How can a Punggol student understand number patterns and nth-term expressions before Secondary 1 in 2028? Begin by telling the difference between a term’s position and its value. Observe what changes from one term to the next, test whether that change remains constant, then write a rule that produces the correct value for any specified position. A good nth-term expression is an explanation that can be tested, not a mysterious formula copied from the final line of a worksheet.
Pupils completing PSLE in 2027 will normally enter Secondary 1 in 2028, studying Mathematics at the relevant G1, G2 or G3 level under Full Subject-Based Banding. The published MOE G2/G3 Mathematics syllabus includes recognising patterns and writing an algebraic expression for the nth term within the Secondary One algebra progression. The exact expectations and extension depth vary by subject level and school. Non-linear sequences, formal proofs and deeper recurrence rules in this guide are enrichment where appropriate, not a compulsory list for every pupil.
This original Punggol learning guide uses forty-four worked sequences, tables, shape-growth stories and error checks. Its bead and tile arrangements are imaginary classroom models, not measured features of Punggol parks or schools. It connects the Primary habit of extending patterns to the Secondary habit of expressing a general mathematical relationship with a letter.
Term number and term value are not the same thing
Write the sequence 5, 8, 11, 14, 17. The first term has value five, the second term has value eight, and the third has value eleven. The term number n counts the position. The term value is the number at that position. Mixing these up leads to expressions that calculate the wrong quantity even when every addition is correct.
The difference between consecutive terms is three. To reach term n from the first term, we take n−1 steps of size three. Hence term n equals 5+3(n−1), or 3n+2. A strong pupil can explain why the expression contains n−1 differences even though the term number is n. The expression is not simply ‘difference times n’ unless the starting value fits that form.
An immediate check is to substitute n=1: 3(1)+2=5. A second check uses n=4 and gives fourteen. When a proposed rule reproduces several known terms and reflects the construction of the sequence, it is more credible. One matching term alone does not prove the rule fits the intended pattern.
Three representations for one growing pattern
Imagine a line of squares built with matchsticks. One separate square needs four sticks. If the next square shares a side, each extra square adds three more sticks. The first three counts become 4, 7 and 10. The drawing, the table of term number and stick count, and the formula 3n+1 all describe the same underlying structure.
A learner can derive the formula by saying ‘start with one stick on the end and add three sticks for every square.’ That gives 3n+1 directly. Another explanation starts at four and adds three for each of the next n−1 squares: 4+3(n−1). Both simplify to 3n+1. Comparing these perspectives helps the child understand that two algebraic expressions can be equivalent.
The representation should remain tied to the actual geometry. If squares do not share sides, each adds four sticks and the nth count is 4n instead. A sketch with a different connection pattern changes the rule. Students should not infer a formula from the word ‘square’ alone.
A constant first difference gives a simple linear rule
| Term number n | 1 | 2 | 3 | 4 | 5 |
| Sequence A | 5 | 8 | 11 | 14 | 17 |
| Difference from previous | — | +3 | +3 | +3 | +3 |
| Sequence B | 12 | 9 | 6 | 3 | 0 |
| Difference from previous | — | −3 | −3 | −3 | −3 |
For the first sequence, nth term is 3n+2. For the second it is 15−3n, because the first term must be twelve and each next term falls by three. These are linear patterns with constant signed differences. A negative common difference does not make the position n negative; it means the values decrease as the index increases.
The method for finding a linear nth term from first value a and common difference d is a+(n−1)d. This can be rearranged to dn+(a−d). Teach the meaning of each part, then use the condensed expression when the pupil is ready. A memorised rule without the n−1 explanation is vulnerable when sequences start at unusual values or decrease.
A table is evidence, but it does not make every pattern linear
The sequence 1, 4, 9, 16 has differences 3, 5 and 7, so the first differences are not constant. The first four values happen to be 1², 2², 3² and 4², suggesting the nth term n² under that defined pattern. A pupil who forces the common difference three across every term gets the fourth and later terms wrong.
