Secondary Mathematics · Worked Repair Guide 22
A sequence is more than a list of numbers. It is an ordered structure. The position matters, the rule matters, and different rules can produce the same first few values before separating later. That is why pattern questions are not solved by guessing one continuation that looks reasonable; they are solved by identifying and testing a rule.
This guide develops one central habit: separate observation from generalisation. First describe what changes. Then build a candidate rule. Finally test that rule against known terms and, where possible, an independently generated term.
All examples are original teaching constructions. The conceptual owners remain How Secondary 1 Patterns and Generalisation Work, How Mathematical Generalisation Works and Why Do Mathematics Sequences and Patterns Become Difficult?. This page adds extensive worked repair practice.
1. Position is part of the information
In the sequence 4,7,10,13,… the first term is 4, the second is 7 and the nth term is the value at position n. A rule such as 3n+1 generates 4 when n=1, 7 when n=2 and 10 when n=3.
Do not confuse “term number” with “term value”. The fifth term may equal 16, but that does not make 16 the fifth term number.
Entry check: for 5,9,13,17,… the 6th term is 25. A correct nth-term rule is 4n+1. Check n=6: 24+1=25.
2. Constant first differences signal a linear sequence
The sequence 2,7,12,17,… has first differences 5,5,5. A constant first difference means the term changes at a constant rate with position.
For an arithmetic sequence with first term a and common difference d, the nth term is a+(n−1)d.
Here a=2 and d=5, so term n=2+5(n−1)=5n−3.
Check n=1 and n=4. The rule gives 2 and 17, matching the list.
3. The quick linear rule comes from gradient and intercept thinking
A linear sequence can be viewed as points (n,Tn) on a straight line. The common difference is the gradient.
For 8,11,14,17,… the common difference is 3, so begin with 3n. At n=1, 3n gives 3 but the sequence needs 8, so add 5. The rule is 3n+5.
This shortcut is not magic. It is the line equation Tn=dn+c, where d is the rate of change per one step in n.
Use the Coordinate Geometry guide to see the same structure as gradient and intercept.
4. A negative common difference is still linear
Consider 20,16,12,8,… The first difference is −4.
The nth term is 20+(n−1)(−4)=24−4n.
Check n=1: 20. Check n=5: 4. The sequence is decreasing but still linear in position.
Do not use the absolute difference 4 and then try to repair the sign later. The signed difference records direction.
5. Recursive rules and nth-term rules answer different questions
The recursive description T₁=4, Tₙ₊₁=Tₙ+3 tells us how to get the next term from the current one.
The explicit rule Tₙ=3n+1 tells us how to calculate any term directly from its position.
Both describe the same arithmetic sequence, but they are not interchangeable forms in every task. A question asking for the 100th term is usually more efficient with the nth term; a process description may naturally be recursive.
Always include the starting condition with a recursive rule. “Add 3 each time” alone does not distinguish 1,4,7,… from 5,8,11,…
6. Constant second differences signal a quadratic pattern
Consider 3,8,15,24,35,… First differences are 5,7,9,11. Second differences are 2,2,2.
A constant second difference suggests a quadratic nth term an²+bn+c.
For a quadratic sequence sampled at consecutive integer positions, the constant second difference equals 2a. Since the second difference is 2, a=1.
So start with n². The values of n² are 1,4,9,16,25. Subtract from the sequence: 2,4,6,8,10=2n. Therefore the rule is n²+2n=n(n+2).
7. Derive the 2a rule instead of memorising it
Let Tn=an²+bn+c.
The first difference Tₙ₊₁−Tₙ equals a[(n+1)²−n²]+b = a(2n+1)+b = 2an+a+b.
This is linear in n with gradient 2a. Taking one more difference therefore gives the constant 2a.
The derivation explains why the second difference diagnoses the leading quadratic coefficient.
8. Work a quadratic sequence with a larger second difference
Sequence: 6,17,34,57,86,… First differences 11,17,23,29; second differences 6.
Since 2a=6, a=3. Compare with 3n²: 3,12,27,48,75. The residuals are 3,5,7,9,11=2n+1.
So Tn=3n²+2n+1.
Check n=4: 48+8+1=57. Check n=5: 75+10+1=86.
9. A table can reveal the rule more clearly than a decorative shape
Suppose figure 1 uses 5 tiles, figure 2 uses 8, figure 3 uses 11. The visual pattern may be drawn in many ways, but the numerical relationship is currently linear: +3 per figure.
