Number Patterns

Finding the nth term: linear and quadratic patterns

By Mr Gan · O-Level E Maths · Updated August 2026 · 8 min read

For a linear sequence, the nth term is dn + c, where d is the common difference and c = (first term) − d. For a sequence generated by a growing pattern (like dots forming a shape), check the second differences: if they are constant, the nth term contains n², with coefficient equal to half the second difference. Find the remaining linear part by subtraction.

Why students lose marks here

The most common error in linear sequences is writing the nth term as n + d instead of dn + c. A sequence like 7, 11, 15, 19 does not go up by adding 4 to n. It goes up by 4 each time n increases by 1, which means n is multiplied by 4, not added to it. Students who spot the "goes up by 4" pattern but then write n + 4 get the direction right and the structure wrong.

The second common failure is with quadratic patterns. Students correctly compute the first differences, notice they are not constant, and stop there, unsure what to do next. The pattern is not linear, but it is not random either: the second differences (the differences between the first differences) reveal a quadratic nth term, and there is a fixed way to read them off. Recognising a familiar special sequence such as squares or triangular numbers can also save you from working out the second differences at all.

The method

Linear sequence

Find d, the common difference. nth term = dn + c, where c = 1st term − d.

Quadratic sequence

Find 1st differences, then 2nd differences. Coefficient of n² = 2nd difference ÷ 2.

The link to coordinate geometry: a linear sequence dn + c behaves exactly like the straight line y = mx + c, with d playing the role of the gradient m and n playing the role of x. If you already know how to read a gradient off two points, you already know how to find d.

Watch the whole method in about a minute.

Worked example 1: linear sequence, then testing a large number

A sequence begins 7, 11, 15, 19, ... Find the nth term, then determine whether 251 is a term in the sequence.

Solution

Find the common difference: 11 − 7 = 4, 15 − 11 = 4, 19 − 15 = 4. So d = 4.

Write the nth term as 4n + c. Substitute n = 1, where the term is 7: 4(1) + c = 7, so c = 3.

The nth term is 4n + 3. Check against n = 2: 4(2) + 3 = 11 ✓. Check against n = 4: 4(4) + 3 = 19 ✓.

To test whether 251 is a term, set 4n + 3 = 251, so 4n = 248, giving n = 62.

Since n = 62 is a positive whole number, 251 is a valid term: it is the 62nd term of the sequence.

Quick check: after solving for n, always check it is a positive whole number. If n comes out as a fraction or a negative number, the target value is not in the sequence at all, which is itself a valid exam answer.

Worked example 2: quadratic pattern using second differences

A pattern of dots grows as follows: 2, 6, 12, 20, 30, ... Find the nth term.

Solution

Find the first differences: 6 − 2 = 4, 12 − 6 = 6, 20 − 12 = 8, 30 − 20 = 10. The first differences are 4, 6, 8, 10, not constant, so this is not a linear sequence.

Find the second differences: 6 − 4 = 2, 8 − 6 = 2, 10 − 8 = 2. The second differences are constant at 2, which confirms a quadratic (n²) pattern.

The coefficient of n² is half the constant second difference: 2 ÷ 2 = 1. So the nth term starts with n².

Subtract n² from each term to find what is left: n = 1: 2 − 1² = 1. n = 2: 6 − 2² = 2. n = 3: 12 − 3² = 3. n = 4: 20 − 4² = 4. The remainder is exactly n.

The nth term is n² + n. Check against n = 5: 5² + 5 = 25 + 5 = 30 ✓.

The step students get wrong

Stopping at the first differences and concluding "there is no pattern" because they are not constant, or writing a linear nth term anyway to force an answer. If the first differences are changing by a constant amount each time, the sequence is quadratic, not patternless: take the second differences before giving up. Also common: forgetting to halve the second difference, which doubles the coefficient of n² and throws off every term.

Frequently asked questions

Why is the nth term dn + c and not n + d?

The common difference d tells you how much the sequence grows per step, and n counts the steps. Growth over n steps is d multiplied by n, not d added to n. This is the same reason a straight line is written as y = mx + c and not y = x + m: m is the rate of change, and it multiplies x.

What if the second differences are not constant?

Then the sequence is not quadratic. O-Level E-Math only requires linear and quadratic nth terms, so if the second differences keep changing, re-check your arithmetic on the original sequence before assuming a more complex pattern is intended.

Can I use the same method to find a specific term without listing every value?

Yes. Once you have the nth term formula, substitute the term number for n directly. This is faster than extending the sequence term by term, and it is exactly what you should do when a question asks for, say, the 50th term.

How does this connect to the equation of a straight line?

A linear sequence and a straight line share the same structure: dn + c mirrors y = mx + c, with d as the gradient and the first term fixing the intercept. If you can find the equation of a straight line from two points, you already have the skill needed to find a linear nth term.

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