Arithmetic pricing model

#hsc-maths#sequences#basic

In this problem, we explore how an arithmetic sequence can model a simple pricing structure. This connects directly to the HSC Sequences topic, where recognising linear growth patterns unlocks quick formula-building and inequality solving. By working through this example, you’ll practise translating a word problem into a sequence formula and then using it to answer both evaluation and budget-limit questions.

Problem Statement

An installer charges $400 for the first window and $100 for each additional window.

  • Write the total cost for 1, 2, 3, and 4 windows.
  • Find a formula for the total cost AnA_n for nn windows.
  • Find A15A_{15}.
  • If the budget is strictly less than $10,000, what is the maximum number of windows?

Hints

Model AnA_n as an arithmetic sequence with first term 400400 and difference 100100.


Solutions

We begin by listing the costs for the first few numbers of windows. The first window costs $400, and every extra window adds $100:

C1=400,C2=400+100=500,C3=600,C4=700.C_1 = 400,\qquad C_2 = 400+100 = 500,\qquad C_3 = 600,\qquad C_4 = 700.

The cost of nn windows follows an arithmetic progression with first term a=400a = 400 and common difference d=100d = 100. The nnth term of an arithmetic sequence is An=a+(n1)dA_n = a + (n-1)d, so

An=400+(n1)100=100n+300.A_n=400+(n-1)100=100n+300.

This compact formula now lets us answer any number of windows. Substituting n=15n = 15 gives the total cost directly: So A15=1800A_{15}=1800.

Finally, we translate the budget restriction into an inequality. The total cost must be strictly less than $10,000, so

100n+300<10000    n<97,100n+300<10000 \implies n<97,

hence the maximum integer number of windows is n=96n=96.


Takeaways

  • Recognise that a constant addition ($100 per window) signals an arithmetic sequence; the first term and common difference are all you need.
  • Build the explicit formula An=a+(n1)dA_n = a + (n-1)d and then use it for both simple evaluations and inequality constraints.
  • When an inequality gives a non-integer bound (like n<97n < 97), always round down to the nearest integer to stay within the strict budget limit.

Further Readings

HSC Sequences, HSC Vectors, HSC Differential Equations, HSC Probability

Written by Vu Hung Nguyen

Mathematics Educator · LinkedIn · About