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Year 12 Maths Advanced (2027) Sequences and series

Arithmetic Sequences

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Sequences and series

Arithmetic sequences increase by a constant common difference \(d\), with \(n\)th term \(T_n=a+(n-1)d\).

Part of the NSW Year 12 Mathematics Advanced course, in the Sequences and series focus area of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

An arithmetic sequence (AP) changes by the same amount each step — the common difference \(d\). This Year 12 Mathematics Advanced topic (MAV-12-03) uses \(a_n=a+(n-1)d\) to find terms and solve problems, and recognises that the terms are a linear function of \(n\).

In an arithmetic sequence each term differs from the one before by a fixed common difference \(d=a_n-a_{n-1}\).

With first term \(a\), the \(n\)th term is \(a_n=a+(n-1)d\). Since this is linear in \(n\) (slope \(d\)), the terms plot as equally spaced points on a straight line. \(d>0\) increases; \(d<0\) decreases.

Arithmetic sequence 3,7,11,15,19The terms of an AP with common difference 4 lie on a straight line. na_n 1 2 3 4 5
The AP \(3,7,11,15,19\) (\(d=4\)) lies on a straight line.
\[a_n=a+(n-1)d\]
a n equals a plus (n minus 1) d

Method

  1. Find the common difference \(d\) (subtract consecutive terms) and the first term \(a\).
  2. Substitute into \(a_n=a+(n-1)d\).
  3. Solve for \(n\) if you are told the value of a term.
Example 1 — Find a term
Find \(d\) and \(a_{10}\) for \(3,7,11,\ldots\)
Solution
\(d\)\(=\)\(4\)
\(a_{10}\)\(=\)\(3+9\times4=39\)
Example 2 — Rule for \(a_n\)
Find a rule for \(5,2,-1,\ldots\) and hence \(a_{20}\).
Solution
\(a_n\)\(=\)\(5+(n-1)(-3)=8-3n\)
\(a_{20}\)\(=\)\(-52\)
Example 3 — Which term?
Which term of \(4,9,14,\ldots\) equals \(99\)?
Solution
\(5n-1\)\(=\)\(99\)
\(n\)\(=\)\(20\)
Example 4 — Salary
Salary \(\$50\,000\) rising \(\$2000\)/year: state \(d\), year-5 salary, and \(a_n\).
Solution
\(d\)\(=\)\(2000\)
\(a_5\)\(=\)\(50000+4(2000)=58000\)
\(a_n\)\(=\)\(48000+2000n\)

Common pitfalls

It is \((n-1)d\), not \(nd\). The first term uses none of the difference.
Keep the sign of \(d\). For \(5,2,-1,\ldots\), \(d=-3\).
Check it is arithmetic. The difference between terms must be constant.

Frequently asked questions

What is the formula for the nth term of an arithmetic sequence?

It is a n equals a plus (n minus 1) times d, where a is the first term and d is the common difference.

How do you find the common difference?

Subtract any term from the term after it. If the sequence is 3, 7, 11 then d equals 7 minus 3, which is 4.

How do you find which term equals a given value?

Set the nth term formula equal to the value and solve for n. If n is a whole number, that position is the term.

Why is an arithmetic sequence linear?

Because a n equals a plus (n minus 1) d is a linear function of n with slope d, so the terms plot as equally spaced points on a straight line.