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algebraDifficulty: Basicalgebrasequences
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等差数列děngchā shùliè

arithmetic sequence
5 min read
Updated October 29, 2025
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Core Concept

An arithmetic sequence (等差数列) is one of the most fundamental types of sequences in mathematics. Starting from the second term, the difference between any term and its preceding term equals the same constant, called the common difference (公差), typically denoted by dd.

Mathematical Definition

For a sequence {an}\{a_n\}, if there exists a constant dd such that:

an+1−an=d(n∈N∗)a_{n+1} - a_n = d \quad (n \in \mathbb{N}^*)

Then the sequence {an}\{a_n\} is called an arithmetic sequence, with dd as the common difference.

General Term Formula

The nn-th term of an arithmetic sequence can be expressed using the first term a1a_1 and common difference dd:

an=a1+(n−1)da_n = a_1 + (n-1)d

Derivation:

  • a2=a1+da_2 = a_1 + d
  • a3=a2+d=a1+2da_3 = a_2 + d = a_1 + 2d
  • a4=a3+d=a1+3da_4 = a_3 + d = a_1 + 3d
  • ...
  • an=a1+(n−1)da_n = a_1 + (n-1)d

Sum Formula

The sum of the first nn terms has two common formulas:

Formula 1 (using first and last terms): Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}

Formula 2 (using first term and common difference): Sn=na1+n(n−1)2d=n[2a1+(n−1)d]2S_n = na_1 + \frac{n(n-1)}{2}d = \frac{n[2a_1 + (n-1)d]}{2}

Important Properties

Property 1: Arithmetic Mean

If aa, bb, cc form an arithmetic sequence, then: b=a+c2b = \frac{a + c}{2}

That is, bb is the arithmetic mean of aa and cc.

Property 2: Index Property

If m+n=p+qm + n = p + q (where m,n,p,q∈N∗m, n, p, q \in \mathbb{N}^*), then: am+an=ap+aqa_m + a_n = a_p + a_q

In particular, if m+n=2pm + n = 2p, then: am+an=2apa_m + a_n = 2a_p

Property 3: Sum Property

The sum of first nn terms can be viewed as nn times the average of the first and last terms: Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}

Real-World Applications

Application 1: Bank Savings

Problem: Ming saves 500 yuan every month for 12 months. What is the total savings after 12 months?

Analysis: Monthly deposits form an arithmetic sequence with a1=500a_1 = 500, d=500d = 500, n=12n = 12.

Solution: S12=12×(500+6500)2=42,000 yuanS_{12} = \frac{12 \times (500 + 6500)}{2} = 42,000 \text{ yuan}

Application 2: Theater Seating

Problem: A theater has 20 seats in the first row. Each subsequent row has 2 more seats than the previous row. If there are 30 rows, how many seats are there in total?

Analysis:

  • First term a1=20a_1 = 20
  • Common difference d=2d = 2
  • Number of terms n=30n = 30

Solution: S30=30×20+30×292×2=600+870=1,470 seatsS_{30} = 30 \times 20 + \frac{30 \times 29}{2} \times 2 = 600 + 870 = 1,470 \text{ seats}

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 {an}\{a_n\}, a3=7a_3 = 7 and a7=15a_7 = 15. Find a10a_{10}.

Options:

  • A. 19
  • B. 21
  • C. 23
  • D. 25

Detailed Solution:

Method 1: Using the general term formula

  1. From the general term formula:

    • a3=a1+2d=7a_3 = a_1 + 2d = 7 ... ①
    • a7=a1+6d=15a_7 = a_1 + 6d = 15 ... ②
  2. ② - ① gives: 4d=84d = 8, so d=2d = 2

  3. Substituting into ①: a1+4=7a_1 + 4 = 7, so a1=3a_1 = 3

  4. a10=a1+9d=3+18=21a_{10} = a_1 + 9d = 3 + 18 = 21

Answer: B


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

In an arithmetic sequence {an}\{a_n\}, S5=25S_5 = 25 and S10=100S_{10} = 100. Find S15S_{15}.

Options:

  • A. 175
  • B. 200
  • C. 225
  • D. 250

Detailed Solution:

Method: Using the property that S5S_5, S10−S5S_{10} - S_5, and S15−S10S_{15} - S_{10} also form an arithmetic sequence:

  • S5=25S_5 = 25
  • S10−S5=100−25=75S_{10} - S_5 = 100 - 25 = 75
  • Common difference: 75−25=5075 - 25 = 50

Therefore: S15−S10=75+50=125S_{15} - S_{10} = 75 + 50 = 125

S15=100+125=225S_{15} = 100 + 125 = 225

Answer: C

Common Mistakes

❌ Mistake 1: Can the common difference be zero?

Answer: Mathematically, when d=0d = 0, the sequence is a constant sequence, which is technically an arithmetic sequence. However, in CSCA exams, "arithmetic sequence" usually implies d≠0d \neq 0 unless specified otherwise.

❌ Mistake 2: Is an arithmetic sequence always increasing?

Answer: Not necessarily!

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

❌ Mistake 3: Confusing with geometric sequences

Key Difference:

  • Arithmetic sequence: difference between consecutive terms is constant (an+1−an=da_{n+1} - a_n = d)
  • Geometric sequence: ratio between consecutive terms is constant (an+1an=r\frac{a_{n+1}}{a_n} = r)

Study Tips

  1. ✅ Master the formulas - General term and sum formulas are fundamental
  2. ✅ Understand common difference - It can be positive, negative, or zero
  3. ✅ Practice properties - Especially arithmetic mean and index properties
  4. ✅ Solve various problems - Arithmetic sequences often combine with functions and inequalities
  5. ✅ Compare with geometric sequences - Understand the differences clearly

💡 Exam Tip: Arithmetic sequences are a high-frequency topic in CSCA math exams, accounting for about 60% of sequence questions. Practice 2-3 related problems daily to ensure mastery.

How the real exam tests it

Questions from real CSCA papers that test this term. Work each one out before you open the answer.

  1. CSCA Math December 2025 Past Paper
    If {an}\{a_n\} is an arithmetic sequence with a1=2a_1=2 and common difference d=3d=3, then a100=a_{100}= ( ).
    • A.299
    • B.305
    • C.302
    • D.298
    Show answer & solution

    Answer: A

    Answer: A a100=a1+99d=2+99×3=299a_{100}=a_1+99d=2+99\times3=299. Choose A.
    See it in the full CSCA Math December 2025 Past Paper
  2. CSCA Math April 2026 Past Paper
    If −1,a,15- 1 , a , 15 is an arithmetic sequence, then a=a =
    • A.9
    • B.5
    • C.7
    • D.11
    Show answer & solution

    Answer: C

    Middle term: a=−1+152=7a = \frac { - 1 + 15 } { 2 } = 7.
    Take the full CSCA Math April 2026 Past Paper — timed & graded

Practice Exercises

Practice to reinforce your understanding