📚 Example 3.4.2: Geometric Sequences and Series | 例题3.4.2：等比数列与级数
This worked example focuses on a classic AQA A-Level Mathematics question: given two terms of a geometric sequence, find the first term, the common ratio, and the sum of the first six terms. We will solve the problem step by step, explain the underlying theory, and highlight common pitfalls.
本例题聚焦于一道经典的 AQA A-Level 数学问题：已知等比数列中的两项，求首项、公比以及前六项之和。我们将一步步解题，解释相关理论，并指出常见易错点。
1. Problem Statement | 题目条件
Let \(\{a_n\}\) be a geometric sequence with positive common ratio \(r\). Given that \(a_2 = 6\) and \(a_5 = 48\), find \(a_1\), \(r\), \(a_6\), and the sum of the first six terms \(S_6\).
设 \(\{a_n\}\) 是一个公比 \(r\) 为正数的等比数列。已知 \(a_2 = 6\)，\(a_5 = 48\)，求 \(a_1\)、\(r\)、\(a_6\) 以及前六项的和 \(S_6\)。
2. Recall the Definitions | 回顾定义
A geometric sequence is one in which each term after the first is obtained by multiplying the previous term by a fixed constant \(r\), called the common ratio. The \(n\)th term is given by:
等比数列是指数列中从第二项起，每一项与其前一项的比等于同一个常数 \(r\) 的数列，\(r\) 称为公比。其通项公式为：
aₙ = a₁rⁿ⁻¹
The sum of the first \(n\) terms of a geometric sequence, denoted \(S_n\), is:
等比数列前 \(n\) 项和记作 \(S_n\)，其公式为：
Sₙ = a₁(1 − rⁿ) / (1 − r) (r ≠ 1)
We will use these formulas to solve the problem.
我们将应用这些公式来解决问题。
3. Write Equations from Given Information | 根据已知条件列方程
From the given terms, we can write:
根据已知的项，我们可以列出：
a₂ = a₁r = 6
a₅ = a₁r⁴ = 48
These two equations will allow us to find both \(a_1\) and \(r\).
这两个方程将帮助我们求出 \(a_1\) 和 \(r\)。
4. Eliminate a₁ to Find r | 消去 a₁ 求公比
Divide the equation for \(a_5\) by the equation for \(a_2\). Since \(a_1\) is nonzero (otherwise both terms would be zero), cancellation is valid:
用 \(a_5\) 的方程除以 \(a_2\) 的方程。由于 \(a_1\) 不为零（否则这两项都为零），因此可以约去 \(a_1\)：
a₅ / a₂ = (a₁r⁴) / (a₁r) = r³
Substituting the numerical values gives:
代入数值得到：
48 / 6 = r³ ⇒ 8 = r³
Since \(r > 0\), the real cube root of 8 is 2. Hence \(r = 2\).
因为 \(r > 0\)，8 的实数立方根为 2，所以 \(r = 2\)。
5. Find the First Term a₁ | 求首项 a₁
Now substitute \(r = 2\) back into the first equation \(a₁r = 6\):
现在将 \(r = 2\) 代回第一个方程 \(a₁r = 6\)：
a₁ × 2 = 6 ⇒ a₁ = 3
Therefore, the first term is 3. We can quickly check using the second equation: \(a₁r⁴ = 3 × 2⁴ = 3 × 16 = 48\), which matches the given information.
因此，首项为 3。我们可以用第二个方程快速验证：\(a₁r⁴ = 3 × 2⁴ = 3 × 16 = 48\)，与已知条件一致。
6. Write the General Term | 写出通项公式
With \(a₁ = 3\) and \(r = 2\), the general term is:
由 \(a₁ = 3\)，\(r = 2\)，通项公式为：
aₙ = 3 × 2ⁿ⁻¹
This compact expression allows us to find any term of the sequence without listing all previous terms.
这个简洁的表达式让我们不必逐一列出前面的项就能求出任意一项。
7. Find the Sixth Term a₆ | 求第六项 a₆
Using the general term with \(n = 6\):
利用通项公式，令 \(n = 6\)：
a₆ = 3 × 2⁵ = 3 × 32 = 96
Thus, the sixth term is 96.
因此，第六项为 96。
8. Calculate the Sum of the First Six Terms | 计算前六项之和
Use the sum formula with \(n = 6\), \(a₁ = 3\), and \(r = 2\):
使用求和公式，其中 \(n = 6\)，\(a₁ = 3\)，\(r = 2\)：
S₆ = a₁(1 − r⁶) / (1 − r) = 3(1 − 2⁶) / (1 − 2)
Since \(2⁶ = 64\), we have:
因为 \(2⁶ = 64\)，我们有：
S₆ = 3(1 − 64) / (1 − 2) = 3(−63) / (−1) = 189
So the sum of the first six terms is 189.
因此，前六项之和为 189。
9. Verify by Explicit Summation | 通过显式求和验证
We can verify the result by listing the first six terms: 3, 6, 12, 24, 48, 96. Their sum is 3 + 6 + 12 + 24 + 48 + 96 = 189. This confirms our formula-based calculation.
我们可以通过列出前六项来验证：3, 6, 12, 24, 48, 96。它们的和为 3 + 6 + 12 + 24 + 48 + 96 = 189。这验证了基于公式的计算结果。
Notice that each term doubles the previous one, so the sequence grows rapidly. This is characteristic of geometric sequences with \(|r| > 1\).
注意每一项都是前一项的两倍，因此数列增长迅速。这是 \(|r| > 1\) 时等比数列的典型特征。
10. Extensions and Common Mistakes | 拓展与常见错误
One common mistake is confusing the exponent in the general term. For example, \(a₅ = a₁r⁴\), not \(a₁r⁵\). Always recall that the first term corresponds to \(n = 1\), so the exponent of \(r\) is \(n – 1\).
一个常见错误是混淆通项中的指数。例如，\(a₅ = a₁r⁴\)，而不是 \(a₁r⁵\)。一定要记住首项对应 \(n = 1\)，因此 \(r\) 的指数为 \(n – 1\)。
Another mistake is applying the sum formula when \(r = 1\). In that case, every term equals \(a₁\), and \(Sₙ = n a₁\). The formula \(a₁(1−rⁿ)/(1−r)\) is undefined for \(r = 1\), so it must be handled separately.
另一个错误是在 \(r = 1\) 时使用求和公式。此时每一项都等于 \(a₁\)，\(Sₙ = n a₁\)。公式 \(a₁(1−rⁿ)/(1−r)\) 在 \(r = 1\) 时无定义，必须单独处理。
If we were asked for the sum to infinity, we would need \(|r| < 1\). In this example, \(r = 2 > 1\), so the sum to infinity diverges. For instance, if \(r = 1/2\) and \(a₁ = 4\), then \(S_∞ = a₁ / (1 − r) = 4 / (1 − 1/2) = 8\).
如果题目要求无穷级数和，则需要 \(|r| < 1\)。在本例中，\(r = 2 > 1\)，因此无穷级数发散。例如，若 \(r = 1/2\)，\(a₁ = 4\)，则 \(S_∞ = a₁ / (1 − r) = 4 / (1 − 1/2) = 8\)。
Finally, always check the sign of \(r\) when taking odd roots. Since we are given \(r > 0\), we take the positive root. Without that condition, \(r = 2\) would be the only real root of \(r³ = 8\), but in general, signs matter.
最后，在取奇次根时要注意 \(r\) 的符号。本题给出 \(r > 0\)，因此取正根。若无此条件，\(r³ = 8\) 只有实根 \(r = 2\)，但在一般情况下，符号很重要。
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