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Nth Term of Arithmetic Sequences

A term-to-term rule only reaches the next term. The nth term rule is a formula that jumps straight to any term you want, even the hundredth, without listing all the ones before it.

MYP 4Standard MathsCoordinate Geometry & SequencesCriteria A · C · D~10 min read

Written by Fayek · MYP student

Arithmetic sequences

An arithmetic sequence goes up (or down) by the same fixed amount each step. That fixed amount is the common difference, \(d\), and the very first term is called \(a\).

Common difference
The fixed amount added to get from one term to the next: \(d = \text{term} - \text{previous term}\). If it is the same for every pair, the sequence is arithmetic.

For \(5,\ 8,\ 11,\ 14,\ \ldots\) the first term is \(a = 5\) and the common difference is \(d = 3\). A difference can be negative: \(20,\ 17,\ 14,\ \ldots\) has \(d = -3\).

The nth term formula

To reach the \(n\)th term from the first, you add \(d\) a total of \(n-1\) times. That gives the familiar rule, which we then tidy up.

\[ n\text{th term} = a + (n-1)d = dn + (a - d) \]

The tidied form \(dn + (a - d)\) is handy because it reads like the equation of a straight line: the coefficient of \(n\) is the common difference \(d\), and the constant is \(a - d\) (the term you would get "at \(n = 0\)").

Finding a term

To use the formula, work out \(a\) and \(d\), substitute them in, and simplify to a rule in \(n\). Then put in whatever term number you need.

Worked example

Find the nth term rule for \(5,\ 8,\ 11,\ 14,\ \ldots\) and use it to find the 10th term.

1
Read off the values: first term \(a = 5\), common difference \(d = 8 - 5 = 3\).
2
Substitute into \(dn + (a - d)\): \(3n + (5 - 3) = 3n + 2\).
3
Check it: \(n = 1\) gives \(3(1) + 2 = 5\), which is the first term. Good.
4
For the 10th term put \(n = 10\): \(3(10) + 2 = 32\).
The rule is \(3n + 2\) and the 10th term is \(32\).

Is a number in the sequence?

The formula also tells you whether a given number appears in the sequence. Set the rule equal to that number and solve for \(n\). If \(n\) is a positive whole number, the value is in the sequence; if not, it is not.

n must be a positive whole number

A term number counts position, so \(n\) has to be a positive integer like \(1, 2, 3, \ldots\). If solving gives \(n = 6\) the number is the 6th term; if it gives \(n = 6.5\) the number is not in the sequence at all. Stating that condition and justifying your yes-or-no answer is exactly the reasoning credited under Criteria C and D.

Check yourself

1. Find the nth term rule for \(7,\ 12,\ 17,\ 22,\ \ldots\) and the 20th term +

Here \(a = 7\) and \(d = 12 - 7 = 5\). The rule is \(dn + (a - d) = 5n + (7 - 5) = 5n + 2\). For the 20th term put \(n = 20\): \(5(20) + 2 = \mathbf{102}\).

2. Write the nth term rule for \(3,\ 9,\ 15,\ 21,\ \ldots\) +

Here \(a = 3\) and \(d = 9 - 3 = 6\). The rule is \(6n + (3 - 6) = \mathbf{6n - 3}\). Check: \(n = 1\) gives \(6 - 3 = 3\), the first term.

3. Which term of \(4,\ 7,\ 10,\ 13,\ \ldots\) equals \(61\)? +

Here \(a = 4\), \(d = 3\), so the rule is \(3n + 1\). Set it equal to \(61\): \(3n + 1 = 61\), so \(3n = 60\) and \(n = 20\). Since \(n\) is a positive whole number, \(61\) is the \(\mathbf{20\text{th}}\) term.

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