📚 Geometric Sequences: Definition, Formulas, Sums and Exam Applications | 几何数列：定义、公式、求和与考试应用
Geometric sequences appear throughout the Edexcel A Level Mathematics specification, particularly in the Pure Mathematics strand, and they also underpin the compound interest and depreciation models used in the Statistics and Mechanics applications of the course. They describe multiplicative patterns in which each term is obtained by multiplying the previous term by a fixed non-zero constant called the common ratio. This article revises the key definitions, the nth term formula, the sum formulas, proof ideas, convergence conditions and typical exam applications.
几何数列在 Edexcel A Level 数学考纲中反复出现，尤其在纯数学部分；它同时是统计与经济应用中复利、折旧等模型的数学基础。几何数列（又称等比数列）描述的是乘法规律：每一项都由前一项乘以一个固定的非零常数（公比）得到。本文梳理关键定义、第 n 项公式、求和公式、证明思路、收敛条件及典型考试应用。
1. Definition and Common Ratio | 定义与公比
A geometric sequence, also called a geometric progression (GP), is a list of numbers such that the ratio between any term and the previous term is constant. If the first term is a and the common ratio is r, the sequence has the form a, ar, ar², ar³, ... , where every term after the first is produced by multiplying the term before it by r.
几何数列又称等比数列（GP），是一列数，其中任意一项与前一项的比值恒定。若首项为 a，公比为 r，则数列形式为 a, ar, ar², ar³, ...，其中除首项之外每一项都由前一项乘以 r 得到。
The common ratio can be found by dividing any term by the term immediately before it:
公比可由任意一项除以其前一项求得：
r = uₙ₊₁ / uₙ = uₙ / uₙ₋₁
m after the first would be zero, which destroys the multiplicative structure and means the sequence is no longer genuinely geometric.
公比可以为正数、负数、整数或分数，但不能为零。若 r = 0，则首项之后的所有项都为零，乘法结构被破坏，数列也就不再是真正的几何数列。
The value of r controls the long-term behaviour of the sequence, and examiners often test this idea directly. The table below summarises the standard cases you should recognise instantly.
r 的取值决定了数列的长期走势，考官常直接考查这一判断。下表总结了必须一眼识别的情形。
| Value of r | r 的取值 | Behaviour | 数列走势 |
|---|---|
| r > 1 | Terms grow without bound | 各项无限增大 |
| 0 < r < 1 | Terms decrease towards zero | 各项递减趋于零 |
| r = 1 | Every term is equal to a | 每项都等于 a |
| r = −1 | Terms alternate a, −a | 各项正负交替 |
| r < 0, r ≠ −1 | Alternating signs, growing or shrinking | 正负交替，数值增减 |
20, −40 has r = −2. Recognising this pattern quickly is worth marks in multiple-choice and "state whether" questions.
公比为负时必定产生正负交替，例如 5, −10, 20, −40 的公比 r = −2。在选择题或"判断并说明"类题目中，快速识别这一模式可以直接得分。
2. The nth Term Formula | 第 n 项公式
For a geometric sequence with first term a and common ratio r, the nth term is given by the formula:
对于首项为 a、公比为 r 的几何数列，第 n 项由以下公式给出：
uₙ = a rⁿ⁻¹
This formula comes from observing the pattern of the sequence term by term. The first term is a with no factor of r, the second term is ar with one factor of r, the third term is ar² with two factors of r, and so on. The number of factors of r is always one less than the term number, so the index on r is n − 1 and never n.
该公式来自逐项观察：首项 a 不含 r，第二项 ar 含一个 r，第三项 ar² 含两个 r，以此类推。r 的个数总是比项号少一，所以 r 的指数是 n − 1，绝不是 n。
In Edexcel questions you may be required to use this formula forwards or backwards. Given a and r you can find any term directly; given two terms you can form simultaneous equations to find a and r; given a, r and a value of uₙ you can solve for the term number n, often using logarithms.
在 Edexcel 考题中，你可能需要正用或反用这个公式：已知 a 和 r，可直接求出任意一项；已知两项，可以建立方程组求 a 与 r；已知 a、r 和某一项的数值，则可以解出项号 n，通常要用对数。
exponent you typed into your calculator is n − 1; this is the single most common slip in the whole topic.
