DWG NO. 9.1 — Lesson 1 of 6

Arithmetic Sequences and Series

Unit 9: Sequences, Series, and Combinatorics · ~20 min

Objective Find any term of an arithmetic sequence from its explicit formula, and evaluate the sum of a finite arithmetic series.

A sequence is a function whose domain is restricted to the positive integers: instead of writing f(1), f(2), f(3), … sequence notation writes \(a_{1}\), \(a_{2}\), \(a_{3}\), …, or \(a_{n}\) for the general (nth) term. An arithmetic sequence is the simplest pattern a sequence can follow — each term is the one before it plus a fixed number d, the common difference. Since \(a_{n}\) changes by the same amount d every step, plotting \(a_{n}\) against n always produces points that fall on a straight line — an arithmetic sequence is a linear function of n, restricted to integer inputs.

Explicit and recursive formulas

Starting from a first term \(a_{1}\) and adding d a total of (n − 1) times to reach the nth term gives the explicit formula:

Explicit formula \(a_{n}\) = \(a_{1}\) + (n − 1)d

The same relationship can be written recursively, defining each term in terms of the one before it: \(a_{n}\) = \(a_{n-1}\) + d, with \(a_{1}\) given to start the chain. The recursive form is often more natural to write down from a pattern; the explicit form is what you need to jump straight to, say, the 200th term without generating the first 199.

Arithmetic series

A series is the sum of a sequence's terms. To sum the first n terms of an arithmetic sequence, pair the first term with the last, the second term with the second-to-last, and so on — each pair adds up to the same total, \(a_{1}\) + \(a_{n}\), because what one term loses going up the sequence, its partner gains going down. There are n/2 such pairs, giving:

Finite arithmetic series \(S_{n}\) = \(\frac{n}{2}\) (\(a_{1}\) + \(a_{n}\)) = \(\frac{n}{2}\) (2\(a_{1}\) + (n−1)d)

The first form is faster when you already know the last term \(a_{n}\); the second substitutes the explicit formula in for \(a_{n}\) so you can work from \(a_{1}\) and d alone.

\(a_{n}\) = 3 + 4(n−1) plotted for n = 1…6 — evenly spaced points fall on a straight line

Worked Example 1 · Finding a term

ProblemAn arithmetic sequence has \(a_{1}\) = 8 and common difference d = −3. Find \(a_{15}\).

1Use \(a_{n}\) = \(a_{1}\) + (n−1)d with n = 15: \(a_{15}\) = 8 + (15−1)(−3).

2\(a_{15}\) = 8 + 14(−3) = 8 − 42.

\(a_{15}\) = −34

Worked Example 2 · Sum of a finite series

ProblemFind the sum of the first 20 terms of the arithmetic sequence with \(a_{1}\) = 7 and d = 4.

1First find \(a_{20}\): \(a_{20}\) = 7 + (20−1)(4) = 7 + 76 = 83.

2Use \(S_{n}\) = \(\frac{n}{2}\)(\(a_{1}\) + \(a_{n}\)): \(S_{20}\) = \(\frac{20}{2}\)(7 + 83) = 10(90).

\(S_{20}\) = 900

Guided practice

1.Find \(a_{12}\) for the arithmetic sequence with \(a_{1}\) = 5 and d = 6.
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\(a_{12}\) = 5 + (12−1)(6) = 5 + 66 = 71.
2.The first term of an arithmetic sequence is −4 and the common difference is 7. Write the explicit formula for \(a_{n}\).
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\(a_{n}\) = −4 + 7(n−1), which simplifies to \(a_{n}\) = 7n − 11.
3.A sequence begins 40, 33, 26, 19, … Find \(a_{10}\).
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d = 33−40 = −7. \(a_{10}\) = 40 + (10−1)(−7) = 40 − 63 = −23.
4.Find the sum of the first 30 terms of the arithmetic sequence with \(a_{1}\) = 2 and d = 5.
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\(a_{30}\) = 2 + (30−1)(5) = 2 + 145 = 147. \(S_{30}\) = \(\frac{30}{2}\)(2 + 147) = 15(149) = 2235.
5.Find the sum 6 + 11 + 16 + … + 96 (an arithmetic series).
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d = 5. Find n from \(a_{n}\) = 96: 96 = 6 + (n−1)(5) → 90 = 5(n−1) → n = 19. \(S_{19}\) = \(\frac{19}{2}\)(6 + 96) = 9.5(102) = 969.

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