📚 Mastering Linear Sequences: Finding the nth Term | 掌握线性数列：求第n项
A sequence is a list of numbers arranged in a specific order. Each number in the list is called a term. Sequences appear everywhere in mathematics, from simple counting patterns to the Fibonacci numbers found in nature. In KS3 Cambridge Mathematics, you learn to recognise different types of sequences, describe how they grow using rules, and find a formula for the nth term. This skill helps you predict any term in the sequence without having to write out all the preceding ones.
数列是按特定顺序排列的一组数。数列中的每一个数称为项。数列在数学中无处不在，从简单的计数规律到自然界中的斐波那契数列。在KS3剑桥数学中，你将学习识别不同类型的数列，用规则描述它们的增长方式，并求通项公式。这项技能能帮助你预测数列中的任意项，而无需写出前面所有的项。
1. What Is a Sequence? | 什么是数列？
A sequence is an ordered collection of numbers. For example, 2, 4, 6, 8, 10 is a sequence of the first five even numbers. The order matters: 2, 6, 4, 8, 10 would be a different sequence. Sequences can be finite (with a fixed number of terms) or infinite (continuing forever). Each number is called a term, and we often refer to the first term as a₁, the second as a₂, and the nth term as aₗ.
数列是有序的数的集合。例如，2, 4, 6, 8, 10是前五个偶数的数列。顺序很重要：2, 6, 4, 8, 10会是一个不同的数列。数列可以是有限的（有固定数量的项）或无限的（一直延续下去）。每个数称为一项，我们通常将第一项记作a₁，第二项记作a₂，第n项记作aₗ。
2. Term-to-Term Rules | 项与项之间的规则
A term-to-term rule tells you how to get from one term to the next. For the sequence 3, 7, 11, 15, the rule is 'add 4' because each term is 4 more than the one before. This type of rule is useful for generating the next few terms, but it becomes inefficient if you need the 100th term – you would have to apply the rule 99 times. Term-to-term rules can also involve subtraction, multiplication, or division.
项间规则告诉你如何从一项得到下一项。对于数列3, 7, 11, 15，规则是‘加4’，因为每一项比前一项多4。这种规则在生成后续几项时很有用，但如果你需要第100项，效率就很低——你必须重复应用规则99次。项间规则也可以涉及减法、乘法或除法。
3. Position-to-Term Rules | 位置与项的规则
g through the sequence, you simply plug the position into a formula. For example, the rule 'multiply n by 5 and subtract 2' written as a&# 8, 13, 18, ... This is much more powerful because you can find the 50th term instantly: a₅₀ = 5×50 − 2 = 248.
位置与项的规则直接将项与其位置编号n联系起来。你不必逐步计算数列，只需将位置代入公式即可。例如，规则‘n乘以5再减去2’写成aₗ = 5n − 2，得到数列3, 8, 13, 18, ... 这种方法强大得多，因为你可以立即求出第50项：a₅₀ = 5×50 − 2 = 248。
4. Understanding Linear Sequences | 理解线性数列
e, d. Because the terms increase by a steady amount, the nth term is always of the form a&# = 2, the sequence goes up in steps of 2; if d = -3, the sequence decreases by 3 each time.
线性数列是指相邻两项之差恒定的数列。这个恒定的差称为公差d。因为项以稳定的量增加，所以第n项总具有aₗ = dn + c的形式，其中c是一个固定数。d的值决定步长；c的值调整起点。例如，如果d = 2，数列以2为步长递增；如果d = -3，数列每次减少3。
5. Finding the nth Term of a Linear Sequence | 求线性数列的第n项
few terms: 5, 8, 11, 14. The difference is +3, so d = 3. The nth term can be written as aₗ = 3n + c. To find c, use the first term: when n = 1, a = 5. Substitute into the formula: 5 = 3(1) + c, so c = 2. Therefore the nth term is aₗ = 3n + 2. You can check with n = 2: 3(2)+2 = 8, which matches the second term.
要找到通项公式，首先确定公差d。观察前几项：5, 8, 11, 14。差是+3，因此d = 3。通项可写作aₗ = 3n + c。为求c，利用第一项：当n = 1时，a₁ = 5。代入公式：5 = 3(1) + c，得c = 2。因此通项为aₗ = 3n + 2。你可以用n = 2检验：3(2)+2 = 8，与第二项吻合。
aₗ = 3n + 2
tant term c = 2 directly. Both methods are reliable and help you cross-check your formula.
第二种方法利用第零项。在项的上方标出位置编号：n = 1对应项5，n = 2对应8，n = 3对应11。如果往回退一步到n = 0，你需减去公差：5 − 3 = 2。这就直接得到了常数项c = 2。两种方法都可靠，并可用来交叉验证你的公式。
6. Using the nth Term to Find Any Term | 利用通项求任意项
ormula, you can calculate the value of any term without listing all previous terms. For aₗ = 4n − 1, what is the 25th term? Substitute n = 25: a₂₅ = 4×25 − 1 = 99. You can also work backwards: if a term is 103, which position is it? Set 4n − 1 = 103, solve for n: 4n = 104, n = 26. So 103 is the 26th term.
