WACE Mathematics Applications sequences, finance and networks deep dive: the 2026 exam

WACEMathematics ApplicationsStudy guide18 min readUpdated 23 Sep 2026

Revision deep dive for the sequences, finance and networks content of WACE Mathematics Applications: arithmetic and geometric sequences, recurrence relations, compound interest, depreciation, loans, annuities and perpetuities, graph terminology, Euler and Hamilton routes, spanning trees, shortest paths, critical paths, flow and assignment, with worked examples and dot point links.

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Two thirds of the paper
Part A: growth and decay in sequences
Part B: graphs, networks and decision mathematics
Common mistakes
Check your knowledge

Two thirds of the paper

Sequences and finance, and graphs and networks, together account for about two thirds of the WACE Mathematics Applications exam under the content ranges in the SCSA design brief. Both reward the same habit: write the model down (the recurrence relation, the solver inputs, the network table) before you calculate. This deep dive links each related dot point on the site. Exam format and timing are in the exam strategy guide.

Part A: growth and decay in sequences

1. Arithmetic sequences and linear growth

Dot points: arithmetic sequences, recurrence relations and sequences.

Arithmetic sequence

T_{n+1} = T_n + d \qquad T_n = a + (n - 1)d

Flat-rate depreciation

A $32 000 ute depreciates by $2500 per year. Its value after $n$ years is $V_n = 32\,000 - 2500n$ , or recursively $V_{n+1} = V_n - 2500$ with $V_0 = 32\,000$ .

When does it first fall below $10 000? $32\,000 - 2500n < 10\,000$ gives $n > 8.8$ , so after 9 years.

Be careful with indexing. If the question starts at $T_1$ the explicit rule is $a + (n - 1)d$ ; if it starts at $V_0$ it is $V_0 + nd$ .

2. Geometric sequences and compound growth

Dot points: geometric sequences, exponential growth and decay, reducing-balance depreciation.

Geometric sequence

T_{n+1} = rT_n \qquad T_n = ar^{n - 1}

Growth when

r > 1

, decay when

0 < r < 1

Reducing-balance depreciation

The same $32 000 ute instead loses 15 percent of its value each year: $V_{n+1} = 0.85V_n$ , so $V_n = 32\,000 \times 0.85^n$ .

After 5 years: $V_5 = 32\,000 \times 0.85^5 \approx \$14\,198.57$ .

It first falls below $10 000 when $0.85^n < 0.3125$ . Testing values (or solving on the calculator) gives $n = 8$ , since $V_7 \approx \$10\,258$ and $V_8 \approx \$8720$ .

Compare with flat rate: reducing balance loses more in the early years and less later.

3. Compound interest with recursion

Dot points: compound interest with recursion, consumer and financial mathematics.

Balance and effective rate

$10 000 is invested at 4.8 percent per annum compounding monthly. The period rate is $4.8 \div 12 = 0.4$ percent, so $A_{n+1} = 1.004A_n$ with $A_0 = 10\,000$ .

After 3 years (36 months): $A_{36} = 10\,000 \times 1.004^{36} \approx \$11\,545.52$ .

Effective annual rate: $1.004^{12} - 1 \approx 0.0491$ , or 4.91 percent, slightly above the nominal 4.8 percent because interest earns interest within the year.

4. Loans, annuities and perpetuities

Dot points: reducing-balance loans and amortisation, annuities and superannuation, perpetuities.

All three are the same recurrence with different signs:

A_{n+1} = (1 + i)A_n \pm P

Reducing-balance loan: subtract the repayment. The balance falls to 0.
Annuity investment (regular deposits): add the deposit. The balance grows.
Annuity paying you an income: subtract the payment from an invested balance until it runs out.
Perpetuity: the payment equals the interest, $P = iA_0$ , so the balance stays constant forever.

Annuity investment and perpetuity

Regular saving. Depositing $200 at the end of each month at 6 percent per annum compounding monthly: $A_{n+1} = 1.005A_n + 200$ , $A_0 = 0$ . After 10 years (120 deposits) the balance is about $32 775.87, of which $24 000 is deposits and about $8775.87 is interest.

Perpetuity. A $500 000 fund earns 4.2 percent per annum, paid monthly. The monthly payment that keeps the balance constant is $500\,000 \times \dfrac{0.042}{12} = \$1750$ .

In an amortisation table, each row shows the balance, interest for the period (balance times period rate), the repayment and the new balance. The interest part of each repayment shrinks as the balance falls, so later repayments pay off more principal than earlier ones.

Part B: graphs, networks and decision mathematics

5. Graph language and matrices

Dot points: graph terminology and adjacency matrices, networks and decision mathematics.

