Sequences
Arithmetic sequences form the backbone of Year 7 algebra, where pupils first encounter the power of pattern recognition in mathematics. Whether students are finding the 15th term of 3, 7, 11, 15... or identifying common differences, sequences bridge the gap between basic number work and advanced algebraic thinking.
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Why it matters
Sequences appear everywhere in real-world applications, from calculating mortgage payments to understanding population growth. In Britain, GCSE Foundation papers regularly feature sequence questions worth 15-20 marks across multiple papers. Students use arithmetic sequences to model saving patterns — if Charlotte saves £5 in January, £8 in February, and £11 in March, she can predict having £23 in June using the nth term formula. Construction workers use sequences to calculate materials: if a staircase needs 12 bricks for step 1, 15 for step 2, and 18 for step 3, the formula helps determine exactly 39 bricks for step 10. Understanding sequences develops logical reasoning skills essential for A-level mathematics, where students encounter more complex geometric and recursive sequences worth significant UCAS points.
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 × rn−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: 5, 9, 13, __, __, __
Answer: 17, 21, 25
- Find the common difference → d = 4 — 9 − 5 = 4. Each term increases by 4.
- Continue the pattern → 17, 21, 25 — 13 + 4 = 17, 17 + 4 = 21, 21 + 4 = 25.
Find the 15th term of: 2, 4, 6, 8, ...
Answer: 30
- Identify first term and common difference → a₁ = 2, d = 2 — First term is 2. Difference: 4 − 2 = 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_15 = 2 + (15 − 1) × 2 — Replace a₁ with 2, n with 15, d with 2.
- Calculate → 30 — 2 + 14 × 2 = 2 + 28 = 30.
Find the common difference and the 20th term: 2, 8, 14, 20, ...
Answer: d = 6, 20th term = 116
- Find the common difference → d = 8 − 2 = 6 — Subtract consecutive terms: 8 − 2 = 6.
- Use the nth term formula → a₂₀ = 2 + (20 − 1) × 6 — aₙ = a₁ + (n − 1)d with n = 20.
- Calculate → 116 — 2 + 19 × 6 = 2 + 114 = 116.
Common mistakes
- Students often confuse the position number with the term value. For sequence 4, 7, 10, 13..., they might say the 5th term is 5 instead of 16, mixing up n with aₙ.
- Pupils frequently calculate the common difference incorrectly by subtracting in the wrong direction. In sequence 15, 11, 7, 3..., they write d = 4 instead of d = -4.
- Many students substitute incorrectly into the nth term formula. For a₁ = 3, d = 4, n = 8, they calculate 3 + 8 × 4 = 35 instead of 3 + (8-1) × 4 = 31.
- Students often forget to identify the sequence type first. Given 2, 6, 18, 54..., they try finding a common difference instead of recognising the common ratio of 3.