Another example is 2, 4, 8, 16, where each term is twice the previous term. The successive differences are 2,4,8, not constant; the simple rule 2n would give six at n=3 instead of eight. These are useful enrichment comparisons, but the central Secondary One objective here is to recognise simple algebraic patterns and translate them into an expression.
A finite list alone can be consistent with many mathematical rules. The problem should supply a pattern description or enough context to support the intended generalisation. When a pupil derives a rule, ask how the next term is built and whether the same construction is being maintained. Mathematical reasoning includes recognising the limits of what has been given.
Eighteen original worked sequences: the first learning ladder
Original pattern question 1 · increasing by three
Question. 5, 8, 11, 14, … Find the nth term.
Working. 3n+2.
Reasoning. The first term is five; add three for each of n−1 moves.
Check. n=1 gives five, n=4 gives fourteen.
Original pattern question 2 · even positive numbers
Question. 2, 4, 6, 8, … Find the nth term.
Working. 2n.
Reasoning. Term n contains n groups of two.
Check. n=5 gives ten.
Original pattern question 3 · odd positive numbers
Question. 1, 3, 5, 7, … Find the nth term.
Working. 2n−1.
Reasoning. Odd values are one less than twice their positive index.
Check. n=4 gives seven.
Original pattern question 4 · starting above zero
Question. 7, 12, 17, 22, … Find term n.
Working. 5n+2.
Reasoning. Difference five and first value seven give 7+5(n−1).
Check. n=3 gives seventeen.
Original pattern question 5 · decreasing
Question. 20, 16, 12, 8, … Find the nth term.
Working. 24−4n.
Reasoning. Begin at twenty, subtract four for each new position.
Check. n=1 gives twenty; n=4 gives eight.
Original pattern question 6 · constant sequence
Question. 9, 9, 9, 9, … Find term n.
Working. 9.
Reasoning. The difference is zero, so the value does not depend on n.
Check. Any positive n has value nine.
Original pattern question 7 · find fifth term
Question. Sequence has rule 4n+3. Find term five.
Working. 23.
Reasoning. Substitute n=5 into the formula.
Check. 4×5+3=23.
Original pattern question 8 · find eighth term
Question. Sequence follows 6n−2. Find term eight.
Working. 46.
Reasoning. Substitute term number eight, not the previous value.
Check. 6×8−2=46.
Original pattern question 9 · recover the position
Question. A sequence has rule 3n+2. Which term has value 32?
Working. n=10.
Reasoning. Solve 3n+2=32, so 3n=30.
Check. At n=10, value is thirty-two.
Original pattern question 10 · another position
Question. Sequence follows 5n−1. Which term equals 44?
Working. n=9.
Reasoning. Solve 5n−1=44.
Check. 5×9−1=44.
Original pattern question 11 · a descending term index
Question. For 18−2n, which term first has value zero?
Working. n=9.
Reasoning. Set 18−2n=0.
Check. 18−2×9=0.
Original pattern question 12 · sign matters
Question. A sequence has rule 7−3n. Find its fourth value.
Working. −5.
Reasoning. 7−12 is negative five; the term can be negative.
Check. n=1 gives four, then one, −2, −5.
Original pattern question 13 · matchsticks in joined squares
Question. n squares share sides in a straight chain; first needs four sticks, each extra three. Find count.
Working. 3n+1.
Reasoning. Start with four and add 3(n−1).
Check. Two joined squares need seven sticks.
Original pattern question 14 · separate squares
Question. n separate squares each need four sticks. Write total.
Working. 4n.
Reasoning. No sides are shared.
Check. Two independent squares need eight sticks.
Original pattern question 15 · repeated triangle panels
Question. n separate triangles each use three sticks. Write total.
Working. 3n.
Reasoning. Each panel is a fixed independent group of three.
Check. n=6 gives eighteen.
Original pattern question 16 · fixed border added
Question. A decoration has three tiles per stage n plus five fixed corner tiles. Write count.