例如 a = 3、r = 2 时，u₇ = 3 × 2⁶ = 3 × 64 = 192。计算器输入指数时务必确认是 n − 1，这是本专题最高频的失误。
3. Finding a and r from Two Terms | 由两项求 a 与 r
using uₙ = a rⁿ⁻¹ and then divide one equation by the other, which eliminates a immediately.
许多考题给出几何数列的两项，要求求 a、r 或某一项。高效做法是用 uₙ = a rⁿ⁻¹ 分别写出这两个已知项，再把两式相除，a 会立即被消去。
ion by the first gives r³ = 27, so r = 3. Substituting back gives a × 3 = 12, hence a = 4.
设 u₂ = 12、u₅ = 324，则 ar = 12、ar⁴ = 324。第二式除以第一式得 r³ = 27，所以 r = 3；代回得 a × 3 = 12，因此 a = 4。
r example, u₂ = 6 and u₄ = 24 give r² = 4, so r = 2 or r = −2, leading to the two possible sequences 3, 6, 12, 24, ... and −3, 6, −12, 24, ... .
若两个项号之差为偶数，相除后得到 r 的偶次幂，此时必须同时考虑正负两个根。例如 u₂ = 6、u₄ = 24 给出 r² = 4，故 r = 2 或 r = −2，对应两个可能数列：3, 6, 12, 24, ... 与 −3, 6, −12, 24, ...。
ed; if the question says the sequence alternates in sign, then r = −2 must be kept instead.
良好的考试习惯是明确写出你拒绝了哪个 r 值以及理由。若题目说明所有项为正，则 r = −2 必须舍去；若说明数列正负交替，则应保留 r = −2。
4. Recurrence Relations and Sigma Notation | 递推关系与求和符号
A geometric sequence is often written as a recurrence relation, which defines each term from the one before it:
几何数列常用递推关系表示，即用前一项定义后一项：
uₙ₊₁ = r uₙ , u₁ = a
This form is useful for questions about modelling or for sequences generated step by step, and it makes the "multiply by the same number each time" idea explicit. It also allows you to check whether a given sequence is geometric by calculating successive ratios.
这种形式适合建模类问题或需要逐步生成的数列，也把"每次都乘以同一个数"这一核心思想表达得更清楚。你也可以通过计算相邻项的比值来判断某数列是否为几何数列。
Series are written using sigma notation, so that a geometric series with n terms is:
级数用求和符号书写，含 n 项的几何级数写作：
Σₖ₌₁ⁿ a rᵏ⁻¹ = a + ar + ar² + ... + arⁿ⁻¹
Recognising that the upper limit tells you how many terms there are is essential. In the expression Σₖ₌₁ⁿ a rᵏ⁻¹ there are exactly n terms, not n − 1, because the counter k runs from 1 to n inclusive.
必须认识到上限表示项数。表达式 Σₖ₌₁ⁿ a rᵏ⁻¹ 中恰好有 n 项，而不是 n − 1 项，因为计数变量 k 从 1 取到 n（含 n）。
5. Sum of a Geometric Series | 几何级数求和
The sum of the first n terms of a geometric series with first term a and common ratio r is given by two equivalent forms, valid whenever r ≠ 1:
首项为 a、公比为 r 的几何级数前 n 项和有两个等价形式，只要 r ≠ 1 就成立：
Sₙ = a(1 − rⁿ) / (1 − r) = a(rⁿ − 1) / (r − 1)
inator positive. The formula fails for r = 1, in which case the series is simply n copies of a and Sₙ = na.
两式结果相同；当 r < 1 时通常用第一个更顺手，当 r > 1 时用第二个，因为可以让分子分母都保持正数。r = 1 时公式失效，此时级数是 n 个 a 之和，Sₙ = na。
For example, to sum 3 + 6 + 12 + ... to 10 terms, take a = 3, r = 2 and n = 10:
例如求 3 + 6 + 12 + ... 前 10 项之和，取 a = 3、r = 2、n = 10：
S₁₀ = 3(2¹⁰ − 1) / (2 − 1) = 3 × 1023 = 3069
+ 20 + ... + 1280, solve 5 × 2ⁿ⁻¹ = 1280, giving 2ⁿ⁻¹ = 256 = 2⁸, so n − 1 = 8 and n = 9.
常见陷阱是当级数写出末项时数错项数。求 5 + 10 + 20 + ... + 1280 的项数，应解 5 × 2ⁿ⁻¹ = 1280，得 2ⁿ⁻¹ = 256 = 2⁸，故 n − 1 = 8，n = 9。
You may also be asked to find how many terms are needed for a sum to exceed a given value. In that case, set up the inequality Sₙ > k, rearrange it into the form rⁿ > something or rⁿ < something, and solve using logarithms.
