Investing with Regular Instalments

Expert reviewed • 22 November 2024 • 6 minute read

HSC Maths Advanced Syllabus

  • solve compound interest problems involving financial decisions, including a home loan, a savings account, a car loan or superannuation
    • use a table of interest factors to perform annuity calculations, eg calculating the present or future value of an annuity, the contribution amount required to achieve a given future value or the single sum that would produce the same future value as a given annuity
  • calculate the future value or present value of an annuity by developing an expression for the sum of the calculated compounded values of each contribution and using the formula for the sum of the first 𝑛 terms of a geometric sequence
  • use geometric sequences to model and analyse practical problems involving exponential growth and decay
    • recognise a reducing balance loan as a compound interest loan with periodic repayments, and solve problems including the amount owing on a reducing balance loan after each payment is made

Note:

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How Compound Interest Forms a GP

The importance of understanding that compound interest forms a GP, is brought to light when invested money that is being compounded, receives continuous instalments at regular intervals.

To find the value of an investment, when regular instalments are made, we must follow the listed steps:

  • Find what each instalment will equal. This can be completed using the formula for compound interest, and left in a form that is easily applicable to geometric series formulae.
  • Create a geometric series using instalment values found.
  • Use the summing geometric series formula to add up all these amounts.

Just to review, the formula to sum a GP is as follows:

Sn=a(rn−1)r−1S_n=\frac{a(r^n-1)}{r-1}

Practice Question 1

Initially there is \1000investedintoabankaccount.Everyyearafterthefirstprincipalpaymentwasputintotheaccount1000 invested into a bank account. Every year after the first principal payment was put into the account \\1000 was regularly installed. At the end of the 10th year since the account had been opened the entire amount was withdrawn. If the money was compounded annually at a rate of 8%, how much money was withdrawn from the account.

First we must identify the variables being used in the compound interest formula:

P=1000R=8%=0.08n=1t=10P=1000\\R=8\%=0.08\\n=1\\t=10

Now we must identify the first few instalment expressions using the compound interest formula, to get a general idea of geometric series they will create:

A1=10000(1+0.08)1×10=10000×1.0810A2=10000(1+0.08)1×9=10000×1.089...A_1=10000(1+0.08)^{1\times10}=10000\times1.08^{10}\\A_2=10000(1+0.08)^{1\times9}=10000\times1.08^9\\...

Thus, creating a geometric series we can see:

A10=1000(1.08)+1000(1.08)2+...+1000(1.08)10A_{10}=1000(1.08)+1000(1.08)^2+...+1000(1.08)^{10}

Now we have a series, we can determine the values need to be used in the summing geometric series formula:

a=1000(1.08)r=1.08n=10a=1000(1.08)\\r=1.08\\n=10

We must note that there are 11 terms of the series, as the original $1000 in the account is included.

Now summing all instalments we find the final value of the account to be:

S10=1000×1.08(1.0810−1)1.08−1=1158.920.08=$14486.56S_{10}=\frac{1000\times1.08(1.08^{10}-1)}{1.08-1}\\=\frac{1158.92}{0.08}\\=\$14486.56

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