Working. 3n+5.
Reasoning. The corner tiles occur once, not once per group.
Check. n=2 gives eleven.
Original pattern question 17 · alternative equivalent form
Question. Simplify 4+3(n−1).
Working. 3n+1.
Reasoning. Distribute three and collect constants.
Check. At n=4 both forms equal thirteen.
Original pattern question 18 · a false formula
Question. A pupil claims nth term for 5,8,11 is 3n. Test the first term.
Working. False; 3n gives 3 at n=1.
Reasoning. The missing fixed offset is two.
Check. Correct 3n+2 produces five at n=1.
Thirty more pattern investigations: translation, generalisation and limits
The first eighteen examples established straightforward nth terms. The following investigations change what is given: sometimes the first value is known, sometimes the difference is negative, sometimes the term index is unknown, and sometimes the story has a fixed portion alongside a repeated part. Not every example should be assigned to every G-level learner; square and geometric sequences are enrichment comparisons, while the constant-difference and algebraic-translation work is the core bridge where appropriate.
For each question, a trustworthy explanation should identify the first term, the amount added per stage or the precise construction rule, and what n represents. The pupil should verify at least one position from the original sequence. When a new example is only a finite list, the stated intended pattern matters: more than one formula can fit the same finite observations.
Original extended pattern case 19 · difference four
Question. 2,6,10,14,… Find term n.
Working. 4n−2.
Reason. Begin at two, add four for each of n−1 gaps.
Independent check. At n=1 gives two and n=4 gives fourteen.
Original extended pattern case 20 · difference six
Question. 8,14,20,26,… Find nth term.
Working. 6n+2.
Reason. First value eight with six per step gives 8+6(n−1).
Independent check. At n=3 the value is twenty.
Original extended pattern case 21 · difference negative five
Question. 15,10,5,0,−5,… Find rule.
Working. 20−5n.
Reason. Each successive position removes five.
Independent check. At n=1 gives fifteen; at n=5 gives −5.
Original extended pattern case 22 · first term negative
Question. −4,−1,2,5,… Find nth term.
Working. 3n−7.
Reason. First term negative four and each step positive three.
Independent check. n=1 gives −4, n=4 gives five.
Original extended pattern case 23 · pattern starts at zero
Question. 0,4,8,12,… Find nth term.
Working. 4n−4.
Reason. Stage one begins at zero; there have been n−1 increases.
Independent check. At n=1 gives zero.
Original extended pattern case 24 · nth value of a decreasing sequence
Question. For term rule 20−3n, find term seven.
Working. −1.
Reason. Substitute seven into 20−3×7.
Independent check. Term one is seventeen, then decrease by three six times.
Original extended pattern case 25 · locate a known value
Question. For 7n+1, find n when value is 50.
Working. n=7.
Reason. Solve 7n+1=50 to find 7n=49.
Independent check. 7×7+1=50.
Original extended pattern case 26 · locate another known value
Question. For 24−2n, find the index with value 4.
Working. n=10.
Reason. Solve 24−2n=4; divide twenty by two.
Independent check. 24−2×10=4.
Original extended pattern case 27 · find first term given difference
Question. A sequence increases by 4 each step and its fifth term is 23. Find first term.
Working. 7.
Reason. Four intervals precede term five, adding sixteen to first.
Independent check. 7+4×4=23.
Original extended pattern case 28 · find difference from first and fifth
Question. First term is 7 and fifth is 27 in an arithmetic pattern. Find common difference.
Working. 5.
Reason. Four equal intervals gain twenty, so each gains five.
Independent check. 7,12,17,22,27.
Original extended pattern case 29 · matchsticks in triangles sharing one edge
Question. A first triangle uses three sticks; each next joined triangle shares one side and adds two sticks. Find nth count.
Working. 2n+1.
Reason. Start with three and add 2(n−1).
Independent check. At n=2 there are five sticks.
Original extended pattern case 30 · add identical polygons
Question. n separate pentagonal outlines use five sticks each. Find count.
