Common Mistakes with Nth Term in GCSE Maths

A student can understand sequences perfectly well and still lose marks on the final line. The difference is correct, the method feels familiar, but the expression is one number out. This is why the common mistakes with nth term in GCSE Maths matter: they are usually small, repeatable and fixable.

For a linear sequence, find the common difference, use it as the coefficient of $n$, then adjust the expression so it produces the first term. Finally, substitute $n=1$, $n=2$ and $n=3$ to check it. That last step catches most nth term mistakes before they cost marks.

Your nth term checklist

For an arithmetic, or linear, sequence:

• Find the common difference.
• Write that difference in front of $n$.
• Generate the first few terms of this temporary sequence.
• Compare it with the sequence in the question.
• Add or subtract the required adjustment.
• Check your rule using at least three term numbers.

The standard form is:

$$an+b$$

Here, $a$ is the common difference and $b$ is the adjustment. For instance, if the difference is $4$, begin with $4n$, not simply $4$.

A student realises they have overcomplicated a sequence pattern

Mistake: giving the term-to-term rule

Suppose the sequence is:

$$5,8,11,14,\dots$$

A student may answer “add $3$”. That describes how to move from one term to the next, but it is not the nth term. An nth-term rule must calculate any term directly from its position.

Correct method

The common difference is $3$, so begin with:

$$3n$$

This generates:

$$3,6,9,12,\dots$$

Each term is $2$ below the required sequence, so add $2$:

$$3n+2$$

Check the first term:

$$3(1)+2=5$$

The answer is therefore $3n+2$. The key question is simple: does your answer contain $n$? If it does not, you have probably written a term-to-term instruction rather than a position-to-term rule.

You can reinforce this distinction in the Sequences Nth Term lesson for Edexcel GCSE.

Mistake: using the first term as the coefficient

Consider:

$$7,12,17,22,\dots$$

A common wrong answer is $7n+5$. This happens because $7$ is visually prominent: it is the first number students see. But the coefficient of $n$ comes from the common difference, not the first term.

Correct method

The sequence increases by $5$, so start with $5n$:

$$5,10,15,20,\dots$$

The required terms are always $2$ greater. Therefore:

$$5n+2$$

Check several terms:

$$5(1)+2=7$$ $$5(2)+2=12$$ $$5(3)+2=17$$

The rule works. If AQA is your exam board, the matching AQA sequences and nth term lesson provides further guided practice.

Once this method makes sense, use the Maths Genie Lessons and Question Bank to practise spotting the difference and checking the adjustment without setting up examples yourself.

Mistake: getting the adjustment sign wrong

Now take:

$$10,16,22,28,\dots$$

The difference is $6$, so the rule begins with $6n$. Students sometimes compare $6$ with $10$, notice a gap of $4$, and write $6n-4$ without considering the direction.

Correct method

Write the temporary and required sequences together:

$$6n:6,12,18,24,\dots$$ $$Required:10,16,22,28,\dots$$

The required sequence is $4$ greater each time, so the rule is:

$$6n+4$$

A reliable habit is to complete the sentence: “I must add $4$ to get from $6$ to $10$.” This is safer than choosing a sign from memory.

Mistake: mishandling a negative difference

Descending sequences often create unnecessary doubt. Consider:

$$18,14,10,6,\dots$$

The common difference is $-4$, not $4$. Therefore, begin with:

$$-4n$$

This gives:

$$-4,-8,-12,-16,\dots$$

To turn the first term, $-4$, into $18$, add $22$. The nth term is:

$$-4n+22$$

Checking confirms it:

$$-4(1)+22=18$$ $$-4(2)+22=14$$

The negative sign belongs to the coefficient because the sequence decreases as $n$ increases.

Mistake: checking only one term

One correct substitution is reassuring, but it is not enough. For the sequence

$$4,9,14,19,\dots$$

a student might incorrectly propose $4n$. It gives the correct first term because $4(1)=4$, but fails immediately afterwards.

The actual difference is $5$, so start with $5n$:

$$5,10,15,20,\dots$$

Subtracting $1$ gives:

$$5n-1$$

Check three positions:

$$5(1)-1=4$$ $$5(2)-1=9$$ $$5(3)-1=14$$

Three substitutions take only a few seconds and test both parts of the expression. You can apply this checking routine to marked questions in the nth term exam booklet.

Mistake: assuming every sequence is linear

A constant first difference indicates a linear sequence. If the first differences change but the second differences are constant, the sequence may be quadratic.

For example:

$$3,8,15,24,35,\dots$$

The first differences are:

$$5,7,9,11$$

These are not constant, so a rule of the form $an+b$ is unsuitable. The second differences are all $2$, indicating a quadratic sequence. In fact, the rule is:

$$n^2+2n$$

Check it with $n=3$:

$$3^2+2(3)=9+6=15$$

Quadratic nth terms require a separate method. Higher-tier students can study it through the quadratic nth term revision guide.

Exam-style practice

Try these without looking back at the examples:

1. Find the nth term of $9,13,17,21,\dots$
2. Find the nth term of $20,17,14,11,\dots$
3. The nth term is $7n-5$. Find the $12$th term.
4. Is $83$ a term in the sequence with nth term $6n+5$?

Answers:

1. The difference is $4$, so the nth term is $4n+5$.
2. The difference is $-3$, so the nth term is $-3n+23$.
3. $7(12)-5=79$.
4. Solve $6n+5=83$, giving $6n=78$ and $n=13$. Since $13$ is a positive integer, $83$ is a term.

Notice that the fourth question needs more than substitution. You form and solve an equation, then check that the resulting value of $n$ is a valid term position.

Turn the method into marks

Knowing the checklist is different from applying it accurately when a question is unfamiliar. Keep following the student’s corrected routine: identify the difference, build the $n$ term, adjust it and check three positions. Start the matching Maths Genie nth term Lessons and Question Bank and practise fixing these exact errors before your next GCSE paper.