Degree of a vertex: the number of edge ends at it (a loop adds 2).
Handshake rule: the sum of the degrees equals twice the number of edges, so the number of odd-degree vertices is always even.
An adjacency matrix has the number of edges between vertex $i$ and vertex $j$ in row $i$ , column $j$ . For an undirected graph it is symmetric.

6. Planar graphs and Euler's formula

Dot point: planar graphs and Euler's formula.

A graph is planar if it can be redrawn with no edges crossing. For a connected planar graph, $v - e + f = 2$ , where the outside region counts as a face.

7. Walks, trails, paths and circuits

Dot point: walks, paths, Euler and Hamilton.

Route	Rule
Trail	No edge repeated
Path	No vertex repeated
Euler trail	Uses every edge exactly once; exactly two odd-degree vertices, start and finish there
Euler circuit	Every edge once, finishing where it started; all vertices even
Hamilton path or cycle	Every vertex exactly once; no simple degree test

8. Minimum spanning trees and shortest paths

Dot points: minimum spanning trees, shortest path problems.

Consider a network with edges AB 4, AC 3, BC 2, BD 5, CD 7, CE 6 and DE 3.

Prim's algorithm, then a shortest path

Minimum spanning tree (Prim's, from A). Add the cheapest edge joining the tree to a new vertex each time: AC (3), then CB (2), then BD (5), then DE (3). Total weight $3 + 2 + 5 + 3 = 13$ . A tree on 5 vertices always has 4 edges.

Shortest path from A to E. Compare routes: A-C-E $= 3 + 6 = 9$ , A-B-D-E $= 4 + 5 + 3 = 12$ , A-C-D-E $= 3 + 7 + 3 = 13$ . The shortest path is A-C-E, length 9.

The two problems give different answers: the spanning tree connects every vertex as cheaply as possible, while the shortest path only links two vertices.

9. Critical path analysis

Dot point: critical path analysis.

Forward and backward scan

Activity	Duration (days)	Immediate predecessors
A	3	none
B	4	none
C	2	A
D	4	A
E	3	B, C
F	2	D, E

Forward scan (earliest start, EST): A 0, B 0, C 3, D 3, E $\max(0 + 4, 3 + 2) = 5$ , F $\max(3 + 4, 5 + 3) = 8$ . Minimum completion time: $8 + 2 = 10$ days.
Backward scan (latest start, LST): F 8, E $8 - 3 = 5$ , D $8 - 4 = 4$ , C $5 - 2 = 3$ , B $5 - 4 = 1$ , A $\min(3, 4) - 3 = 0$ .
Float = LST minus EST: B has 1 day and D has 1 day. All others have zero float, so the critical path is A-C-E-F.

10. Flow networks and assignment

Dot points: flow networks and maximum flow, assignment problems and the Hungarian algorithm.

Maximum flow equals the capacity of the minimum cut. When you add up a cut's capacity, count only edges that flow from the source side to the sink side; edges pointing back across the cut count as zero.

Hungarian algorithm, 3 by 3

Costs (hundreds of dollars) for workers W1, W2, W3 on jobs J1, J2, J3:

	J1	J2	J3
W1	8	4	7
W2	5	2	3
W3	9	4	8

Row reduce (subtract 4, 2, 4), then column reduce (subtract 3 from J1 and 1 from J3):

	J1	J3
W1	1	2
W2	0	0
W3	2	3

The zeros can be covered by 2 lines (row W2, column J2), fewer than 3, so adjust: the smallest uncovered value is 1. Subtract it from uncovered entries and add it where lines cross:

	J1	J2	J3
W1	0	0	1
W2	0	1	0
W3	1	0	2

Now 3 lines are needed. W3 has only one zero, so W3 to J2; then W1 to J1 and W2 to J3. Minimum cost: $8 + 3 + 4 = 15$ , that is, $1500.

Common mistakes

Where finance and network marks go missing

Using the annual rate as the period rate, or years as the number of periods.
Signs in the recurrence: a loan subtracts the repayment, an investment adds the deposit.
Stopping Prim's algorithm early: a spanning tree on $n$ vertices has $n - 1$ edges.
In a cut, counting edges that flow back from the sink side.
Forgetting that the backward scan takes the minimum over the following activities.

Check your knowledge

A car worth $40 000 depreciates 20 percent per year (reducing balance). Find its value after 3 years. (Answer: $20 480.)
A connected planar graph has 10 edges and 6 faces. How many vertices? (Answer: 6.)
A $240 000 fund pays a perpetuity at 5 percent per annum, paid annually. Find the payment. (Answer: $12 000 per year.)

Then try the finance and networks quiz.

Sources & how we know this

Mathematics Applications ATAR course Year 12 syllabus — School Curriculum and Standards Authority (SCSA)
Mathematics Applications past ATAR course examinations — School Curriculum and Standards Authority (SCSA)
Mathematics Applications ATAR course: syllabus and support materials — SCSA

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