Keep the two previous positions and the constant distinct.
Check bounded recurrences at every transition, including equivalent rule descriptions. Six practice answers supply no norm, IQ or hiring prediction.
The six questions each contribute one correct answer. The count describes this fixed rule family and edition, without a speed adjustment, IQ conversion or comparison with other people. Every worked answer and this complete guide is available at every count.
A fixed formula needs initial values
Each question gives the first two terms and restricts subsequent terms to p times the previous term plus q times the term before it plus a constant c. The permitted coefficients p and q are integers from −2 to 3; c is an integer from −3 to 3. One set of coefficients must fit the whole displayed row from the third term onward. These limits make the candidate family explicit instead of inviting an unlimited search for any formula that can fit a short prefix.
Generate a new row one term at a time
Start with 2 and 5 and use previous plus the term before it plus 2. The third term is 5 + 2 + 2 = 9, the fourth is 9 + 5 + 2 = 16, and the fifth is 16 + 9 + 2 = 27. The next is 27 + 16 + 2 = 45. After each calculation, advance the pair of previous values; keeping the original starting pair forever would not implement a recurrence.
Coefficient order changes the answer
For a new example starting 3 and 8, let the rule be twice the previous term minus the term before it plus 1. This gives 14, 21, 29 and then 38. Check the first step: 2 × 8 − 3 + 1 = 14. Reversing the two positions would use 2 × 3 − 8 + 1 and produce −1 instead. Write the immediately previous value beside p and the earlier value beside q before performing the arithmetic.
One successful transition does not verify a rule
Many coefficient sets can reproduce a single transition from two starting terms to a third value. Test a proposed formula against every displayed transition before using it at the blank. A candidate that gets the third term right but misses the fourth is rejected, even if its next number appears among the choices. The answer-bank check enumerates every coefficient set in the stated bounds for each complete candidate row. This verifies an option under that finite family, not the psychometric quality of the practice.
An answer can be unique when rule descriptions are not
Some rows admit more than one coefficient description that behaves identically on all their terms. For 1, 3, 7, 15, 31, ?, “twice the previous plus one” yields 63. The rule “three times the previous minus twice the earlier term” also gives 7, 15, 31 and 63. Both lie inside the stated bounds. The practice checks that only one offered missing value works, while allowing such equivalent descriptions. It does not claim that every row identifies a unique triple of coefficients.
Audit a calculation before expanding the model
If a recurrence seems to fail, check coefficient order, signs, the constant and the two values you carried forward. The constant is added once per new term, not once to each earlier term. Then test the formula on an earlier transition again. If it still fails, reject it rather than inventing a one-time exception. For independent practice, change the two initial values while keeping a chosen valid formula and generate a fresh row; explain each transition without referring to the original answer letters.
A finite-family exercise is not a general reasoning diagnosis
The six-item result is a count of selected answers under these published coefficient limits. It supplies no speed estimate, intelligence score, percentile, employer pass line or certified mathematical competence. It also does not show that the same recurrence governs a real measured process. All keys, rationales and this full guide remain free at every count. Paper and a calculator may be used. Read a missed item as a specific opportunity to inspect the model or arithmetic, rather than as a fixed label about your capacity to learn.