📚 A-Level Mathematics: Geometric Sequences: Concepts and General Term | A-Level 数学：等比数列的概念与通项公式
A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. This simple rule creates patterns that appear throughout mathematics, physics, finance, and biology. In this article, we will explore the definition, the general term formula, properties, and practical applications of geometric sequences, with clear examples designed for A-Level students.
等比数列是一种数列，其中从第二项起，每一项与它的前一项的比等于同一个固定且非零的常数，这个常数称为公比。这一简单规则所生成的模式广泛出现在数学、物理、金融和生物学中。本文将围绕等比数列的定义、通项公式、性质以及实际应用展开，结合清晰的例子，帮助 A-Level 学生掌握核心考点。
1. What Is a Geometric Sequence? | 什么是等比数列？
A geometric sequence is a list of numbers such that the ratio between consecutive terms is constant. For example, the sequence 2, 6, 18, 54, … is geometric, because dividing any term by the previous term always gives 3. This constant ratio is usually denoted by the letter r.
等比数列是一列数，其中任意相邻两项之比为常数。例如数列 2, 6, 18, 54, … 是等比数列，因为任意项除以前一项都等于 3。这个常数比称为公比，通常用字母 r 表示。
The key feature of a geometric sequence is multiplicative growth. Unlike an arithmetic sequence, which grows by adding a fixed number, a geometric sequence grows by multiplying by a fixed factor. This means that geometric sequences can increase rapidly or decrease rapidly, depending on whether the common ratio is greater than 1 or between 0 and 1.
等比数列的关键特征是“倍增”式的变化。与通过加上固定数值增长的等差数列不同，等比数列通过乘以固定因子来增长。因此，当公比大于 1 时，数列会快速增大；当公比在 0 和 1 之间时，数列会快速减小。
2. Formal Definition and Common Ratio | 形式定义与公比
A sequence \(a_1, a_2, a_3, \ldots\) is called a geometric sequence if there exists a constant \(r\) such that for every integer \(n \ge 1\),
如果一个数列 \(a_1, a_2, a_3, \ldots\) 满足：对于每个整数 \(n \ge 1\)，都存在常数 \(r\)，使得
aₙ₊₁ = aₙ × r
for all \(n \ge 1\), then the sequence is geometric. The value \(r\) is called the common ratio. We can calculate \(r\) by dividing any term by its preceding term:
对所有 \(n \ge 1\) 成立，那么该数列就是等比数列，其中 \(r\) 称为公比。我们可以通过任一项除以其前一项来计算公比：
r = aₙ₊₁ ÷ aₙ
It is important to note that the initial term \(a_1\) cannot be zero for a meaningful geometric sequence, because if \(a_1 = 0\), all subsequent terms will also be zero, and the sequence would be trivial. In most A-Level questions, \(a_1 \neq 0\) and \(r \neq 0\).
需要特别注意，等比数列的首项 \(a_1\) 不能为零。如果 \(a_1 = 0\)，那么后续所有项都为零，数列将失去研究意义。在大多数 A-Level 考题中，默认 \(a_1 \neq 0\) 且 \(r \neq 0\)。
3. The General Term Formula | 通项公式
The general term of a geometric sequence can be derived by observing the pattern. Starting from the first term \(a_1\):
等比数列的通项公式可以通过观察规律来推导。从首项 \(a_1\) 开始：
a₂ = a₁r, a₃ = a₁r², a₄ = a₁r³, …
We see that the exponent of \(r\) is always one less than the term number. Therefore, the \(n\)th term of a geometric sequence is given by:
我们发现 \(r\) 的指数总是比项号少 1。因此，等比数列的第 \(n\) 项为：
aₙ = a₁ × rⁿ⁻¹
where \(n\) is a positive integer. This formula is extremely powerful: it allows us to find any term directly without listing all previous terms. For example, if \(a_1 = 5\) and \(r = 2\), then the 10th term is \(a_{10} = 5 \times 2^{9} = 2560\).
