Loan Repayments

Expert reviewed • 08 January 2025 • 5 minute read

HSC Maths Advanced Syllabus

  • solve problems involving financial decisions, including a home loan, a savings account, a car loan or superannuation

Note:

Video coming soon!

How is a Loan paid off?

Now that we have learned about calculating interest, and the involvement of regular instalments, we must explore how a loan is paid off. Problems that are related to paying off a loan are often very similar to problems involving interest. As such, we can simply solve these types of problems by using the following steps:

  • Determine the principal and the compound interest it would generate for a given period of time (Do not include instalments).
  • Determine the instalments and the compound interest they generate for a given period of time.
  • Once the previous steps are completed, we can simply apply them to the following formula:
An=(P+interest)−(instalments+interest)A_n=(P+interest)-(instalments+interest)

where,

  • An=0A_n=0, means the loan is paid off
  • PP is the Principal or "borrowed" amount

Practice Question 1

Martha and John take out a loan of \250,000 on 1st January 2023 to buy a home. They will repay the loan in monthly instalments of \2,600. Interest is charged at 10% per annum, compounded monthly. Find the formula for the amount owing after nn months and thus determine the amount owing after 4 years.

First we must determine the monthly interest rate:

r=10%12=0.008333...r=\frac{10\%}{12}=0.008333...

Writing out a few of the first terms of the instalments we get:

1. 2600×1.00833n−12.  2600×1.00833n−2...1.\:2600\times 1.00833^{n-1}\\2.\;2600\times 1.00833^{n-2}\\...

Now using the formula given above, and previously learnt knowledge from this module, we must find a formula for AnA_n:

An=(P+interest)−(instalments+interest)=250000×1.00833n−(2600+2600×1.00833+...+2600×1.00833n−1)A_n=(P+interest)-(instalments+interest)\\=250000\times 1.00833^n-(2600+2600\times1.00833+...+2600\times1.00833^{n-1})

Thus, we can see the formation of a geometric series for the instalments. This formula can be simplified using the summing geometric series formula, where a=2600a=2600 and r=1.00833r=1.00833

An=250000×1.00833n−a(rn−1)r−1=250000×1.00833n−2600(1.00833n−1)0.00833A_n=250000\times 1.00833^n-\frac{a(r^n-1)}{r-1}\\=250000\times 1.00833^n-\frac{2600(1.00833^n-1)}{0.00833}

Thus, to find the amount owing after 4 years (48 months) we must substitute n=48n=48 into the derived formula:

A48=250000×1.0083348−2600(1.0083348−1)0.00833=372279.45−152665.82=219613.63A_{48}=250000\times 1.00833^{48}-\frac{2600(1.00833^{48}-1)}{0.00833}\\=372279.45-152665.82\\=219613.63

∴\therefore The amount owing after 4 years is \219,613.63$

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Investing with Regular Instalments

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