Working. 5n.
Reason. No edges are shared by the stated separate shapes.
Independent check. n=4 needs twenty sticks.
Original extended pattern case 31 · arrangement with fixed starter
Question. A row has n blocks of three tiles each and two fixed end tiles. Find total.
Working. 3n+2.
Reason. Three per repeated block, but two end tiles only once.
Independent check. At n=3 total eleven.
Original extended pattern case 32 · three fixed tiles for each row
Question. Each of n rows contains x patterned tiles and 3 decorations. Write total.
Working. n(x+3).
Reason. The decoration is repeated in every row rather than added once.
Independent check. For n=4, x=2 total twenty.
Original extended pattern case 33 · one fixed decoration
Question. n groups have x objects each, and one group-independent decoration of 3 items. Write total.
Working. nx+3.
Reason. Only the x objects repeat; the extra three happens once.
Independent check. For n=4,x=2 total eleven.
Original extended pattern case 34 · find rule from a table
Question. n:1,2,3,4 and value:11,16,21,26. Find rule.
Working. 5n+6.
Reason. First value eleven, common difference five.
Independent check. At n=4 value 26.
Original extended pattern case 35 · another table
Question. n:1,2,3,4; values 19,16,13,10. Find rule.
Working. 22−3n.
Reason. Each step subtracts three from previous.
Independent check. At n=1 19 and at n=4 ten.
Original extended pattern case 36 · calculate a later term
Question. Sequence 11,16,21,… Find the twentieth term.
Working. 106.
Reason. Rule 5n+6; substitute n=20.
Independent check. 5×20+6=106.
Original extended pattern case 37 · compare two patterns
Question. Rules A=3n+2 and B=2n+6. Which is larger at n=5?
Working. A=17 and B=16, so A by one.
Reason. Use the same index five for both sequences.
Independent check. 3×5+2=17; 2×5+6=16.
Original extended pattern case 38 · where two rules agree (extension)
Question. Find positive n for 3n+2=2n+6.
Working. n=4.
Reason. Subtract 2n and two; one n equals four.
Independent check. Both give fourteen at n=4.
Original extended pattern case 39 · an incorrect shortcut
Question. A pupil says nth term of 7,12,17 is 5n. Find the first wrong value.
Working. n=1: the proposed formula gives five rather than seven.
Reason. Difference five does not by itself specify the initial value.
Independent check. Correct rule is 5n+2.
Original extended pattern case 40 · constant first differences
Question. Sequence 5,10,15,20: does y increase by a constant difference as term index increases one?
Working. Yes, by five.
Reason. Each consecutive subtraction gives five.
Independent check. Rule 5n at n=4 gives twenty.
Original extended pattern case 41 · nonconstant first differences
Question. Sequence 1,4,9,16: are successive differences constant?
Working. No: differences are 3,5,7.
Reason. A linear nth-term rule cannot use one constant d here.
Independent check. Under square-number construction, term n is n².
Original extended pattern case 42 · doubling pattern (extension)
Question. 2,4,8,16,… Find next two terms under a doubling rule.
Working. 32,64.
Reason. Multiply each earlier value by two, not add a fixed amount.
Independent check. A proposed 2n would give six at n=3, incorrectly.
Original extended pattern case 43 · triangular arrangement (extension)
Question. A triangle of dots grows as 1,3,6,10,… Find next term under the row-addition rule.
Working. 15.
Reason. The successive additional rows have 2,3,4, then 5 dots.
Independent check. 1+2+3+4+5=15; general term n(n+1)/2.
Original extended pattern case 44 · triangular number formula (extension)
Question. Find sixth triangular number from n(n+1)/2.
Working. 21.
Reason. 6×7/2=21.
Independent check. The successive row totals 1+2+3+4+5+6=21.
Original extended pattern case 45 · square-number comparison (extension)
Question. A pattern has n² tiles at stage n. Find the fifth stage.
Working. 25.
