Sequences
Students encounter sequences daily without realizing it—from seat numbers 2, 4, 6, 8 in an auditorium to page numbers in a textbook. Teaching sequences through CCSS.HSF.BF and CCSS.HSF.LE builds the foundation for advanced functions and exponential growth models that appear throughout high school mathematics.
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Why it matters
Sequences model real-world patterns that students encounter constantly. Population growth follows geometric sequences—if a town of 5,000 people grows by 3% annually, the sequence becomes 5,000, 5,150, 5,304.5, creating the pattern aₙ = 5000 × (1.03)ⁿ⁻¹. Loan payments follow arithmetic sequences when fixed amounts reduce principal monthly. A $12,000 car loan with $300 monthly payments creates the remaining balance sequence: 11,700, 11,400, 11,100. Sports tournaments use sequences for seeding—64 teams become 32, then 16, then 8, following the geometric pattern 64, 32, 16, 8, 4, 2, 1. Understanding sequence formulas helps students predict outcomes without calculating every term, developing mathematical reasoning skills essential for algebra, calculus, and statistics.
How to solve sequences
Sequences
- Arithmetic sequence: constant difference (d) between terms. aₙ = a₁ + (n−1)d.
- Geometric sequence: constant ratio (r) between terms. aₙ = a₁ × r^(n−1).
- To identify: check differences first, then ratios.
- Sum of arithmetic series: S = n/2 × (first + last).
Example: 2, 6, 18, 54: ratio = 3, geometric. a₅ = 2 × 3⁴ = 162.
Worked examples
Write the next 3 terms: 4, 9, 14, __, __, __
Answer: 19, 24, 29
- Find the common difference → d = 5 — 9 − 4 = 5. Each term increases by 5.
- Continue the pattern → 19, 24, 29 — 14 + 5 = 19, 19 + 5 = 24, 24 + 5 = 29.
Find the 8th term of: 3, 5, 7, 9, ...
Answer: 17
- Identify first term and common difference → a₁ = 3, d = 2 — First term is 3. Difference: 5 − 3 = 2.
- Use the nth term formula → aₙ = a₁ + (n − 1)d — The nth term of an arithmetic sequence is a₁ + (n − 1)d.
- Substitute → a_8 = 3 + (8 − 1) × 2 — Replace a₁ with 3, n with 8, d with 2.
- Calculate → 17 — 3 + 7 × 2 = 3 + 14 = 17.
Find the common difference and the 20th term: 5, 11, 17, 23, ...
Answer: d = 6, 20th term = 119
- Find the common difference → d = 11 − 5 = 6 — Subtract consecutive terms: 11 − 5 = 6.
- Use the nth term formula → a₂₀ = 5 + (20 − 1) × 6 — aₙ = a₁ + (n − 1)d with n = 20.
- Calculate → 119 — 5 + 19 × 6 = 5 + 114 = 119.
Common mistakes
- ✗Students confuse arithmetic and geometric sequences. For 3, 6, 12, 24, they might add 3 each time, writing the next term as 27 instead of recognizing the doubling pattern that gives 48.
- ✗When finding the nth term, students forget to subtract 1 from n. For sequence 5, 8, 11, 14 with d = 3, they calculate a₁₀ = 5 + 10 × 3 = 35 instead of the correct a₁₀ = 5 + (10-1) × 3 = 32.
- ✗Students mix up the sum formula, calculating Sₙ = n × (first + last) instead of Sₙ = n/2 × (first + last). For 5 terms of 2, 4, 6, 8, 10, they get S = 5 × 12 = 60 instead of S = 5/2 × 12 = 30.
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