What is the nth-term rule for 2, 5, 8, 11?

It is 3n − 1. The common difference is 3, and the difference becomes the coefficient of n, so the rule starts as 3n. Then you adjust the constant to hit the first term: 3 × 1 = 3, which is one too big, so you subtract 1 to land on 2. Check it: n = 1 gives 3 × 1 − 1 = 2, and n = 2 gives 3 × 2 − 1 = 5. Writing n + 3 instead adds the difference as a constant and fails the first term, since n = 1 gives 4, not 2.

Why is the rule not n + 3 when the sequence goes up by 3?

Because the common difference tells you how much the rule grows for each step of n, which is a multiplier, not a number to add on. If you add 3 as a constant, the gap between n + 3 for n = 1 and for n = 2 is only 1, not 3, so the sequence would rise by 1 each time. Using 3n makes the rule rise by 3 each step, which matches the sequence; then a constant is added or subtracted once to fix the starting value.

How do you find the constant in an nth-term rule?

First take the common difference as the coefficient of n. For a difference of 3 the rule starts 3n. Then work out 3n for n = 1, which is 3, and compare it with the actual first term, 2. Here 3 is one too big, so subtract 1, giving 3n − 1. The constant is whatever you must add or subtract to turn the coefficient times 1 into the real first term.

GCSE Maths Foundation · AQA · Sequences

Finding the nth-term rule: the difference multiplies n, so 2, 5, 8, 11 is 3n − 1

Asked for the nth-term rule of $2,\ 5,\ 8,\ 11$ , students spot the common difference of 3 and write it as a constant to add — $n + 3$ — when the difference is the coefficient of n. Test it: at $n = 1$ , $n + 3$ gives 4, not the 2 the sequence starts on. The right rule is $3n - 1$ .

The thirty-second fix: the common difference becomes the number in front of n, then adjust the constant so the rule hits the first term. Difference 3 gives $3n$ ; at $n = 1$ that is 3, one too big, so subtract 1 for $3n - 1$ . Check: $3 \times 1 - 1 = 2$ and $3 \times 2 - 1 = 5$ .

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 wrote the difference as a constant, e.g. $n + 3$ for $2,\ 5,\ 8,\ 11$ , when the difference is the coefficient of n, so the rule is $3n - 1$ .
Your rule failed the first term: at $n = 1$ , $n + 3 = 4$ , not the 2 the sequence starts on.
You used the difference as the coefficient but forgot to adjust the constant, writing $3n$ (which gives 3, 6, 9, …) instead of $3n - 1$ .
You never checked the rule against $n = 1$ and $n = 2$ to confirm it reproduces the sequence.

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.

$2,\quad 5,\quad 8,\quad 11$

Find an expression for the nth term.

The misconception is to see the gap of 3 and bolt it on as a constant — $n + 3$ — as if “up by 3” meant “add 3”. But at $n = 1$ that gives 4, not 2, so it is the wrong rule.

Put the difference in front of n: $3n$ . At $n = 1$ that is 3, one too big, so subtract 1 to land on 2. The rule is $3n - 1$ .

Why students fall for this

“The sequence goes up by 3” sounds like an instruction to add 3, so students write $n + 3$ . But the position number n already steps up by 1 each term, and adding a fixed 3 makes the expression rise by only 1 each step, not 3 — so $n + 3$ gives $4,\ 5,\ 6,\ 7$ , the wrong sequence entirely. The difference has to multiply n so that the rule grows by 3 for every step of n.

Once the coefficient is right, the rule still needs anchoring to the correct start. $3n$ alone gives $3,\ 6,\ 9,\ 12$ — the right gaps but starting one too high. Comparing $3n$ at $n = 1$ (which is 3) with the real first term (2) shows you must subtract 1, giving $3n - 1$ .

AQA Foundation papers exploit this directly: a linear sequence where you must give the nth-term rule, with $n + d$ sitting as the tempting wrong answer that fails the very first term.