你还可能被问到需要多少项才能使和超过某值。此时应列出不等式 Sₙ > k，整理成 rⁿ > 某数 或 rⁿ < 某数 的形式，再用对数求解。
6. Proof of the Sum Formula | 求和公式的证明
Edexcel regularly asks for a proof of the sum formula, so you should be able to reproduce it from memory without hesitation. The standard proof uses a subtraction trick, and the same method underlies many later results in the specification.
Edexcel 经常要求证明求和公式，因此你必须能毫不犹豫地默写出来。标准证明使用相减技巧，这一方法也是后续许多结论的基础。
Start by writing the sum and then multiplying it by r:
先写出和式，再把它乘以 r：
Sₙ = a + ar + ar² + ... + arⁿ⁻¹
r Sₙ = ar + ar² + ar³ + ... + arⁿ
Now subtract the second line from the first. All the middle terms cancel in pairs, which is the whole point of the trick, and you are left with a difference of just two terms:
现在用第一式减去第二式。中间各项成对消去，这正是该技巧的关键，最后只剩下两项之差：
Sₙ − r Sₙ = a − arⁿ
es, giving Sₙ(1 − r) = a(1 − rⁿ), and divide by (1 − r), which is permitted because r ≠ 1:
两边因式分解得 Sₙ(1 − r) = a(1 − rⁿ)，再除以 (1 − r)，因为 r ≠ 1 所以允许这样做：
Sₙ = a(1 − rⁿ) / (1 − r)
When you write this proof in an examination, state explicitly that the middle terms cancel and that the division requires r ≠ 1. These two explanatory sentences are usually worth method marks even if an algebra slip occurs later.
考试中书写此证明时，要明确写出"中间项消去"以及"除法要求 r ≠ 1"。即使后面代数出错，这两句说明通常也能拿到方法分。
7. Sum to Infinity and Convergence | 无穷级数与收敛性
If the terms of a geometric series become smaller and smaller, the total of infinitely many terms can still be a finite number. This is only possible when the common ratio satisfies the convergence condition:
如果几何级数的各项越来越小，那么无穷多项之和仍可能是有限值。这只有在公比满足收敛条件时才可能发生：
|r| < 1 , that is −1 < r < 1
Under this condition rⁿ tends to zero as n tends to infinity, so the term rⁿ in the sum formula disappears and the sum to infinity is:
在此条件下，当 n 趋于无穷时 rⁿ 趋于零，求和公式中的 rⁿ 项消失，无穷级数之和为：
S∞ = a / (1 − r)
If |r| the sum to infinity does not exist; the series is said to diverge. A common exam instruction is "given that the series converges", which is your signal that |r| < 1 may be used to reject a value of r.
若 |r| ≥ 1，各项不会趋于零，无穷级数之和不存在，称级数发散。考题常写"已知该级数收敛"，这正是你可以用 |r| < 1 来排除某个 r 值的信号。
.4444... is geometric with a = 0.4 and r = 0.1, so S∞ = 0.4 / (1 − 0.1) = 0.4 / 0.9 = 4/9.
无穷级数求和还可把循环小数化为精确分数。例如 0.4444... 是首项 a = 0.4、公比 r = 0.1 的几何级数，故 S∞ = 0.4 / (1 − 0.1) = 0.4 / 0.9 = 4/9。
1000 and r = 0.5 still converges, while a series with a = 0.001 and r = 2 still diverges.
一个值得记住的细节：收敛性只取决于 r，与 a 无关。a = 1000、r = 0.5 的级数仍然收敛，而 a = 0.001、r = 2 的级数仍然发散。
8. Solving for n Using Logarithms | 用对数求 n
Take logs of both sides and use the power law log(xᵏ) = k log x, which turns the exponent into a multiplier.
当未知量是项号 n 而不是某项的值时，方程 uₙ = a rⁿ⁻¹ 就是指数方程，需要用到对数。两边取对数并使用幂法则 log(xᵏ) = k log x，把指数变成乘数。
Rearranging the general case gives an explicit formula for the term number:
整理一般情形，可得到项号的显式公式：
n = log(uₙ / a) / log r + 1
iving n − 1 > 9.846 and therefore n = 11. Checking: u₁₀ = 39366 and u₁₁ = 118098, so the
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