其中 \(n\) 为正整数。这个公式非常强大：它使我们无需逐项列举，就能直接求出任意一项。例如，若 \(a_1 = 5\)，\(r = 2\)，则第 10 项为 \(a_{10} = 5 \times 2^{9} = 2560\)。
Sometimes we need to find the general term from two given terms. Suppose \(a_3 = 20\) and \(a_6 = 160\). We can write two equations in terms of \(a_1\) and \(r\):
有时我们需要从已知的两项中求出通项。假设 \(a_3 = 20\)，\(a_6 = 160\)。我们可以列出两个方程，用 \(a_1\) 和 \(r\) 表示：
a₃ = a₁r² = 20, a₆ = a₁r⁵ = 160
By dividing the second equation by the first, we eliminate \(a_1\):
用第二个方程除以第一个方程，可消去 \(a_1\)：
r³ = 160 ÷ 20 = 8 ⇒ r = 2
Then substituting back gives \(a_1 = 20 ÷ 2² = 5\). So the general term is \(a_n = 5 \times 2^{n-1}\).
代回得 \(a_1 = 20 ÷ 2² = 5\)。因此通项为 \(a_n = 5 \times 2^{n-1}\)。
4. Finding Terms and the Common Ratio | 求项与公比
In many exam questions, you are given one or two terms and asked to find the common ratio, a specific term, or the position of a term. The key is to use the general term formula \(a_n = a_1 r^{n-1}\) and solve the resulting equation.
在 많은考题中，你会被给出一项或两项，然后要求求公比、某一项或某项的位置。关键是使用通项公式 \(a_n = a_1 r^{n-1}\) 并解所得方程。
Example: The 4th term of a geometric sequence is 54 and the 7th term is 1458. Find the common ratio and the first term.
例：某等比数列的第 4 项为 54，第 7 项为 1458。求公比和首项。
a₄ = a₁r³ = 54, a₇ = a₁r⁶ = 1458
Dividing the second equation by the first gives \(r^3 = 1458 ÷ 54 = 27\), so \(r = 3\). Then \(a_1 \times 27 = 54\), so \(a_1 = 2\). The general term is \(a_n = 2 \times 3^{n-1}\).
第二个方程除以第一个方程得 \(r^3 = 1458 ÷ 54 = 27\)，所以 \(r = 3\)。于是 \(a_1 \times 27 = 54\)，故 \(a_1 = 2\)。通项为 \(a_n = 2 \times 3^{n-1}\)。
When \(r\) is unknown but the ratio of two terms is given, we often have to calculate \(r\) by considering the difference in term positions. If \(a_m\) and \(a_n\) are known, then \(a_m / a_n = r^{m-n}\). This is a faster method than writing out all equations.
当公比未知但已知两项之比时，我们通常通过两项位置之差来求 \(r\)。若已知 \(a_m\) 和 \(a_n\)，则 \(a_m / a_n = r^{m-n}\)。这比列出所有方程更快。
5. Key Properties of Geometric Sequences | 等比数列的重要性质
Geometric sequences have several useful properties that often appear in A-Level problems.
等比数列具有若干有用性质，这些性质经常出现在 A-Level 题目中。
Property 1: Geometric mean. If \(x\), \(y\) and \(z\) are three consecutive terms of a geometric sequence, then \(y^2 = xz\). More generally, \(y\) is called the geometric mean of \(x\) and \(z\). For example, in the sequence 4, 12, 36, we have \(12^2 = 4 \times 36 = 144\).
性质 1：等比中项。 如果 \(x\), \(y\), \(z\) 是等比数列的连续三项，则 \(y^2 = xz\)。更一般地，\(y\) 称为 \(x\) 和 \(z\) 的等比中项。例如，在数列 4, 12, 36 中，有 \(12^2 = 4 \times 36 = 144\)。
Property 2: Multiplication property. For any positive integers \(m\) and \(n\), we have \(a_m \times a_n = a_1^2 r^{m+n-2}\). This is not a simple single-term expression, but it is often easier to use the formula \(a_m = a_1 r^{m-1}\) directly in calculations.
