📚 Arithmetic Sequences: nth Term, Sum and Applications | 等差数列：通项、求和与应用
Arithmetic sequences are one of the core building blocks of A-level Pure Mathematics. In an arithmetic sequence, the difference between consecutive terms is constant, which allows us to find any term and the sum of the first n terms using simple formulas. This article covers the key definitions, formulas, proofs and problem types required for Edexcel A-level Mathematics.
等差数列是 A-level 纯数学的核心基础内容之一。在等差数列中，相邻两项的差保持不变，因此我们可以用简洁的公式求出任意项以及前 n 项的和。本文涵盖 Edexcel A-level 数学要求的关键定义、公式、证明和常见题型。
1. Definition and Notation | 定义与记号
11, 15, ... is arithmetic with a = 3 and d = 4 because each term is obtained by adding 4.
等差数列是一列数，其中任意一项与前一项的差保持不变。这个固定的差称为公差，通常记作 d。首项通常记作 a。例如，3, 7, 11, 15, ... 就是一个等差数列，其中 a = 3，d = 4，因为每一项都是通过加 4 得到的。
as a. The common difference can be calculated by d = aₙ₊₁ - aₙ for any positive integer n.
我们通常把各项写作 a₁, a₂, a₃, ..., aₙ。首项 a₁ 就是 a。公差可以用 d = aₙ₊₁ - aₙ 计算，其中 n 为任意正整数。
2. The nth Term Formula | 通项公式
. This formula works because to reach the nth term from the first term, we add the common difference d exactly n - 1 times.
等差数列的第 n 项由公式 aₙ = a + (n - 1)d 给出。这个公式成立是因为从首项到第 n 项，需要恰好加上 n - 1 次公差 d。
aₙ = a + (n - 1)d
For example, if a = 5 and d = 3, then the 20th term is a₂₀ = 5 + (20 - 1)×3 = 5 + 57 = 62.
例如，若 a = 5，d = 3，则第 20 项为 a₂₀ = 5 + (20 - 1)×3 = 5 + 57 = 62。
You can also use this formula in reverse: if you know two terms or one term and d, you can find a or n. We often substitute known values and solve the resulting linear equation.
你也可以反向使用这个公式：如果已知两项，或已知一项与公差 d，就可以求出 a 或 n。我们通常代入已知数值，然后解所得的一元一次方程。
3. Finding the Common Difference and First Term | 求公差与首项
ch known term produces a linear equation in a and d. Solving the two equations simultaneously gives both unknowns.
考试中常见的任务是给出等差数列的两个特定项，要求求出 a 和 d。由于 aₙ = a + (n - 1)d，每个已知项都会产生一个关于 a 和 d 的一次方程。联立求解这两个方程即可得到两个未知量。
d = 32. Subtracting gives 6d = 18, so d = 3. Substituting back gives a + 9 = 14, so a = 5.
例如：第 4 项为 14，第 10 项为 32。则有 a + 3d = 14 和 a + 9d = 32。两式相减得 6d = 18，因此 d = 3。代回可得 a + 9 = 14，所以 a = 5。
efficient of d is 3, not 4. Writing the equations clearly reduces sign and index mistakes.
仔细标注各项会很有帮助：a₄ 表示 n = 4，因此 d 的系数是 3，而不是 4。清晰地写出方程可以减少符号和下标错误。
4. Arithmetic Mean | 等差中项
c mean of its neighbouring terms. That is, if a, b, c are consecutive, then b = (a + c)/2.
两个数 x 和 y 的等差中项是 (x + y)/2。在等差数列中，除首项和末项外，任何一项都是其相邻两项的等差中项。也就是说，如果 a, b, c 是连续三项，那么 b = (a + c)/2。
thmetic mean between 8 and 20, the mean is (8 + 20)/2 = 14, and the sequence is 8, 14, 20.
这个性质在要求在两个给定数之间插入一个或多个等差中项时非常有用。例如，要在 8 和 20 之间插入一个等差中项，中项为 (8 + 20)/2 = 14，因此数列为 8, 14, 20。
erms of an arithmetic sequence with k terms, the common difference is d = (b - a)/(k - 1).
更一般地，如果 a 和 b 是一个有 k 项的等差数列的首项和末项，那么公差为 d = (b - a)/(k - 1)。
5. Sum of the First n Terms | 前 n 项和公式
o standard forms: Sₙ = n/2 [2a + (n - 1)d] and Sₙ = n/2 (a + l), where l is the last term.
等差数列前 n 项的和记作 S
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