What is the nth-term rule for 5, 11, 17, 23?

It is 6n − 1. The common difference is 6, so 6 is the coefficient of n; then 6 × 1 = 6 is one too big for the first term 5, so subtract 1, giving 6n − 1. Check it: n = 1 gives 6 × 1 − 1 = 5 and n = 2 gives 6 × 2 − 1 = 11. The term-to-term rule for the same sequence is just add 6, which is a different answer for a different question.

Why can't I answer with add 6 when asked for the nth term?

Because add 6 is the term-to-term rule: it only tells you how to step from one term to the next, so to reach the 50th term you would have to add 6 forty-nine times. The nth-term rule, 6n − 1, lets you find any term directly from its position, so the 50th term is 6 × 50 − 1 = 299 in one step. If a question asks for the nth term or an expression for the nth term, it wants the rule in terms of n, not the step.

GCSE Maths Foundation · AQA · Sequences

Term-to-term vs nth-term rules: add 6 is not 6n − 1

For $5,\ 11,\ 17,\ 23$ students who are asked for the nth-term rule often answer add 6 — the term-to-term step — because it is the first pattern they see. But these are two different rules answering two different questions: add 6 tells you how to get the next term, while $6n - 1$ lets you jump straight to any term from its position.

The thirty-second fix: decide which the question wants — the step to the next term (term-to-term) or an expression in n that reaches any term (nth-term) — and never give a listed term as if it were the step. Here add 6 is term-to-term; $6n - 1$ is the nth-term rule, with $n = 1 \to 5$ and $n = 2 \to 11$ .

Already know this is your gap? Skip the diagnostic and jump straight into the targeted lesson.

How to spot it in your own work

You answered add 6 when the question asked for the nth-term rule, giving the term-to-term step instead of $6n - 1$ .
You wrote an expression in n when the question wanted the term-to-term rule — giving $6n - 1$ instead of plain add 6.
You treated a listed term as the step, e.g. taking the 6 inside $6n - 1$ as a term of the sequence rather than the common difference.
You tried to reach a far-off term by stepping rather than substituting, adding 6 over and over instead of using $6n - 1$ directly.

An exam question that triggers it

Here is a canonical AQA Foundation trigger (non-calculator paper, find the rule):

Here are the first four terms of a sequence.

$5,\quad 11,\quad 17,\quad 23$

Find an expression for the nth term.

The misconception is to answer add 6, because the step between terms is the most obvious pattern. But that is the term-to-term rule; the question asked for an expression in n.

The difference 6 becomes the coefficient of n, then adjust: at $n = 1$ , $6n = 6$ is one too big, so subtract 1. The nth-term rule is $6n - 1$ .

Why students fall for this

The step between terms is the first thing you notice in a sequence, so add 6 feels like “the rule”. It is a rule — the term-to-term rule — but it only describes how to move from one term to the next. It needs the previous term to do anything, so to find a distant term you would have to step through every one before it.

The nth-term rule answers a different question: given a position n, what is the term? That lets you jump straight there. For $5,\ 11,\ 17,\ 23$ the difference 6 is the coefficient of n, and adjusting the constant gives $6n - 1$ : $n = 1 \to 5$ , $n = 2 \to 11$ , and the 50th term is $6 \times 50 - 1 = 299$ in one calculation.

AQA Foundation papers exploit this directly: they may ask for the nth term (wanting the expression in n) or for the next term / the rule to continue (wanting the step), and the wrong kind of rule scores nothing even when the arithmetic is right.

Worked example — two questions, two rules. For $5,\ 11,\ 17,\ 23$ , give both the term-to-term rule and the nth-term rule.

The step is 6; the expression in n adjusts the constant:

\text{term-to-term: } +6 \qquad \text{nth term: } 6n - 1

Both are correct — for different questions. Offering add 6 when the nth term was asked for, or $6n - 1$ when the step was asked for, answers the wrong question.

The fix: Decide which rule the question wants: the step to the next term, or an expression in n that reaches any term

Read what is asked. “Find the nth term” or “an expression for the nth term” wants a rule in n. “Write the next term” or “describe the rule to continue” wants the term-to-term step.

Term-to-term is the step. For $5,\ 11,\ 17,\ 23$ it is add 6 — it needs the previous term and only reaches the next one.

nth-term is the position rule. The difference 6 is the coefficient of n; adjust the constant to hit the first term, giving $6n - 1$ . It reaches any term directly.

Never read a term as the step. The 6 in $6n - 1$ is the common difference, not a term of the sequence — keep the roles separate.

Worked example

For $5,\ 11,\ 17,\ 23$ give the right rule for each question. The trap is to offer the term-to-term step when the nth-term rule was asked for.

State the term-to-term rule. Each step adds 6, so to get the next term you add 6: $23 + 6 = 29$ .
Start the nth-term rule from the difference. The difference 6 is the coefficient of n, so the rule starts $6n$ , which gives $6,\ 12,\ 18,\ 24$ — one too high each time.
Adjust the constant. Compare $6n$ at $n = 1$ (which is 6) with the first term (5): subtract 1.
$\text{nth term} = 6n - 1$
Check, then jump.
$6 \times 1 - 1 = 5 \qquad 6 \times 2 - 1 = 11$
And the 50th term is $6 \times 50 - 1 = 299$ straight away — something add 6 could only reach by stepping 49 times.

So the two rules are not interchangeable: add 6 answers “how do I get the next term?” and $6n - 1$ answers “what is the term in position n?” — give the one the question asked for.

Find out if this is costing you marks

The 10-minute diagnostic checks for this pattern (and four others) using AQA-style GCSE Higher items. Free, no signup, anonymous.

Take the diagnostic →

Common questions

What is the difference between a term-to-term rule and an nth-term rule?: They answer different questions. A term-to-term rule tells you how to get from one term to the next, for example add 6 in $5,\ 11,\ 17,\ 23$ . An nth-term rule lets you jump straight to any term from its position number, for example $6n - 1$ , where $n = 1$ gives 5 and $n = 2$ gives 11. Term-to-term needs the previous term to work; the nth-term rule needs only the position.
What is the nth-term rule for $5,\ 11,\ 17,\ 23$ ?: It is $6n - 1$ . The common difference is 6, so 6 is the coefficient of n; then $6 \times 1 = 6$ is one too big for the first term 5, so subtract 1, giving $6n - 1$ . Check it: $n = 1$ gives $6 \times 1 - 1 = 5$ and $n = 2$ gives $6 \times 2 - 1 = 11$ . The term-to-term rule for the same sequence is just add 6, which is a different answer for a different question.
Why can't I answer with add 6 when asked for the nth term?: Because add 6 is the term-to-term rule: it only tells you how to step from one term to the next, so to reach the 50th term you would have to add 6 forty-nine times. The nth-term rule, $6n - 1$ , lets you find any term directly from its position, so the 50th term is $6 \times 50 - 1 = 299$ in one step. If a question asks for the nth term or an expression for the nth term, it wants the rule in terms of n, not the step.

Related misconceptions

Finding the nth-term rule of a sequenceHow the nth-term rule is built: the common difference is the coefficient of n, then adjust the constant to hit the first term, so 2, 5, 8, 11 is 3n − 1.
Geometric and Fibonacci sequencesWhen the term-to-term rule is not adding a constant at all: geometric sequences multiply by a fixed ratio and Fibonacci sequences add the two previous terms.

← All GCSE Maths Higher misconceptions