性质 2：乘积性质。 对于任意正整数 \(m\) 和 \(n\)，有 \(a_m \times a_n = a_1^2 r^{m+n-2}\)。这并不直接变成一项，但在计算中通常直接使用 \(a_m = a_1 r^{m-1}\) 更方便。
Property 3: Sub-sequence. If we take every \(k\)th term of a geometric sequence, the resulting sequence is also geometric, with common ratio \(r^k\). For example, if we take terms 2, 4, 8, … (multiples of 2?) Actually from a geometric sequence with ratio \(r\), the terms at positions 1, 3, 5, … form a new geometric sequence with common ratio \(r^2\).
性质 3：子数列。 等比数列中每隔 \(k\) 项取一项，所得到的新数列仍为等比数列，公比为 \(r^k\)。例如，从公比为 \(r\) 的等比数列中取出第 1、3、5、… 项，它们构成公比为 \(r^2\) 的等比数列。
Property 4: Taking logarithms. If all terms of a geometric sequence are positive, then taking logarithms of all terms produces an arithmetic sequence. Indeed, \(\log(a_n) = \log(a_1) + (n-1)\log(r)\), which is linear in \(n\). This property is often used in applications to transform multiplicative data into additive data.
性质 4：取对数。 若等比数列各项均为正，则对各项取对数后，所得新数列是等差数列。事实上，\(\log(a_n) = \log(a_1) + (n-1)\log(r)\)，它是 \(n\) 的线性函数。该性质常用于应用问题中，将乘法关系转化为加法关系。
6. Applications in Real Life | 实际应用
Geometric sequences model many real-world situations where a quantity changes by a constant percentage. Here are three classic examples.
等比数列常用于建模那些以固定百分比变化的量。以下是三个经典例子。
Population growth: If a population of bacteria doubles every hour, then after \(n\) hours the population is given by \(P_n = P_0 \times 2^n\), where \(P_0\) is the initial population. This is a geometric sequence with common ratio 2.
种群增长： 如果细菌种群每小时翻倍，那么经过 \(n\) 小时后，种群数量为 \(P_n = P_0 \times 2^n\)，其中 \(P_0\) 是初始数量。这是一个公比为 2 的等比数列。
Compound interest: When money is invested at an annual interest rate of \(i\) (expressed as a decimal), the amount after \(n\) years is \(A_n = A_0(1+i)^n\). Here the common ratio is \(1+i\). For example, an initial investment of £1000 at 5% interest yields after 10 years: \(1000 \times 1.05^{10} \approx 1628.89\) pounds.
复利： 当资金以年利率 \(i\)（小数形式）投资时，\(n\) 年后的金额为 \(A_n = A_0(1+i)^n\)，公比为 \(1+i\)。例如，初始投资 1000 英镑，年利率 5%，10 年后所得为 \(1000 \times 1.05^{10} \approx 1628.89\) 英镑。
Radioactive decay: The mass of a radioactive substance decays by a constant fraction per unit time. If the half-life is \(h\) years, then after \(t\) years the remaining mass is \(M_t = M_0 \times (1/2)^{t/h}\). This is a geometric sequence with common ratio \(1/2\) when measured at each half-life.
放射性衰变： 放射性物质的质量在单位时间内按恒定比例减少。若半衰期为 \(h\) 年，则经过 \(t\) 年后剩余质量为 \(M_t = M_0 \times (1/2)^{t/h}\)。若按半衰期逐期测量，它就是一个公比为 \(1/2\) 的等比数列。
7. Solving for n Using Logarithms | 利用对数求项数 n
Sometimes a question asks: “Which term of the sequence is equal to a given number?” In such cases, the unknown \(n\) appears as an exponent, and we need to use logarithms.