Reason. Five rows and five columns have 25 cells.
Independent check. 5×5=25.
Original extended pattern case 46 · difference of squares (extension)
Question. Square tiles grow from n² to (n+1)². How many new tiles are added?
Working. 2n+1.
Reason. Expand (n+1)²−n² to get 2n+1.
Independent check. At n=3 the growth is 16−9=7.
Original extended pattern case 47 · stage index is not tile count
Question. A rule says 4n+3 tiles at stage n. How many tiles at stage 8?
Working. 35 tiles.
Reason. The variable n names stage number, not a tile.
Independent check. 4×8+3=35.
Original extended pattern case 48 · reverse an index from a count
Question. Under 4n+3, which stage has 43 tiles?
Working. n=10.
Reason. Solve 4n+3=43; 4n=40.
Independent check. 4×10+3=43.
Original extended pattern case 49 · pattern only applies to whole stages
Question. A stage n uses 3n+1 objects. Does n=2.5 describe an actual stage if stages are counted in whole steps?
Working. No, not under the given stage definition.
Reason. The algebraic expression has a numerical value at 2.5, but an invented half-stage may not exist.
Independent check. The permissible input domain is positive whole term indices.
Original extended pattern case 50 · check two formulas
Question. Are 5+3(n−1) and 3n+2 equal for every n?
Working. Yes.
Reason. Distribute three: 5+3n−3=3n+2.
Independent check. At n=1 both give five; symbolic simplification establishes equality.
Four mistakes that make a pattern formula seem correct when it is not
| Written misconception | Why it fails | Better teaching question |
| Difference equals 3, so the rule must be 3n | Common difference does not fix the starting term | What value must the rule give at n=1? |
| A term equals 32, so n=32 | Term value mistaken for position | Which input to the rule produces value 32? |
| All growing patterns are linear | Some changes are not constant | Are the consecutive differences equal? |
| Two rules match at one stage, so they are identical | Coincidence at one input is not an algebraic identity | Do the expressions simplify to the same form for every permitted n? |
A student may copy a successful formula 3n+2 into a new question even when the pattern begins at eight or changes by five. The underlying confusion is choosing an expression from appearance instead of constructing it from the given relationship. Ask the pupil to state the first value and step size aloud before multiplying. Then write the rule in the transparent form a+(n−1)d, test term one and only after that simplify.
A second student might understand the difference pattern perfectly but use the term value as the input n. That misunderstanding can be exposed by a table with separate columns titled ‘term number’ and ‘number of objects’. A learner who uses the table correctly should be able to move between ‘What is term ten?’ and ‘Which term has value thirty-two?’ without confusing the two problems.
A third pupil may force all patterns into constant differences because that is what the previous worksheet practised. Show a non-linear sequence, ask for its differences, and allow the learner to say that this particular linear rule does not apply. Recognising that a model’s conditions are absent is valuable mathematical reasoning, even when a more advanced pattern could be derived later.
A 12-minute diagnostic that parents can use at home
Start with 4,7,10,13. Ask for the next term and then for the tenth term. If the pupil repeatedly adds three to reach term ten, they may have the pattern meaning but not the general expression. Ask for the rule 3n+1 and evaluate it at one and ten. Then change the first term to five while keeping the increase three. Does the rule change? A secure learner should notice the offset changes to 3n+2.
Next give the decreasing sequence 18,14,10,6. Ask what the next term is, what the difference is and whether a future term could be negative. The correct rule 22−4n shows that values continue below zero if the same numerical rule is maintained. This combines signed-number understanding with nth-term structure, but the term index still counts a positive whole position.
Finally present four separate pentagons and four pentagons joined along one complete side each in a chain. Ask whether they use the same number of edges in a stick model. If the count differs, the student has noticed that geometry, not only the word ‘pentagon’, determines the construction rule. The specific number of shared sides must be defined before the formula is selected.
