can also express the sum in terms of a₁ and d only:

因为 aₙ = a₁ + (n-1)d，所以我们也可以用仅含 a₁ 和 d 的形式来表示和：

Sₙ = n/2 [2a₁ + (n-1)d]

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5. Deriving the Sum Formula | 求和公式的推导

A classic and elegant derivation uses the “reverse-and-add” method. Write the sum forward, then write it backward, and add the two results. Each corresponding pair adds to exactly (a₁ + aₙ), and there are n such pairs, giving 2Sₙ = n(a₁ + aₙ). Dividing by 2 yields the sum formula.

一个经典而优雅的推导方法是“倒序相加法”。先把和正写一次，再倒写一次，然后将两个式子相加。每一对对应项相加都等于 (a₁ + aₙ)，一共有 n 对，所以得到 2Sₙ = n(a₁ + aₙ)，再除以 2 就得到了求和公式。

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6. Worked Examples | 典型例题

Example 1: Find the sum of all integers from 1 to 100. This is an AP with a₁ = 1, aₙ = 100 and n = 100. Using Sₙ = n/2(a₁ + aₙ), we get S₁₀₀ = 100/2 × (1+100) = 50 × 101 = 5050.

例1：求 1 到 100 所有整数之和。这是首项 a₁ = 1、末项 aₙ = 100、项数 n = 100 的等差数列。根据 Sₙ = n/2(a₁ + aₙ)，可得 S₁₀₀ = 100/2 × (1+100) = 50 × 101 = 5050。

Example 2: An AP has first term a₁ = 3 and common difference d = 2. Find S₁₀. Use Sₙ = n/2[2a₁ + (n-1)d], so S₁₀ = 10/2 × [2(3) + 9(2)] = 5 × (6+18) = 120.

例2：等差数列首项 a₁ = 3，公差 d = 2，求 S₁₀。用 Sₙ = n/2[2a₁ + (n-1)d]，得 S₁₀ = 10/2 × [2(3) + 9(2)] = 5 × (6+18) = 120。

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7. Word Problems Involving AP | 等差数列应用题

Arithmetic progressions appear in many real-life situations. For example, suppose you save £50 in the first month and increase your monthly saving by £10 each month. The amounts saved each month form an AP with a₁ = 50, d = 10. The total saved after 12 months is S₁₂ = 12/2 × [2(50) + 11(10)] = 6 × (100+110) = 1260.

等差数列在生活中有很多应用。例如，假设你第一个月存 50 英镑，之后每月多存 10 英镑。每月存款金额构成首项 a₁ = 50、公差 d = 10 的等差数列。12 个月后的总存款为 S₁₂ = 12/2 × [2(50) + 11(10)] = 6 × (100+110) = 1260。

Another classic problem involves stacked objects: a pile of apples has 5 apples on the top layer, and each lower layer contains 2 more apples than the layer above it. If there are 20 layers, the total number of apples forms an AP with a₁ = 5, d = 2 and n = 20, so S₂₀ = 20/2 × [2(5) + 19(2)] = 10 × (10+38) = 480.

另一类经典问题是物体堆叠：最上层有 5 个苹果，每向下一层比上一层多 2 个。若共有 20 层，苹果总数构成首项 a₁ = 5、公差 d = 2、项数 n = 20 的等差数列，因此 S₂₀ = 20/2 × [2(5) + 19(2)] = 10 × (10+38) = 480。

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8. Common Pitfalls and Tips | 常见错误与答题技巧

Students often confuse the index n with the value of the last term aₙ. Always identify a₁, d and n carefully before substituting. Also remember that d can be negative, in which case the sequence decreases. When using Sₙ = n/2(a₁ + aₙ), you must know the actual last term; if you know only a₁ and d, the variant with 2a₁ is safer.

学生常把项数 n 与末项的值 aₙ 弄混。代入前应先明确 a₁、d 和 n。另外，公差 d 可以为负，此时数列递减。使用 Sₙ = n/2(a₁ + aₙ) 时，必须知道实际末项 aₙ；如果只知道 a₁ 和 d，使用含 2a₁ 的形式更稳妥。

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9. Exam-Style Practice | 模拟考题

Practice 1: An AP has a₁ = 7 and d = -3. Find the 10th term and the sum of the first 10 terms. Answer: a₁₀ = 7 + 9(-3) = -20; S₁₀ = 10/2 × (7+(-20)) = 5 × (-13) = -65.

练习1：等差数列 a₁ = 7，d = -3。求第 10 项及前 10 项和。答案：a₁₀ = 7 + 9(-3) = -20；S₁₀ = 10/2 × (7+(-20)) = 5 × (-13) = -65。

Practice 2: The sum of the first n terms of an AP is given by Sₙ = 3n² + 2n. Find the first term and the common difference. Hint: a₁ = S₁ and S₂ = a₁ + a₂. Answer: S₁ = 5, so a₁ = 5; S₂ = 12, so a₂ = 7, hence d = 2.

练习2：等差数列前 n 项和为 Sₙ = 3n² + 2n。求首项和公差。提示：a₁ = S₁，S₂ = a₁ + a₂。答案：S₁ = 5，所以 a₁ = 5；S₂ = 12，所以 a₂ = 7，因此 d = 2。

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10. Summary | 总结

An arithmetic progression is defined by its first term a₁ and common difference d. The nth term is aₙ = a₁ + (n-1)d, and the sum of the first n terms is Sₙ = n/2(a₁ + aₙ) = n/2[2a₁ + (n-1)d]. Always check that the sequence is genuinely arithmetic before applying these formulas, and identify a₁, d, and n explicitly in every question.

等差数列由首项 a₁ 和公差 d 决定。通项公式为 aₙ = a₁ + (n-1)d，前 n 项和公式为 Sₙ = n/2(a₁ + aₙ) = n/2[2a₁ + (n-1)d]。应用公式前务必判断数列是否为等差数列，并在每道题中明确写出 a₁、d 和 n。

Quantity	Formula
First term	a₁
Common difference	d = aₖ₊₁ – aₖ
nth term	aₙ = a₁ + (n-1)d
Sum of n terms	Sₙ = n/2(a₁ + aₙ) = n/2[2a₁ + (n-1)d]

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