Sequences and the nth Term

La méthode, étape par étape

1. 1

   Find the common difference

   5, 8, 11, 14, … → difference is +3

   Subtract each term from the one after it. If the difference is constant the sequence is linear and the method below applies; if it is not, the sequence is quadratic or something else and needs a different approach.

2. 2

   The difference is the coefficient of n

   difference 3 → the rule starts 3n

   A sequence going up in threes must contain 3n, because 3n is the sequence 3, 6, 9, 12 — which climbs at exactly the right rate. Everything left to do is a constant adjustment.

3. 3

   Adjust to match the first term

   3n gives 3 when n=1, but the sequence starts at 5 → add 2 → 3n + 2

   Compare 3n against the actual sequence at n = 1 and add or subtract the gap. Because both climb at the same rate, one adjustment fixes every term at once rather than just the first.

4. 4

   Check against a later term

   n = 4: 3(4) + 2 = 14 ✓

   Testing only the first term proves nothing about the rate, so always verify a term further along. This catches a right-constant-wrong-coefficient error, which the n = 1 check would pass.

5. 5

   For quadratic sequences, use second differences

   2, 6, 12, 20 → differences 4, 6, 8 → second difference 2 → n² term is 1n²

   Halve the second difference to get the coefficient of n². Subtract that n² sequence from the original, and what remains is a linear sequence you already know how to handle.

The rule uses position, not the previous term

"Add 3 each time" describes the sequence but is not an nth term rule, because using it to find the 100th term requires generating the 99 before it. The nth term rule takes the position directly: 3(100) + 2 = 302, in one step.

This distinction — term-to-term versus position-to-term — is what the topic is actually testing. Students who have not made it will describe the pattern correctly and still be unable to answer the question that was asked.

• Common difference → coefficient of n
• Adjust with a constant to match term 1
• Always verify against a later term
• Second differences → quadratic sequences
• Halve the second difference for the n² coefficient

Sequences that go down

A sequence falling by 4 each time has common difference −4, so its rule contains −4n. For 20, 16, 12, 8 the rule is −4n + 24. The negative coefficient is the whole of the difficulty, and it is a sign-handling problem rather than a sequences problem.

Students who drop the minus produce 4n + something and then cannot make any constant fit. When no constant works, a wrong sign on the n term is almost always the reason.

The sequences worth recognising on sight

Square numbers (1, 4, 9, 16), cube numbers (1, 8, 27, 64), triangular numbers (1, 3, 6, 10) and the Fibonacci sequence (1, 1, 2, 3, 5, 8) appear regularly and are not linear. Trying to find a common difference for any of them wastes time and produces nothing.

The triangular numbers have nth term n(n+1)/2, and Fibonacci has no simple position rule at all — each term is the sum of the two before it. Recognising which family a sequence belongs to is the first decision, before any method is chosen.

How we teach sequences

We insist on the later-term check every time. Verifying only n = 1 passes a rule with the wrong coefficient, and that is exactly the error a student is most likely to have made.

We also teach recognition of the special sequences before the nth term method, because the method does not apply to them. A student who spends four minutes hunting for the common difference of 1, 4, 9, 16 has lost the time and the mark.