What a six-week number-pattern bridge may include
| Week | Learning task | Evidence of independent progress |
| 1 | Term index versus value; extend simple lists | Can explain n=1, n=2 and next term |
| 2 | Constant differences and linear rules | Derives a+(n−1)d from first term and difference |
| 3 | Translate tile and stick constructions into expressions | Models a repeated part and a fixed part |
| 4 | Find later positions and recover indices from values | Checks by substitution |
| 5 | Mix decreasing, signed and unfamiliar contexts | Chooses appropriate algebraic form unaided |
| 6 | Compare non-linear extension patterns with simple linear ones | Explains when a constant-difference model is inappropriate |
This is a suggested teaching sequence, not the official 2028 term plan of a Punggol school. The applicable Mathematics course and teacher determine the level and timing of nth-term expressions. A pupil in G1 may require more concrete numerical patterns and an explicitly differentiated route. A secure G2 or G3 pupil may move more quickly into expressing simple general terms, while advanced non-linear cases remain enrichment where relevant.
How a small-group tutorial can individualise this topic
Three pupils might all struggle with the tenth term for different reasons. One can see the repeated addition but cannot express it with n; another writes 3n but misses the starting offset; the third writes the correct formula and then substitutes ten inaccurately. The visible final score does not identify the cause. A tutor should ask each pupil to explain where n appears in the construction, then assign different follow-up questions.
A 90-minute small-group lesson can begin with counters and a shared table, move to guided formula building, and finish with new sequences the pupils have not seen. A learner who reaches the answer with a teacher’s prompts still needs an independent attempt after a delay. A teacher who adapts the next task to the child’s actual misunderstanding makes the small class meaningful.
Parents in Punggol considering additional Mathematics support should ask for evidence of this diagnosis, not an ungrounded guarantee of an examination grade. A child’s schoolwork and syllabus must guide practice, and home revision should leave time for ordinary post-PSLE rest and activities.
Frequently asked questions
Why is the term number one when the sequence begins with five?
Because term number refers to the position in the list, not to its numerical value. The first term is at index one even if its value is five, negative two or one hundred.
Does a common difference of three mean the nth term is 3n?
Not necessarily. The first value determines the fixed offset. A sequence beginning 5,8,11 uses 3n+2, while 3,6,9 uses 3n.
Is n allowed to be a decimal in a tile-stage problem?
Not when n is defined as a whole stage number. The algebraic formula may be calculated at a decimal, but that can be outside the valid real-world domain.
Must every Secondary 1 student learn square and triangular nth-term formulas?
No. Those examples are extensions and the precise curriculum depends on G-level and the school. Secure recognition of simple numerical patterns and, where required, linear nth terms first.
What is a quick way to check a proposed rule?
Substitute n=1 to check the starting term, then another known index to check the growth. For a stronger general explanation, derive the expression from the stated construction or common difference.
Why does the rule sometimes contain n−1 instead of n?
There are no completed changes before the first term and one change before the second. To reach position n from position one, there are n−1 intervals.
Official guidance and related eduKate lessons
- MOE G2/G3 Mathematics syllabus — Secondary One pattern representation and algebraic terms.
- MOE Full Subject-Based Banding information — subject-level differences.
- Punggol PSLE 2027 to Secondary 1 Mathematics 2028 learning route.
- Algebraic expressions, variables and substitution — the language behind nth-term rules.
- Solving equations and recovering unknown indices — inverse operations.
- eduKate Secondary 1 Mathematics small-group tutorial approach.
The purpose of an nth-term rule is to explain every stage allowed by the construction. A student who can distinguish index from value, identify the repeated change, write the relationship and independently check it will find later algebra more meaningful.
From seeing a number pattern to explaining it
A sequence such as 5, 8, 11, 14 is simple to extend by three, but naming the next number is not the same as finding an nth-term rule. To reach term n from term one, a pupil makes n−1 equal changes. The formula is therefore 5+3(n−1), which simplifies to 3n+2. Both expressions describe the same calculation. One emphasises the starting value and accumulated steps; the other is compact. A learner who can move between those explanations understands what the symbols preserve.