Worked example — checking the trap. Does $n + 3$ generate $2,\ 5,\ 8,\ 11$ ?

Substitute the positions $n = 1, 2, 3, 4$ :

n + 3:\quad 4,\ 5,\ 6,\ 7

Wrong from the first term and rising by only 1 each step. The rule must rise by 3, so the 3 belongs in front of n: $3n - 1$ gives $2,\ 5,\ 8,\ 11$ .

The fix: The common difference is the coefficient of n; then adjust the constant to hit the first term

Find the common difference. For $2,\ 5,\ 8,\ 11$ each step adds 3, so the difference is 3.

Put it in front of n. The difference is the coefficient, so the rule starts $3n$ — never $n + 3$ .

Adjust the constant. Work out $3n$ at $n = 1$ , which is 3, and compare with the first term, 2. It is one too big, so subtract 1: $3n - 1$ .

Check two terms. $3 \times 1 - 1 = 2$ and $3 \times 2 - 1 = 5$ — both match, so the rule is right.

Worked example

Find the nth-term rule for $2,\ 5,\ 8,\ 11$ . The trap is to read “up by 3” as $n + 3$ ; the fix is to make 3 the coefficient and then adjust.

Find the common difference. $5 - 2 = 3$ , $8 - 5 = 3$ , $11 - 8 = 3$ — the sequence rises by 3 each time.
Make the difference the coefficient of n. The rule starts $3n$ , which gives $3,\ 6,\ 9,\ 12$ — the right gaps, one too high.
Adjust the constant. Compare $3n$ at $n = 1$ (which is 3) with the first term (2): subtract 1.
$\text{nth term} = 3n - 1$
Check it.
$3 \times 1 - 1 = 2 \qquad 3 \times 2 - 1 = 5$
Both match, so $3n - 1$ is correct — not the $n + 3$ that gives 4 at $n = 1$ .

So the difference always becomes the number in front of n, and the constant is whatever you must add or subtract to anchor the rule to the first term — here, $3n - 1$ , not $n + 3$ .

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 nth-term rule for $2,\ 5,\ 8,\ 11$ ?: It is $3n - 1$ . The common difference is 3, and the difference becomes the coefficient of n, so the rule starts $3n$ . Then adjust the constant to hit the first term: $3 \times 1 = 3$ , one too big, so subtract 1 to land on 2. Check it: $n = 1$ gives $3 \times 1 - 1 = 2$ , and $n = 2$ gives $3 \times 2 - 1 = 5$ . Writing $n + 3$ adds the difference as a constant and fails the first term, since $n = 1$ gives 4, not 2.
Why is the rule not $n + 3$ when the sequence goes up by 3?: Because the common difference tells you how much the rule grows for each step of n, which is a multiplier, not a number to add on. If you add 3 as a constant, the gap between $n + 3$ at $n = 1$ and at $n = 2$ is only 1, not 3, so the sequence would rise by 1 each time. Using $3n$ makes the rule rise by 3 each step, which matches; then a constant is added or subtracted once to fix the starting value.
How do you find the constant in an nth-term rule?: First take the common difference as the coefficient of n. For a difference of 3 the rule starts $3n$ . Then work out $3n$ at $n = 1$ , which is 3, and compare it with the actual first term, 2. Here 3 is one too big, so subtract 1, giving $3n - 1$ . The constant is whatever you must add or subtract to turn the coefficient times 1 into the real first term.

Related misconceptions

Term-to-term vs nth-term rulesThe neighbouring sequences skill: a term-to-term rule tells you how to get the next term, while an nth-term rule jumps straight to any term, so 5, 11, 17, 23 is add 6 or 6n − 1.
Geometric and Fibonacci sequencesWhen the sequence is not linear at all: check the rule first, because geometric sequences multiply by a fixed ratio and Fibonacci sequences add the two previous terms.

← All GCSE Maths Higher misconceptions