Create a table with columns “figure number n” and “number of tiles T”. If the construction genuinely adds three tiles per new stage, T=3n+2.
The diagram should explain why the +3 occurs. For example, perhaps two horizontal tiles and one vertical tile are added at each stage.
A rule inferred only from three values remains a conjecture until the construction supports it.
10. Several rules can fit a finite beginning
Consider 1,2,4. A simple continuation is doubling, giving 8 next. But another rule can be constructed that agrees for the first three positions and gives a different fourth term.
For example, Tn=2ⁿ⁻¹+(n−1)(n−2)(n−3) agrees with 1,2,4 at n=1,2,3 and also with 8 at n=4 because the added product is zero there; changing the factor structure can make divergence occur later.
The lesson is not that pattern questions are impossible. It is that the intended rule must come from stated structure, a simple recognised family, or additional information—not from certainty that one continuation is logically forced by a short list.
11. Check a proposed nth term against more than one position
A learner proposes Tn=4n−1 for 3,7,11,15,…
Check n=1: 3. Check n=2: 7. Check n=10: 39. The first two establish local agreement; the later check tests whether the same formula continues consistently.
If the sequence was generated by a construction, also derive the term count directly from that construction. Agreement between algebra and structure is stronger than matching the first two terms by coincidence.
12. Determine whether a number is in the sequence
For Tn=5n−2, ask whether 118 is a term. Solve 5n−2=118, giving 5n=120 and n=24.
Since n=24 is a positive integer, 118 is the 24th term.
Ask whether 119 is a term. Solving gives n=121/5=24.2, not an integer position. Therefore 119 is not in the sequence.
The equation can have a real solution while the discrete sequence rejects it because positions are positive integers.
13. Find a missing term from structural information
An arithmetic sequence has 7th term 31 and common difference 4. Since T7=a+6d, a+24=31, so a=7.
The nth term is 7+4(n−1)=4n+3.
Therefore the 20th term is 83.
This reverses the usual problem: the first term is reconstructed from a later position and a rate.
14. Two known terms can determine a linear nth term
Suppose T4=19 and T10=43 in an arithmetic sequence.
Six position steps produce a value increase of 24, so common difference d=24/6=4.
Use T4=a+3d=19 to obtain a=7. Thus Tn=4n+3.
This is exactly a gradient problem on the points (4,19) and (10,43).
15. Geometric sequences change by a constant multiplier
As an extension where relevant, 3,6,12,24,… has common ratio 2. The nth term is 3×2ⁿ⁻¹.
The differences are not constant; they are 3,6,12,… The multiplicative relationship is the useful structure.
For 81,27,9,3,… the ratio is 1/3, so Tn=81(1/3)ⁿ⁻¹.
Do not use an arithmetic-sequence formula merely because the values decrease regularly to the eye. Test additive and multiplicative change separately.
16. Recursive patterns may depend on more than one earlier term
Consider 2,3,5,8,13,… where each new term is the sum of the previous two. A recursive rule needs two starting values:
T1=2, T2=3, Tn=Tₙ₋₁+Tₙ₋₂ for n≥3.
The next terms are 21 and 34.
A first-difference method is not appropriate because the differences 1,2,3,5,… are themselves changing recursively.
17. A visual pattern can be generalised in more than one equivalent way
Suppose figure n is a rectangle n tiles long and 3 tiles high, plus two fixed corner tiles outside the rectangle. One count is 3n+2.
Another decomposition might identify a strip of length n+2 repeated three times and then subtract four overcounted tiles, giving 3(n+2)−4=3n+2.
The two expressions look different but are algebraically equivalent.
Comparing two valid counts is a powerful way to link visual structure with algebraic simplification.
18. A difference table can diagnose degree, but not prove the generating law
Constant first differences fit a linear polynomial in n. Constant second differences fit a quadratic polynomial. Constant third differences fit a cubic polynomial, and so on, when positions are equally spaced.
This is a useful interpolation fact, but a finite list may also come from a non-polynomial rule designed to agree at those positions.
Therefore finite differences identify a natural polynomial model rather than revealing the only logically possible rule.
That distinction becomes important when moving from school pattern recognition to mathematical modelling.
19. Capstone: build and test a quadratic rule from a growing figure
An invented square-frame pattern uses 8,15,24,35 tiles in figures 1–4.