Ask for the twentieth term: the transparent form gives 5+3×19=62, while the compact form gives 3×20+2=62. A student who accidentally uses twenty changes in the first form obtains sixty-five and can catch the mistake by testing the first stage. The first term should contain zero completed changes, not one. That initial-position check is especially powerful in a longer sequence with unfamiliar numbers.
Now change only the starting term to nine. The difference remains three, so the new rule is 9+3(n−1), equivalent to 3n+6. The common difference determines the coefficient of n, while the starting point determines the fixed offset. Two sequences can grow at the same rate without having the same values. This distinction is more important than memorising a formula with blanks labelled a and d.
Words and pictures determine what repeats
Imagine four individual square outlines, each made with four matchsticks. Since no edges are shared, the fourth stage has sixteen sticks and the general count is 4n. Now join n squares in one row so that each new square shares a complete side with the previous one. The first square uses four sticks but each later square adds only three; the rule becomes 4+3(n−1), or 3n+1. The word ‘square’ appears in both questions, yet the connection arrangement changes the number of objects needed.
Next consider an invented row of panels with three red tiles in each panel and two blue decorations fixed at the ends of the entire row. At stage n, the total is 3n+2. If the two decorations are repeated in every panel instead, the expression becomes n(3+2)=5n. At stage four, the totals are fourteen and twenty. The different answers are not an arithmetic dispute; the bracket tells us which pieces belong to each repeated group.
Pupils should draw at least two small stages and physically identify the parts that remain fixed and those that grow. This concrete representation supports generalisation when the pattern becomes too large to draw. Students can then test the formula at a known early stage, identify a distant term by substitution and explain why the same construction rule continues. That is how picture growth becomes algebraic language.
When a pattern is not linear
Compare 2, 5, 8, 11 with 1, 4, 9, 16. The first list has equal successive differences of three and fits 3n−1. The second has differences three, five and seven, so one fixed common difference cannot describe its growth. If the second list comes from n-by-n square arrays, the term n² makes sense. However, a finite list on its own may fit several formulas. The drawing or stated rule gives the justification for choosing one intended model.
A decreasing arithmetic sequence also uses a linear nth-term relationship. For 20,16,12,8, the difference is negative four and the rule is 20−4(n−1), or 24−4n. At stage six the term is zero. The negative sign reports the direction of change in the measured value; the index n remains a positive whole-number stage. Pupils who treat the changing sign and the position number as the same quantity need to revisit their variable definition.
An advanced comparison is the difference between two formulas that happen to match once and two expressions that are always equivalent. The rules 3n+2 and 2n+6 both give fourteen at n=4, but at n=5 they give seventeen and sixteen. By contrast 5+3(n−1) and 3n+2 are algebraically equivalent for every n because they simplify to the same expression. Testing one input can disprove equivalence if the results differ, but a single match does not establish a general identity.
A final ten-minute independent pattern review
Give the pupil the sequence 7,12,17,22, then ask for a formula, its tenth value and the position at which the value reaches forty-seven. The common difference is five, so the formula is 7+5(n−1), or 5n+2. Term ten is fifty-two. To find the term with value forty-seven, solve 5n+2=47, giving n=9. This exercise deliberately switches from finding a value from an index to finding an index from a value; a learner must know which role n plays.
Afterwards change the first term to eleven while keeping the difference five. The correct rule is 5n+6, so term ten becomes fifty-six. Do not announce the formula. Ask what changed and what remained the same. A child who explains that the repeated increase remains five while the starting amount is four higher has understood the construction. The answer can be checked at terms one and two before the child is asked to calculate a distant term.
Conclude with one unseen visual example or a decreasing sequence after a delay of several days. The student should name the starting stage, identify the repeated change, state any fixed pieces, and verify the formula without an adult supplying the operation. A small-group tutor can use the child’s first error to choose the next question; the goal is durable independent generalisation, not a large number of almost-identical pattern pages.