有时题目会问：“该数列的哪一项等于某个给定数？”此时未知数 \(n\) 出现在指数位置，我们需要借助对数来求解。
Example: Given a geometric sequence with \(a_1 = 3\) and \(r = 2\), find the value of \(n\) such that \(a_n = 192\).
例：已知等比数列的首项 \(a_1 = 3\)，公比 \(r = 2\)，求使得 \(a_n = 192\) 的 \(n\) 值。
3 × 2ⁿ⁻¹ = 192 ⇒ 2ⁿ⁻¹ = 64
Since \(64 = 2^6\), we have \(n – 1 = 6\), so \(n = 7\). Here we did not need logarithms because we recognized the power of 2. But if the number is not an integer power, we would take logs:
因为 \(64 = 2^6\)，所以 \(n – 1 = 6\)，即 \(n = 7\)。此处我们无需对数，因为能直接看出是 2 的幂。但如果目标数不是整数次幂，我们就需要取对数：
For example, solve \(3 \times 1.1^{n-1} = 10\). Dividing by 3 gives \(1.1^{n-1} = 10/3\). Taking natural logs: \((n-1)\ln(1.1) = \ln(10/3)\), so \(n = 1 + \ln(10/3)/\ln(1.1) \approx 13.6\). Since \(n\) must be an integer, the 14th term would exceed 10.
例如，解 \(3 \times 1.1^{n-1} = 10\)。两边除以 3 得 \(1.1^{n-1} = 10/3\)。取自然对数：\((n-1)\ln(1.1) = \ln(10/3)\)，所以 \(n = 1 + \ln(10/3)/\ln(1.1) \approx 13.6\)。由于 \(n\) 必须是整数，第 14 项将超过 10。
8. Common Mistakes and Exam Tips | 常见错误与应试技巧
Students often make mistakes when dealing with geometric sequences. Here are some important points to remember.
学生在处理等比数列时经常犯错。以下是一些重要的注意事项。
Mistake 1: Confusing arithmetic and geometric sequences. In an arithmetic sequence you add a common difference \(d\); in a geometric sequence you multiply by a common ratio \(r\). Always check whether the difference between consecutive terms is constant (arithmetic) or the ratio is constant (geometric).
错误 1：混淆等差数列与等比数列。 等差数列逐项加上公差 \(d\)；等比数列逐项乘以公比 \(r\)。务必检查相邻两项之差是否恒定（等差），还是相邻两项之比是否恒定（等比）。
Mistake 2: Incorrect exponent in the general term. The \(n\)th term formula is \(a_n = a_1 r^{n-1}\), not \(a_1 r^n\). Check with small \(n\): when \(n=1\), the exponent should be 0, giving \(a_1\).
错误 2：通项公式中指数写错。 第 \(n\) 项的公式是 \(a_n = a_1 r^{n-1}\)，而不是 \(a_1 r^n\)。可用小 \(n\) 验证：当 \(n=1\) 时，指数应为 0，得到 \(a_1\)。
Mistake 3: Forgetting that \(r\) can be negative or fractional. If \(r\) is negative, terms alternate in sign. If \(0 < r < 1\), the absolute values decrease. Do not ignore these cases in word problems.
错误 3：忘记公比 \(r\) 可以是负数或分数。 若 \(r\) 为负，数列正负交替；若 \(0 < r < 1\)，绝对项递减。在应用题中不要忽略这些情况。
Exam tip: When given two non-consecutive terms, always set up two equations in \(a_1\) and \(r\), then divide to eliminate \(a_1\). When asked to find the number of terms \(n\), convert the equation to a power form or use logarithms. Finally, always check that your value of \(r\) is consistent with both given terms.
应试技巧： 当已知两个不相邻的项时，先列出关于 \(a_1\) 和 \(r\) 的两个方程，然后相除消去 \(a_1\)。当要求项数 \(n\) 时，将方程化为幂的形式或使用对数。最后，务必验证所求的 \(r\) 是否与两项一致。
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