Patterns become algebra when the learner can describe what changes and what stays fixed. A list of terms is only the surface. The deeper task is to identify a relationship that works for any position, then express it compactly and verify it.
This Secondary 3 Mathematics Learning Guide develops nth-term rules, algebraic patterns, substitution, identities, expansion and factorisation. It complements the earlier guides on algebraic control, quadratic equations and algebraic fractions by concentrating on general rules and structural equivalence.
The current 2027 SEC G3 Mathematics syllabus listing from SEAB identifies Mathematics as K310. Its algebraic-expression content includes representing patterns by an nth-term expression, evaluation, brackets, factorisation, identities, formulae and quadratic expressions.
From a Pattern to a General Rule
Consider 5, 8, 11, 14, … . The difference is constantly +3. A linear nth-term rule therefore has coefficient 3. Testing 3n gives 3,6,9,12, so every term is 2 larger. The rule is 3n+2.
The rule must work for all listed positions, not only the first one. Substituting n=4 gives 14, confirming the fourth term.
Worked Example 1: Find an nth Term
Find the nth term of 7, 12, 17, 22, … The common difference is 5, so begin with 5n.
At n=1, 5n gives 5 but the sequence begins at 7, so add 2. Therefore the nth term is 5n+2.
Use the Rule to Test Membership
If a sequence has nth term 5n+2, is 102 a term? Solve 5n+2=102. Then 5n=100 and n=20.
Because n is a positive integer, 102 is the 20th term. If solving gave n=20.4, the number would not occur in the sequence even though it lies between nearby term values.
Patterns Can Come From Geometry
Suppose a row of n adjacent unit squares is built from matchsticks. One square uses 4 sticks. Each extra square shares one side with the previous square and therefore adds only 3 sticks.
The total is 4+3(n−1)=3n+1. The algebra records both the starting structure and the repeating addition.
Worked Example 2: Build a Rule From a Diagram Description
A pattern has 6 tiles in Stage 1 and adds 4 tiles at each new stage. Find a formula for Stage n.
Start with 6 and add 4 for each of the n−1 later steps: 6+4(n−1)=4n+2. Therefore Tₙ=4n+2.
Substitution Is Evaluation, Not Rearrangement
If P=3a²−2b and a=4, b=5, substitute values carefully: P=3(4²)−2(5)=48−10=38.
Brackets matter especially for negative values. If a=−4, then a² means (−4)²=16, not −16.
An Identity Is True for Every Allowed Value
An equation such as x+3=7 is true only for x=4. An identity such as (a+b)²=a²+2ab+b² is true for all real a and b.
The distinction matters because identities can transform expressions without solving for a particular unknown.
Why (a+b)² Is Not a²+b²
(a+b)² means (a+b)(a+b). Expanding gives a²+ab+ab+b² = a²+2ab+b².
The middle term records the two cross-products. Forgetting them changes the expression rather than simplifying it.
Worked Example 3: Use an Identity Efficiently
Evaluate 51² mentally using an identity. Write 51=50+1.
51²=(50+1)²=2500+100+1=2601.
The identity turns a large multiplication into structured arithmetic.
Difference of Two Squares
a²−b²=(a+b)(a−b). The structure is recognisable because both terms are perfect squares separated by subtraction.
Thus x²−49=(x+7)(x−7), and 9p²−16q²=(3p+4q)(3p−4q).
Worked Example 4: Factorise by Structure
Factorise 25x²−81. Recognise (5x)²−9².
Therefore 25x²−81=(5x+9)(5x−9).
Grouping Can Reveal a Common Binomial
Expressions such as ax+bx+kay+kby can be grouped:
ax+bx+kay+kby = x(a+b)+ky(a+b) = (a+b)(x+ky).
The useful skill is not memorising one arrangement. It is spotting a repeated factor after grouping.
Worked Example 5: Factorise by Grouping
Factorise 3x+6y+5xz+10yz. Group the first two and last two terms:
3(x+2y)+5z(x+2y) = (x+2y)(3+5z).
Expansion and Factorisation Are Reverse Routes
Expansion converts a product into a sum. Factorisation converts a sum into a product. A strong check is to reverse the operation.
If x²+7x+12=(x+3)(x+4), expanding the right side gives x²+7x+12 and confirms the factorisation.
Worked Example 6: Quadratic Factorisation
Factorise 2x²+7x+3. Seek factors whose product gives 2x² and 3 while cross-products total 7x.
(2x+1)(x+3)=2x²+6x+x+3=2x²+7x+3. Hence (2x+1)(x+3).
Formulae Are General Relationships
A formula such as A=πr² describes a relationship across every valid radius. Substitution evaluates the formula for a particular case. Rearrangement changes which variable is isolated while preserving the relationship.
For example, from v=u+at, subtract u and divide by a to obtain t=(v−u)/a, provided a≠0.
Worked Example 7: Change the Subject
Make h the subject of V=πr²h. Divide both sides by πr²:
h=V/(πr²), provided r≠0 in the algebraic rearrangement.
A Pattern Rule Must Respect Its Domain
For an nth-term rule, n normally represents positive integer positions 1,2,3,… . Solving an equation may produce a real number, but sequence membership still requires n to be a valid position.
This is the same habit seen in word problems: algebra gives candidates; context decides which candidates are meaningful.
Common Errors
Using common difference as the full nth term: repair by comparing the candidate rule with the first term.
Writing (a+b)²=a²+b²: repair by expanding the two brackets explicitly.
Factorising without a reverse check: repair by expanding the proposed factors.
Accepting a non-integer n as a sequence position: repair by returning to the meaning of n.
Independent Practice
1. Find the nth term of 4,9,14,19,…
2. Is 164 a term of the sequence 6n+2?
3. A tile pattern begins with 9 tiles and adds 5 each stage. Find Tₙ.
4. Expand (x+7)².
5. Expand (2a−3b)².
6. Factorise x²−64.
7. Factorise 4p²−25q².
8. Factorise 2x+6y+5xz+15yz.
9. Factorise 3x²+11x+6.
10. Make t the subject of d=vt+1/2at² only in the special case a=0, and explain why the general case is not a simple linear rearrangement.
Explained Answers
1. Difference 5, so 5n−1.
2. 6n+2=164 gives n=27, so yes.
3. 9+5(n−1)=5n+4.
4. x²+14x+49.
5. 4a²−12ab+9b².
6. (x+8)(x−8).
7. (2p+5q)(2p−5q).
8. (x+3y)(2+5z).
9. (3x+2)(x+3).
10. If a=0, d=vt and t=d/v for v≠0. With a≠0, t appears quadratically and solving requires a quadratic-equation method.
Continue the Secondary 3 Learning Route
Continue with Number Structure, Prime Factorisation, HCF and LCM, Simultaneous Linear Equations and Modelling, and Compound Interest, Repeated Growth and Financial Reasoning.
Algebraic pattern work becomes secure when the student can move from examples to a general rule and then back from the rule to verified cases. Return to the Secondary Mathematics Hub.
