When it is neither arithmetic nor geometric, there is still an order to check things in.
Find the next term
2, 6, 12, 20, 30, ?
The gaps run 4, 6, 8, 10 — rising by 2 each time — so the next gap is 12 and the term is 30 + 12 = 42. (They are n² + n, and 6² + 6 = 42.) D just repeats the last gap instead of continuing the pattern in the gaps.
Under the gaps, not under the terms. It costs five seconds and settles most questions.
2, 6, 12, 20, 30
the sequence
4, 6, 8, 10
the differences — not constant
so it is not arithmetic
but they are marching
Constant differences mean arithmetic and you are finished. Differences that themselves form a pattern mean you have simply moved the problem down one row — which is progress.
Constant second differences mean the rule involves n².
second differences all 2
the bottom row is flat
next difference = 10 + 2 = 12
climb back up
next term = 30 + 12 = 42
no formula needed
1×2, 2×3, 3×4, 4×5, 5×6 — the terms are n(n + 1), so the sixth is 6 × 7 = 42. Spotting that is faster; the difference triangle is what you fall back on when you do not.
1, 2, 6, 24
differences 1, 4, 18 — no help
2 ÷ 1 = 2, 6 ÷ 2 = 3, 24 ÷ 6 = 4
the ratios are climbing
next ratio is 5
×2, ×3, ×4, ×5
24 × 5 = 120
these are the factorials
A constant ratio would have made it geometric. A ratio that itself counts 2, 3, 4, 5 is the signature of factorials — 1!, 2!, 3!, 4!, 5!.
2, 3, 5, 8, 13
differences 1, 2, 3, 5 — the sequence again
2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13
each is the sum of the two before
8 + 13 = 21
Fibonacci, started differently
“Add the two before” has to work for the whole list, not just the last pair. A rule that explains four of five terms is the wrong rule, and there is always a choice built for it.
| Family | Starts | Tell |
|---|---|---|
| squares | 1, 4, 9, 16, 25 | differences are the odd numbers |
| cubes | 1, 8, 27, 64 | jumps get large fast |
| triangular | 1, 3, 6, 10, 15 | differences 2, 3, 4, 5 |
| Fibonacci | 1, 1, 2, 3, 5, 8 | each is the sum of two before |
| factorials | 1, 2, 6, 24, 120 | ratios 2, 3, 4, 5 |
| primes | 2, 3, 5, 7, 11, 13 | no pattern in the gaps at all |
| powers of 2 | 2, 4, 8, 16, 32 | doubling |
| n² + 1, n² − 1 | 2, 5, 10, 17 | a square list, shifted |
Half of all pattern questions are one of these eight, or one of them with the signs alternating. If the terms flip +, −, +, −, the rule carries a (−1)ⁿ.
When nothing works, read every second term.
3, 10, 5, 20, 7, 30, ?
differences make no sense
positions 1, 3, 5: 3, 5, 7
going up by 2
positions 2, 4, 6: 10, 20, 30
going up by 10
next is position 7 → 9
back in the odd stream
The giveaway is a sequence that goes up, down, up, down, or one where the terms are wildly different in size. Split it into two lists before trying anything cleverer.
Write the counts as a sequence, then ask what one more piece costs. A shared side is not bought twice.
3, 5, 7, 9, …
each new triangle adds 2 sticks
sticks = 3 + 2(n − 1) = 2n + 1
arithmetic: a₁ = 3, d = 2
n = 25: 2(25) + 1 = 51
25 × 3 = 75 pays for every shared side twice. Only the first triangle needs 3 new sticks; each later one shares a side with its neighbour.
No family fits, and the term is far away. Find the 15th term of 5, 11, 19, 29, 41, …
gaps 6, 8, 10, 12; second gaps 2
constant, so the rule is n² + bn + c
n² coefficient = 2 ÷ 2 = 1
half the second difference
1 + b + c = 5 and 4 + 2b + c = 11
n = 1 and n = 2
b = 3, c = 1: aₙ = n² + 3n + 1
n = 3 gives 19 ✓
a₁₅ = 225 + 45 + 1 = 271
no table of fifteen terms
What is different: climbing the difference triangle to the 15th term is slow. Turn it into a rule in n, test it on a third term, then substitute.
Taking 2n² leads to 2n² + 3, which fits only the first two terms (it gives 21, not 19, for n = 3). The n² coefficient is half the second difference.
Alternating signs. Find the 25th term of 2, −4, 6, −8, …
sizes 2, 4, 6, 8: 2n
ignore the signs first
odd positions +, even positions −
a factor of (−1)ⁿ⁺¹
a₂₅ = +2(25) = 50
25 is odd
What is different: the sign and the size are two separate patterns. Find the size rule, then decide the sign from whether n is odd or even.
Block k ends at position 1 + 2 + … + k = k(k + 1) ÷ 2 — a triangular number.
block 9 ends at 9 × 10 ÷ 2 = 45
positions 1 to 45 are used up
block 10 ends at 10 × 11 ÷ 2 = 55
positions 46 to 55 are all 10
the 50th term is 10
45 < 50 ≤ 55
9 is the block that ends at position 45. Position 50 is past it, inside the next block.
Pattern items show three or four terms or figures and ask for a far one — the 20th, the 50th, the 100th. Write the differences first, turn the pattern into a rule in n, and test the rule on the first three terms before trusting it. A rule that fits only some of the given terms is the wrong rule.
Find the next term
1, 3, 6, 10, 15, ?
The gaps are 2, 3, 4, 5, so the next is 6: 15 + 6 = 21. These are the triangular numbers n(n + 1)2, and the sixth is 6(7)2 = 21. A reuses the last gap of 5.
