Arithmetic sequences and series
Master arithmetic sequences and series in Year 11 VCE Specialist Mathematics. An arithmetic sequence changes by the same common difference \(d\) at every step, so its \(n\)th term is \(t_n=a+(n-1)d\), and a series is the sum of its terms. It sits in the Discrete mathematics area of study of the VCE Mathematics Study Design (VCAA), within the Sequences and series topic of Unit 1.
You will learn to find any term with the \(n\)th-term rule, work back to the first term \(a\), the common difference \(d\) or the position \(n\), and add the first \(n\) terms with the partial sum \(S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)=\dfrac{n}{2}(a+l)\) — the groundwork for geometric sequences, recurrence relations and real-world modelling.
Theory
An arithmetic sequence increases (or decreases) by the same common difference \(d\) at every step. In Year 11 Specialist Mathematics you use the \(n\)th-term rule \(t_n = a + (n-1)d\) to find any term, the first term \(a\), the common difference \(d\) or the position \(n\), and the partial sum \(S_n = \dfrac{n}{2}\big(2a+(n-1)d\big) = \dfrac{n}{2}(a+l)\) to add the first \(n\) terms. This page shows how, with fully worked examples.
A sequence is an ordered list of numbers called terms, written \(t_1, t_2, t_3, \ldots\) In an arithmetic sequence each term is found by adding a fixed number, the common difference \(d\), to the term before it. The first term is \(a\) (also written \(t_1\)).
Because the same \(d\) is added each time, the difference between any term and the one before it is constant: \(d = t_2 - t_1 = t_3 - t_2 = \cdots\). A positive \(d\) makes the sequence increase; a negative \(d\) makes it decrease. For example \(3,\ 7,\ 11,\ 15,\ \ldots\) is arithmetic with \(a=3\) and \(d=4\).
Adding a step of \(d\) each time means the \(n\)th term is \(a\) plus \((n-1)\) lots of \(d\): \(t_n = a + (n-1)d\). Written as a rule in \(n\), \(t_n\) is a linear expression, so plotting the terms as points \((n, t_n)\) gives points on a straight line of gradient \(d\).
A series is the sum of the terms of a sequence. The partial sum \(S_n = t_1 + t_2 + \cdots + t_n\) adds the first \(n\) terms. If the last term added is \(l = t_n\), the sum can be found from the first and last terms alone, \(S_n = \dfrac{n}{2}(a+l)\), because pairing terms from the two ends gives equal totals.
The \(n\)th term of an arithmetic sequence starts at \(a\) and adds \((n-1)\) steps of the common difference \(d\):
The partial sum of the first \(n\) terms, from the first term \(a\) and common difference \(d\):
When the last term \(l = t_n\) is known, the same sum is found from the first and last terms:
Choosing and applying the rule
- Identify \(a\) and \(d\). The first term is \(a\); the common difference is any term minus the one before it, \(d = t_2 - t_1\).
- Finding a term: substitute \(a\), \(d\) and the position \(n\) into \(t_n = a + (n-1)d\).
- Finding \(a\), \(d\) or \(n\): substitute what you know into \(t_n = a + (n-1)d\) and solve the resulting equation. Two given terms give two equations — subtract them to find \(d\), then back-substitute for \(a\).
- Finding a sum: use \(S_n = \dfrac{n}{2}\big(2a+(n-1)d\big)\), or \(S_n = \dfrac{n}{2}(a+l)\) if the last term \(l\) is known.
- Solving for \(n\) from a sum: substitute into the sum formula and solve — this usually gives a quadratic in \(n\); keep the positive whole-number solution.
For a worded problem, name \(a\), \(d\) and \(n\) from the context first (for example a first row of \(12\) seats with \(3\) more each row gives \(a=12\), \(d=3\)), then apply the matching rule.
Apply the \(n\)th-term rule \(t_n = a + (n-1)d\) with \(n=10\):
| \(t_{10}\) | \(=\) | \(a + (10-1)d\) |
| \(=\) | \(4 + 9\times 3\) | |
| \(=\) | \(4 + 27\) | |
| \(=\) | \(31\) |
The \(10\)th term is \(t_{10} = 31\).
Substitute into \(t_n = a + (n-1)d\) and solve for \(n\):
| \(3 + (n-1)\times 5\) | \(=\) | \(48\) |
| \((n-1)\times 5\) | \(=\) | \(45\) |
| \(n-1\) | \(=\) | \(9\) |
| \(n\) | \(=\) | \(10\) |
The term \(48\) is the \(10\)th term, so \(n = 10\).
Read off \(a=4\) and \(d=3\), then apply \(S_n = \dfrac{n}{2}\big(2a+(n-1)d\big)\) with \(n=10\):
| \(S_{10}\) | \(=\) | \(\dfrac{10}{2}\big(2\times 4 + (10-1)\times 3\big)\) |
| \(=\) | \(5\times(8 + 27)\) | |
| \(=\) | \(5\times 35\) | |
| \(=\) | \(175\) |
The sum of the first \(10\) terms is \(S_{10} = 175\).
The weekly amounts are arithmetic with \(a=50\) and \(d=8\). Week-\(10\) amount, using \(t_n = a + (n-1)d\):
| \(t_{10}\) | \(=\) | \(50 + (10-1)\times 8\) |
| \(=\) | \(50 + 72\) | |
| \(=\) | \(122\) |
Total over \(10\) weeks, using \(S_n = \dfrac{n}{2}\big(2a+(n-1)d\big)\):
| \(S_{10}\) | \(=\) | \(\dfrac{10}{2}\big(2\times 50 + 9\times 8\big)\) |
| \(=\) | \(5\times(100 + 72)\) | |
| \(=\) | \(5\times 172\) | |
| \(=\) | \(860\) |
In week \(10\) she saves \(\$122\), and she saves \(\$860\) in total over the first \(10\) weeks.
Common pitfalls
Frequently asked questions
What is an arithmetic sequence?
It is a sequence in which each term differs from the one before it by the same fixed amount, the common difference \(d\). Starting from the first term \(a\), the terms are \(a,\ a+d,\ a+2d,\ a+3d,\ \ldots\)
What is the formula for the nth term?
The \(n\)th term is \(t_n = a + (n-1)d\), where \(a\) is the first term, \(d\) is the common difference and \(n\) is the position of the term.
How do I find the sum of an arithmetic series?
Use \(S_n = \dfrac{n}{2}\big(2a+(n-1)d\big)\) when you know \(a\), \(d\) and the number of terms \(n\). If you already know the last term \(l\), the shorter form \(S_n = \dfrac{n}{2}(a+l)\) gives the same answer.
How do I find how many terms give a particular value or sum?
Set \(t_n\) or \(S_n\) equal to the target value and solve for \(n\). For a term this is a linear equation; for a sum it is usually a quadratic in \(n\) — keep the positive whole-number solution.
How do I find the common difference from two terms?
Write \(t_n = a + (n-1)d\) for each given term to get two equations, then subtract them so the \(a\) cancels and \(d\) is left. For example \(t_7 - t_3 = 4d\), so if \(t_7=33\) and \(t_3=17\) then \(4d = 16\) and \(d = 4\).
Why do the terms lie on a straight line?
Written as a rule in \(n\), the \(n\)th term \(t_n = a + (n-1)d = dn + (a-d)\) is linear in \(n\). So plotting the terms as points \((n, t_n)\) gives points on a straight line whose gradient is the common difference \(d\).