Arithmetic Sequences - Common Difference and Sum
An arithmetic sequence is a list of numbers in which each term is obtained from the previous one by adding (or subtracting) a fixed amount. That fixed amount is called the common difference, usually written d. Arithmetic sequences are the most fundamental type of sequence in mathematics and appear constantly in real life.
The story of a young Carl Friedrich Gauss instantly summing 1 to 100 is one of the most retold anecdotes in mathematics: around 1787, when Gauss was about nine or ten years old, his teacher J.G. Büttner reportedly set the class the busywork task of adding every number from 1 to 100, expecting it to keep them occupied for the rest of the lesson. Gauss is said to have realised almost immediately that pairing the first and last numbers (1+100, 2+99, 3+98 …) always gives 101, and with 50 such pairs the answer is simply 50 × 101 = 5,050 – the exact sum formula still taught today. Beyond classroom legend, arithmetic sequences quietly run real financial products: a simple-interest savings account, unlike compound interest, earns exactly the same fixed amount every year, so the account balance over time forms a genuine arithmetic sequence.
Key Terms
| Term | Symbol | Meaning |
|---|---|---|
| First term | a | The starting value of the sequence |
| Common difference | d | The fixed amount added each time (can be negative) |
| nth term | an | The value of the term in position n |
| Number of terms | n | How many terms are in the sequence |
The nth Term Formula
an = a + (n − 1)d
Where a = first term, d = common difference, n = position of the term.
Sum of an Arithmetic Sequence
Sn = n/2 × (first term + last term)
or equivalently: Sn = n/2 × (2a + (n−1)d)
This formula was famously used by the young Gauss to add the numbers 1 to 100 in seconds: S = 100/2 × (1 + 100) = 50 × 101 = 5,050.
Worked Examples
a = 7, d = 4. a20 = 7 + (20−1) × 4 = 7 + 76 = 83.
From 5th to 9th is 4 steps: 38 − 22 = 16, so d = 16 ÷ 4 = 4.
a5 = a + 4d → 22 = a + 16 → a = 6.
Sequence: 6, 10, 14, 18, 22, 26, 30, 34, 38 … ✓
a = 3, d = 5, n = 15. S15 = 15/2 × (2(3) + 14(5)) = 15/2 × (6 + 70) = 15/2 × 76 = 570.
a = 15, d = 2, n = 20. Last row: a20 = 15 + 19(2) = 53.
Total = 20/2 × (15 + 53) = 10 × 68 = 680 seats.
Key Takeaways
- Arithmetic sequence: constant difference d between consecutive terms.
- nth term: an = a + (n−1)d.
- Sum of n terms: Sn = n/2 × (first + last) = n/2 × (2a + (n−1)d).
- d positive → increasing sequence; d negative → decreasing sequence.
Practice: nth Term & Sum
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