First differences are 7,9,11; second differences are 2, so a=1.
Subtract n² from the terms: 7,11,15,19. This residual is 4n+3. Therefore Tn=n²+4n+3=(n+1)(n+3).
Check figure 4: 16+16+3=35. Predict figure 5: 25+20+3=48.
If a diagram construction can independently be counted as a rectangle (n+1) by (n+3), that provides a structural explanation for the factorised form.
20. Independent practice
- Find the next three terms of 4,9,14,19,…
- Find the nth term of 4,9,14,19,…
- Find the nth term of 20,17,14,11,…
- Find the 50th term of 7,11,15,19,…
- Decide whether 202 is a term of 6n+4.
- Decide whether 203 is a term of 6n+4.
- An arithmetic sequence has first term 12 and common difference −5. Find Tn.
- An arithmetic sequence has T8=34 and d=3. Find the first term.
- An arithmetic sequence has T3=14 and T9=38. Find d and Tn.
- For 2,7,14,23,34,… find first and second differences.
- Find the nth term of the sequence in Question 10.
- For 5,16,33,56,85,… find the nth term.
- Write a recursive rule for 6,10,14,18,…
- Write a recursive rule for 1,1,2,3,5,8,…
- Find the nth term of 3,6,12,24,… if geometric sequences are in scope.
- Find the 8th term of 81,27,9,3,…
- A linear pattern has T5=21 and T12=49. Find its nth term.
- Explain why matching the first three terms does not logically prove a unique infinite continuation.
- A tile pattern has counts 7,12,17,22. Give a linear rule and check figure 10.
- A pattern has counts 8,15,24,35. Find figure 6 using a quadratic rule.
21. Worked answers
1. 24,29,34. Add 5 each time.
2. 5n−1. Check n=1 gives 4.
3. 23−3n. First term 20 and difference −3.
4. 203. Rule 4n+3; n=50 gives 203.
5. Yes. 6n+4=202 gives n=33, a positive integer.
6. No. n=199/6 is not an integer.
7. Tn=17−5n. Equivalent form 12−5(n−1).
8. 13. T8=a+7(3)=34.
9. d=4 and Tn=4n+2. Six steps change the value by 24.
10. First differences 5,7,9,11; second differences 2,2,2.
11. n²+2n−1. Subtract n² to get 1,3,5,7,9=2n−1.
12. 3n²+2n. Second difference 6 gives a=3; residual after 3n² is 2n.
13. T1=6, Tn+1=Tn+4.
14. T1=1, T2=1, Tn=Tn−1+Tn−2 for n≥3.
15. 3×2ⁿ⁻¹.
16. 1/27. Since 81=3⁴, we have 81(1/3)⁷=3⁴·3⁻⁷=3⁻³=1/27.
17. 4n+1. Seven position steps from 5 to 12 change the value by 28, so d=4; T5=21 gives a=5.
18. A finite prefix can be matched by multiple rules; more structure or assumptions are needed to select an intended continuation.
19. 5n+2; figure 10 has 52 tiles.
20. 63. The rule is n²+4n+3; at n=6, 36+24+3=63.
22. Diagnose pattern errors by the level of generalisation
Common failures include confusing term number with term value, using a positive common difference when the sequence decreases, applying an arithmetic formula to multiplicative growth, treating a recursive description as an nth term, or assuming one visually plausible continuation is logically forced.
A useful repair note identifies the structure: “constant first difference → linear candidate”, “constant second difference → quadratic candidate”, “same multiplier → geometric candidate”, or “rule matches values but still needs structural justification”.
After a repair, ask for a term far from the beginning and one membership test. A rule that survives only the first two positions has not yet demonstrated transferable control.
23. Continue through the BTT learning routes
Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Graphs, Tables and Relationships for representation changes and Quadratic Equations, Factorisation and Roots for quadratic algebra.
Within Batch 06, use Coordinate Geometry for the linear connection between n and Tn, Transformations, Symmetry, Coordinates and Invariants for geometric pattern structure, and Direct and Inverse Proportion, Variation and Rates for multiplicative relationships.
24. Sources and scope
The sequences, difference tables, tile counts and practice questions are original teaching material. The finite-difference results are used as mathematical structure for polynomial sequences sampled at consecutive integer positions.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match geometric, recursive and higher-degree extensions to the learner’s actual subject level and school programme.