公差gōngchā

Core Concept

The common difference is the core parameter of an arithmetic sequence, representing the constant difference between consecutive terms. In an arithmetic sequence, starting from the second term, the difference between each term and its preceding term equals this fixed constant.

Mathematical Definition

For an arithmetic sequence $\{a_n\}$, the common difference $d$ is defined as:

$d = a_{n+1} - a_n \quad (n \in \mathbb{N}^*)$

where $d$ is a constant for all terms in the sequence.

Properties

1. Uniqueness: An arithmetic sequence has only one common difference

2. Can be positive, negative, or zero:

   • $d > 0$ → increasing sequence
   • $d < 0$ → decreasing sequence
   • $d = 0$ → constant sequence

3. Calculation formula: $d = \frac{a_n - a_m}{n - m} \quad (n \neq m)$

Real-World Applications

Application 1: Temperature Changes

Problem: Daily high temperatures for a week are: 20°C, 22°C, 24°C, 26°C, 28°C, 30°C, 32°C. Find the common difference.

Solution: $d = 22 - 20 = 2°C$

Temperature increases by 2°C daily, showing arithmetic growth.

Application 2: Salary Growth

Problem: Ming's first-year salary is $5000, increasing by$500 annually. What is the common difference?

Solution: Common difference $d = 500$ per year

This forms an arithmetic sequence with $a_1 = 5000$, $d = 500$.

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus and Chinese standardized test formats to help students familiarize themselves with question types and problem-solving approaches.

Example 1: Basic (Difficulty ★☆☆☆☆)

In an arithmetic sequence $\{a_n\}$, $a_1 = 3$ and $a_5 = 11$. Find the common difference $d$.

Options:

• A. 1
• B. 2
• C. 3
• D. 4

Detailed Solution:

Using the formula: $d = \frac{a_5 - a_1}{5 - 1} = \frac{11 - 3}{4} = \frac{8}{4} = 2$

Verification:

• $a_2 = 3 + 2 = 5$
• $a_3 = 5 + 2 = 7$
• $a_4 = 7 + 2 = 9$
• $a_5 = 9 + 2 = 11$ ✓

Answer: B

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Example 2: Advanced (Difficulty ★★★☆☆)

In an arithmetic sequence $\{a_n\}$ with $d \neq 0$, given $a_1 + a_3 + a_5 = 15$ and $a_1 \cdot a_3 \cdot a_5 = 105$, find $d$.

Detailed Solution:

Let $a_3 = a$ (middle term), then:

• $a_1 = a - 2d$
• $a_5 = a + 2d$

Condition 1: $(a - 2d) + a + (a + 2d) = 15$ $3a = 15$ $a = 5$

Condition 2: $(5 - 2d) \cdot 5 \cdot (5 + 2d) = 105$ $5(25 - 4d^2) = 105$ $4d^2 = 4$ $d = \pm 1$

Answer: $d = \pm 1$

Common Mistakes

❌ Mistake 1: Common difference must be positive

Correction: The common difference can be positive, negative, or zero.

Example: Sequence 10, 8, 6, 4, 2, ... has $d = -2$ (negative).

❌ Mistake 2: Any two terms differ by d

Correction: Common difference is the difference between consecutive terms only.

For $a_n$ and $a_m$ where $n > m$: $a_n - a_m = (n - m) \cdot d$

❌ Mistake 3: Confusing first term with common difference

Correction: The first term $a_1$ is the starting value; common difference $d$ is the change between terms. They are different concepts.

Study Tips

1. ✅ Understand the essence: Common difference describes uniform change
2. ✅ Sign awareness: Note whether d is positive or negative
3. ✅ Flexible calculation: Master multiple methods to find d
4. ✅ Real-world recognition: Identify arithmetic patterns in daily life

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💡 Exam Tip: Common difference is fundamental to arithmetic sequences. Almost every arithmetic sequence problem involves it. Ensure you can calculate it quickly and accurately.