一旦你有了通项公式，就能求出任意一项的值，而无需列出所有前面的项。对于aₗ = 4n − 1，第25项是多少？代入n = 25：a₂₅ = 4×25 − 1 = 99。你还可以倒推：如果某一项是103，它是第几项？设4n − 1 = 103，解出n：4n = 104，n = 26。所以103是第26项。
7. Generating Sequences from the nth Term | 从通项公式生成数列
ven a formula, you can produce the first few terms by substituting n = 1, 2, 3, ... For a&# he terms are 2, 8, 18, 32, ... This is a quadratic sequence because of the n² term. For a&# n ratio of 2. Working between formula and sequence builds fluency in recognising patterns.
给定一个公式，你可以通过代入n = 1, 2, 3, ...生成前几项。对于aₗ = 2n²，项为2, 8, 18, 32, ... 这是一个二次数列，因为含有n²项。对于aₗ = 3×2ⁿ⁻¹，n = 1时得3，n = 2得6，n = 3得12。这是一个公比为2的等比数列。在公式与数列之间来回练习，可以增强你识别规律的熟练度。
8. Special Sequences | 特殊数列
Some sequences have famous names. The triangle numbers 1, 3, 6, 10, ... are given by aₗ = n(n+1)/2. The square numbers 1, 4, 9, 16, ... are simply a&# e are not linear, recognising them deepens your understanding of how sequences can behave.
一些数列有著名的名称。三角形数1, 3, 6, 10, ... 由aₗ = n(n+1)/2给出。平方数1, 4, 9, 16, ... 就是aₗ = n²。斐波那契数列从1, 1, 2, 3, 5, 8, ...开始，其中每一项是前两项之和。尽管它们不是线性的，认识它们可以加深你对数列变化方式的理解。
9. Real-Life Applications | 实际应用
Sequences model many real-world patterns. For example, if you save £5 in week 1, £10 in week 2, £15 in week 3, your savings form an arithmetic sequence with a&# ks is a sum of the sequence. Another example: the number of matchsticks needed to make a row of n squares follows the rule 1 + 3n (or 4 + 3(n-1)). Understanding the nth term helps you generalise the pattern for any number of squares.
数列为许多现实规律建立模型。例如，如果你第1周存5英镑，第2周存10英镑，第3周存15英镑，你的储蓄形成一个通项为aₗ = 5n的等差数列。n周后的总存款是数列的和。另一个例子：搭建一排n个正方形所需的火柴棍数量遵循规则1 + 3n（或4 + 3(n-1)）。理解通项能帮助你将规律推广到任意数量的正方形。
10. Common Misconceptions | 常见误解
Students often confuse the common difference with the first term. Just because the gap is 3 does not mean the formula is 3n; you must find the constant c. Another mistake is forgetting that n starts at 1, not 0. Using a table with n and term values helps avoid these errors. Always test your formula on the second and third terms to confirm it works.
学生经常将公差与第一项混淆。差为3并不意味着公式就是3n；你必须求出常数c。另一个错误是忘记n从1开始，而非0。使用列出n和项值的表格有助于避免这些错误。始终用第二项和第三项检验你的公式，确认它有效。
11. Practice: Find the nth Term | 练习：求第n项
Try this example: Sequence 4, 10, 16, 22, ...
- Step 1: Find the common difference. 10 − 4 = 6, 16 − 10 = 6, so d = 6.
- Step 2: Write aₗ = 6n + c.
- Step 3: Use the first term n = 1, a₁ = 4. So 4 = 6(1) + c, giving c = −2.
- Step 4: nth term is aₗ = 6n − 2.
- Check: For n = 2, 6×2 − 2 = 10 correct; n = 3, 6×3 − 2 = 16 correct.
尝试这个例子：数列 4, 10, 16, 22, ...
- 步骤1：求公差。10 − 4 = 6, 16 − 10 = 6，所以d = 6。
- 步骤2：写出aₗ = 6n + c。
- 步骤3：利用第一项n = 1, a₁ = 4。因此4 = 6(1) + c，得c = −2。
- 步骤4：通项为aₗ = 6n − 2。
- 检验：n = 2时，6×2 − 2 = 10正确；n = 3时，6×3 − 2 = 16正确。
Now try with a descending sequence: 20, 17, 14, 11, ... The common difference is −3, so aₗ = −3n + c. Using first term: 20 = −3(1) + c gives c = 23. So nth term is a&# n. Always write the formula with the positive constant followed by the n term for clarity.
现在尝试一个递减的数列：20, 17, 14, 11, ... 公差是−3，所以aₗ = −3n + c。利用第一项：20 = −3(1) + c，得c = 23。因此通项为aₗ = 23 − 3n。为了清晰，通常将正常数写在前，n项写在后。
12. Summary and Tips | 总结与技巧
ar sequences, remember these key points: the nth term of a linear sequence has the form a&# table to organise your position and term data. Double-check by testing two different terms. Recognising linear patterns quickly is a fundamental skill that leads on to quadratic sequences and beyond in KS4 and IGCSE Mathematics.
要掌握线性数列，记住这些关键点：线性数列的通项具有aₗ = dn + c的形式，其中d是公差，c是n = 0时的值。使用表格来组织位置和项的数据。通过检验两个不同的项进行复查。快速识别线性规律是一项基本技能，将为KS4和IGCSE数学中的二次数列及更高阶内容铺平道路。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图