An arithmetic sequence has u1 = 11 and d = 6. Find the sum of the first 15 terms.
- Use Sn = (n/2)(2u1 + (n − 1)d) with n = 15.
- Bracket: 2 × 11 + 14 × 6 = 22 + 84 = 106.
- S15 = (15/2) × 106 = 7.5 × 106 = 795.
Answer: S15 = 795
Arithmetic sequences and series
Maths: Applications and Interpretation · Number and algebra. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
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The idea
In an arithmetic sequence you add the same number each time to get the next term. That number is the common difference, d. It is negative when the terms go down. If you know the first term u1 and d, you know every term.
A series is the sum of the terms. Sigma notation is a short way to write it: the numbers below and above Σ tell you where k starts and stops.
Simple interest is arithmetic, because the same amount of interest is added every year. Real data is rarely perfectly arithmetic. If the differences are roughly constant, an arithmetic model is still reasonable for prediction.
Rules to remember
- common difference = any term − the term before it (d = un+1 − un)
- nth term: un = u1 + (n − 1)d
- Sum of the first n terms: Sn = (n/2)(2u1 + (n − 1)d)
- Sum, if you know the last term: Sn = (n/2)(u1 + un)
- Given two terms, subtract them to find d: u8 − u3 = 5d
- Simple interest: total value = P + P × (r/100) × n, with r in percent per year and n in years
- An arithmetic model fits when the differences between consecutive values are roughly constant
Common mistake
Students write un = u1 + nd. There are only n − 1 steps from the first term to the nth term, so the multiplier is (n − 1).
Worked example
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