Sequences & patterns

Term-to-term rules, nth-term, picture sequences.

Every sequence hides a rule. Your job is to catch it — and the fastest way is to write the differences between the terms in the gaps above the sequence. Nine times out of ten, the differences give the rule away instantly.

Find the rule — 6, 10, 14, 18, …

Step 1 · write the gaps

+4, +4, +4

Step 2 · name the rule

add 4 each time

Step 3 · use it

next term: 18 + 4 = 22

Special sequences to recognise on sight

Square numbers
1, 4, 9, 16, 25, 36…
Cube numbers
1, 8, 27, 64, 125
Triangular numbers
1, 3, 6, 10, 15, 21…
Doubling (powers of 2)
2, 4, 8, 16, 32, 64…
Add the previous two
3, 4, 7, 11, 18, 29…

Jumping straight to the 10th term

What is the 10th term of 6, 10, 14, 18, …?

  1. The gap is 4, so the rule starts "4 × position"
  2. 4 × 1 = 4, but the first term is 6 — so the rule is 4n + 2
  3. Check on term 2: 4 × 2 + 2 = 10 ✓
  4. 10th term: 4 × 10 + 2

42

⚠ Watch out

Always CHECK your rule on the second term before you use it. A rule that only works for term 1 is the exam's favourite trap — the check takes five seconds and saves the mark.

⚠ Watch out

If the differences look random, stop adding and try dividing: 3, 6, 12, 24 is ×2 each time. And if the third term equals the first two added together, it is an add-the-previous-two sequence.

✏ Your turn

What comes next: 3, 4, 7, 11, 18, ?

Show the answer
  1. The gaps (+1, +3, +4, +7) look random — so test the other rules
  2. Check: 3 + 4 = 7 ✓ and 4 + 7 = 11 ✓ and 7 + 11 = 18 ✓ — each term is the previous two added
  3. Next: 11 + 18

29

For parents — the full topic guide

Algebra, Sequences & Function Machines — 11+ Topic Guide

Part of the GrammarMock topic guides. Drill with the Algebra & Sequences mini-tests.

What "algebra" means at 11+

Nobody expects a Year 5 child to expand brackets like a GCSE student. 11+ algebra is three specific skills: solving a simple equation, running a function machine forwards and backwards, and finding the rule in a sequence. All three are teachable to full fluency.

Solving simple equations

"If 4b − 5 = 23, what is b?"

The method to teach: undo operations in reverse order. The equation did "×4 then −5" to b; to get b back, do "+5 then ÷4". So 23 + 5 = 28, then 28 ÷ 4 = 7.

Say it as a story: "b was multiplied by 4 and then 5 was taken away, leaving 23. Put the 5 back: 28. Un-multiply: 7."

Harder variant with brackets: 3(x − 2) = 18. Undo the ×3 first: x − 2 = 6, so x = 8. The bracket means "this whole thing was multiplied by 3" — undo the outside before the inside.

Two-variable substitution: "If 5a + 3b = 41 and a = 4, find b." Substitute what you know: 20 + 3b = 41, then solve normally. The only skill is not panicking at seeing two letters.

Function machines

Forwards is easy (input 6 → ×3 → +4 → output 22). The exam money is in backwards:

"A machine multiplies by 5 then subtracts 12. The output is 43. What was the input?"

Rule: go backwards through the machine, inverting each operation. Last operation was −12, so first undo: 43 + 12 = 55. Then undo ×5: 55 ÷ 5 = 11.

The classic error is undoing in the forwards order (÷5 first, then +12: 43÷5 = 8.6 + 12 = 20.6 — nonsense, and the nonsense is the tell). Teach: "walk back through the machine from the exit."

Sequences: finding and using the rule

Term-to-term rules ("each term is double the previous, minus 2") — just run the rule carefully, writing every term. Errors here are pure arithmetic slips; writing each step kills them.

Position-to-term (nth term) — the stretch skill: "What is the 10th term of 6, 10, 14, 18...?" The difference is 4, so the rule is "4n plus-or-minus something". 4×1 = 4, but the first term is 6, so the rule is 4n + 2. Check with term 2: 4×2 + 2 = 10 ✓. Then the 10th term is 4×10 + 2 = 42 — no need to write out ten terms.

Teach the check step as compulsory: build the rule from term 1, verify on term 2, only then use it.

Missing middle terms: "4, ?, 16, 32, 64" — work from the ends. The right side doubles, so test doubling from the left: 4 → 8 → 16 ✓.

Consecutive number puzzles

"Two consecutive odd numbers add to 52. What is the larger?"

Method without algebra: half of 52 is 26, so the numbers straddle 26 → 25 and 27. Larger = 27.

Method with algebra (for children ready for it): n + (n + 2) = 52, so n = 25.

Three consecutive numbers summing to S: the middle one is always S ÷ 3. "Three consecutive odd numbers sum to 87" → middle = 29 → they're 27, 29, 31. This ÷3 shortcut turns a hard-looking question into a five-second one.

Age problems (the hard finisher)

"Adam is 3 times as old as Ben. In 7 years, Adam will be twice as old as Ben. How old is Adam?"

These need a table: ages NOW and ages LATER as separate columns. Now: Ben = b, Adam = 3b. In 7 years: Ben = b + 7, Adam = 3b + 7. The condition: 3b + 7 = 2(b + 7). Solve: 3b + 7 = 2b + 14, so b = 7, Adam = 21. Check in the story: now 21 and 7 (3× ✓); in seven years 28 and 14 (2× ✓). The check-in-the-story step is what separates a right answer from a plausible-looking wrong one.

One-week drill plan

  • Days 1-2: equations (one-step, then two-step, then brackets)
  • Day 3: function machines backwards
  • Days 4-5: nth-term rules with the verify-on-term-2 habit
  • Day 6: consecutive number shortcuts + one age problem
  • Day 7: Algebra & Sequences mini-test, timed

Ready to practise this topic?

Try a full paper — questions on sequences & patterns show up in most 11+ maths and VR papers.