Arithmetic sequences and nth terms · Foundation
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| common difference/ˈkɒmən ˈdɪfrəns/ | 公差 | gōng chāi |
Does the change add or multiply?
- A saving plan adds ¥30 more each week; a population model grows by 5% each year. Equal differences and equal ratios need different models.
- This lesson studies common difference 公差: The constant added between consecutive terms of an arithmetic sequence.
这种变化是相加还是相乘？
- 储蓄计划每周增加¥30；人口模型每年增长5%。等差与等比变化需要不同的数学模型。
- 本课程学习公差：等差数列中相邻两项之间恒定的差值。
Choose the mathematical structure
- Find a constant difference for an arithmetic sequence. Its nth term is a+(n-1)d. A term-to-term rule describes how to reach the next term; a position-to-term rule gives a term directly.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
选择数学结构
- 求等差数列的恒定公差。其第 n 项公式为 a+(n-1)d。递推规则描述如何得到下一项；通项规则可直接给出任意一项的值。
- 计算前请先明确允许的输入项和单位。方程应表达关系本身，而不仅仅是记录计算器按键过程。
Which description correctly defines common difference?
The constant added between consecutive terms of an arithmetic sequence.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For 5,8,11,14,... the common difference is 3. The nth term is 5+3(n-1)=3n+2. At n=8, u₈=26. To find the position of 62, solve 3n+2=62, giving n=20.
通过验证案例进行推导
- 将结果与初始量进行比对。代入原始关系式，或在适当情况下对比图表与数值答案。
对于序列 5,8,11,14,...，公差为 3。第 n 项为 5+3(n-1)=3n+2。当 n=8 时，u₈=26。要求出 62 所在的位置，解方程 3n+2=62，得 n=20。
Arithmetic sequences and nth terms
Find a constant difference for an arithmetic sequence
Compare the model with the worked case and explain one change.
For a=5,d=3, find u_8.
u₈=5+(8-1)×3=26.
Test a tempting shortcut
- The first term has index 1, so the exponent is n-1. A sequence is a list; a series is a sum. A geometric sequence can alternate in sign and still converge.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The first term of a sequence always has index zero. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 首项的下标为1，因此指数为n-1。数列是一列数，级数是这些数的和。等比数列即使符号交替也可能收敛。
- 当捷径失效时，找出其违背的假设。保留精确值直到题目要求的最终舍入步骤。
数列的第一项索引总是零。该说法是错误的。请说明它违反了哪一定义或假设。
At what position does 62 occur in u_n=3n+2?
Solve 3n+2=62: 3n=60, hence n=20.
The first term of a sequence always has index zero.
The first term has index 1, so the exponent is n-1. A sequence is a list; a series is a sum. A geometric sequence can alternate in sign and still converge.
Interpret a new situation
- Generate several terms and check a proposed nth-term rule. Infinite geometric series and advanced sum formulae are excluded from this Foundation/Core lesson.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
解读新情境
- 生成若干项以检验所提出的通项公式是否正确。无穷等比级数及高级求和公式不属于本基础/核心课程内容。
- 完整解答需给出数学结果并解释其含义。检查其在所述背景下是否可行。
Find u_4 for u_n=3n+2.
u₄=3×4+2=14.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The constant added between consecutive terms of an arithmetic sequence. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- 8300 · 基础级 · 3.2。在布置拓展练习前，请核对目标层级与考纲要求。
- 在给出最终答案前先展示解题方法，并遵循试卷的计算器及公式使用规则。通过定位第一个无效步骤来复盘错误答案。
等差数列中相邻两项之间恒定的差值。请选择正确的关系式，展示解题方法，检查其适用条件并解释结果。