Faced with a run of numbers, most people stare at it and hope the rule will present itself. Sometimes it does. When it does not, they lose a full minute with no fallback.

Here is a procedure that solves the large majority of sequences you will meet in a test, in the order it should be applied.

The four attempts

Attempt 1 — differences. Subtract each term from the next. Constant gaps mean an arithmetic progression.

Attempt 2 — differences of the differences. If the gaps vary, do the same to them. A constant second difference signals a quadratic progression, which is very common.

Attempt 3 — ratios. Divide each term by the one before. A constant ratio gives a geometric progression.

Attempt 4 — recurrence and alternation. If none of that worked, test two hypotheses: each term depends on the two before it (Fibonacci-style), or two separate rules alternate across odd and even positions.

Those four attempts take fifteen seconds and, in practice, cover more than nine sequences in ten.

Twelve examples

1. 3, 6, 12, 24, …
Ratios: constant ×2. Answer: 48.

2. 7, 10, 16, 25, 37, …
Differences: 3, 6, 9, 12 — rising by 3. Next gap is 15. Answer: 52.

3. 1, 4, 9, 16, 25, …
The square numbers. Answer: 36. Worth knowing on sight: squares and cubes up to ten should be instant.

4. 2, 3, 5, 8, 13, 21, …
Each term is the sum of the two before it. Answer: 34.

5. 2, 6, 12, 20, 30, …
Differences: 4, 6, 8, 10 — the even numbers. Answer: 42. Equivalently, each term is n × (n+1).

6. 4, 9, 19, 39, 79, …
Neither differences nor ratios are constant. But 4×2+1 = 9 and 9×2+1 = 19. Rule: double then add one. Answer: 159.

7. 1, 2, 6, 24, 120, …
Ratios: ×2, ×3, ×4, ×5. The multiplier climbs by one. Answer: 720.

8. 81, 27, 9, 3, …
Division by 3. Answer: 1. Always read a falling sequence as multiplication by a fraction.

9. 3, 5, 11, 29, 83, …
Gaps explode; ratios are not constant but drift towards 3. Test “×3 minus something”: 3×3−4 = 5, 5×3−4 = 11. Answer: 245.

10. 1, 3, 2, 6, 3, 9, 4, …
None of the first three methods gives anything, because two sequences are interleaved. Odd positions: 1, 2, 3, 4. Even positions: 3, 6, 9. The missing term sits in an even position. Answer: 12.

11. 2, 12, 36, 80, 150, …
Hard. Second differences are not constant either. Decompose: 1³+1², 2³+2², 3³+3²… Answer: 252.

12. 1, 11, 21, 1211, 111221, …
No arithmetic at all. Each term describes the one before, read aloud: 1, then “one 1” (11), then “two 1s” (21), then “one 2, one 1” (1211). Answer: 312211. Rare in formal exams, a staple of technical interviews.

Three traps

Several rules can fit. Over three or four terms, more than one law produces the same sequence. The answer options are what settle it: if your rule yields none of them, your rule is wrong, not the question.

The simplest rule wins. If a one-operation explanation works, that is the intended one. Writers are not trying to defeat you with complexity.

Do not get stuck. Past a minute on one sequence, move on. On a timed paper, a hard item usually costs more than it returns.

The nine number sequences in our full test span exactly this range of difficulty, and each is explained at the end with the reasoning that was expected.