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Sequences

A sequence is a list of numbers with a rule. Finding the nth term gives you a formula for any position at once, so you never have to count all the way to the 100th term.

MYP 4-5ExtendedAlgebraCriteria A · B~12 min read

Written by Fayek Jawad Haider · MYP student

Linear (arithmetic) sequences

A linear sequence goes up or down by the same amount each time. That step is the common difference, \(d\). Because the change is constant, the nth term is always of the form \(dn + \text{(a constant)}\).

\[ n\text{th term} = dn + (\,\text{first term} - d\,) \]
Worked example

Find the nth term of \(5,\ 8,\ 11,\ 14,\ \ldots\)

1
The common difference is \(d = 3\) (each term is 3 more than the last). So the rule starts \(3n\).
2
Compare \(3n\) with the sequence: \(3n\) gives \(3, 6, 9, 12\), which is 2 less than each term.
3
Add the 2 back on to line them up.
\(n\text{th term} = 3n + 2\)

Always test with \(n = 1\)

Put \(n=1\) into your rule and check it gives the first term. Here \(3(1)+2 = 5\) ✓. This one substitution catches most errors on the spot.

Using the nth term

Once you have the formula it does two jobs: find any term, and test whether a number belongs to the sequence.

Worked example

For the sequence \(3n + 2\), is 100 a term?

1
Set the rule equal to 100: \(3n + 2 = 100\).
2
Solve: \(3n = 98 \Rightarrow n = 32.67\ldots\)
3
Because \(n\) is not a whole number, 100 cannot be at a term position.
No, 100 is not a term (it would need \(n = 32.67\ldots\)).

Quadratic sequences

If the differences between terms are not constant, look at the second differences, the differences of the differences. When those are constant, the sequence is quadratic, with an nth term of the form \(an^2 + bn + c\).

The key fact: the constant second difference equals \(2a\), so \(a = \dfrac{\text{second difference}}{2}\).

Worked example

Find the nth term of \(3,\ 8,\ 15,\ 24,\ 35,\ \ldots\)

1
First differences: \(5, 7, 9, 11\), not constant. Second differences: \(2, 2, 2\), constant, so it is quadratic.
2
The second difference is 2, so \(a = \dfrac{2}{2} = 1\). The rule contains \(n^2\).
3
Subtract \(n^2\) from each term: \(3-1,\ 8-4,\ 15-9,\ 24-16 = 2, 4, 6, 8\).
4
That leftover is linear with nth term \(2n\). Add it to the \(n^2\) part.
\(n\text{th term} = n^2 + 2n\)

The method in one line

Constant second difference → halve it for \(a\) → subtract \(an^2\) → find the nth term of what is left → add the two rules. Showing these steps clearly is Criterion B (investigating) and Criterion A together.

Check yourself

1. Find the nth term of \(7,\ 10,\ 13,\ 16,\ \ldots\) +

Common difference \(d = 3\), so start with \(3n = 3,6,9,12\), which is 4 short each time. nth term \(= \mathbf{3n + 4}\). Check \(n=1\): \(3+4=7\) ✓.

2. What is the 10th term of \(2n + 5\)? +

Substitute \(n = 10\): \(2(10) + 5 = \mathbf{25}\).

3. Find the nth term of \(2,\ 5,\ 10,\ 17,\ \ldots\) +

First differences \(3,5,7\); second difference \(2\), so \(a = 1\). Subtract \(n^2\): \(2-1, 5-4, 10-9, 17-16 = 1,1,1,1\), a constant 1. nth term \(= \mathbf{n^2 + 1}\). Check \(n=1\): \(1+1=2\) ✓.

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