Use the rule for term n of a compound-interest or reducing-balance model practice questions | General Mathematics (Year 11, VCE Units 1 & 2)

Geometric money models

A compound-interest or reducing-balance model changes by a multiplier each period. The starting value is term zero, and the multiplier is applied once for each period

V_n=V_0\left(1\pm\frac{r}{100}\right)^n

Start at term zero Use the initial balance or purchase value as V_0, before any periods have passed $\displaystyle V_0=\$800$
Build the multiplier Use 1+r/100 for growth and 1-r/100 for reducing-balance decay $\displaystyle 1+\frac{5}{100}=1.05$
Raise the multiplier Apply a repeated multiplier by raising it to the number of periods $\displaystyle R^3=1.05^3$
Multiply by the start Multiply the powered multiplier by V_0 to obtain the value at term n. $\displaystyle V_3=800\times1.05^3=926.10$

Worked example

A car loan starts at $12,000. No repayments are made, and the amount owed grows at 8% per year. Find the amount owed after 2 years

V_2=12000\left(1+\frac{8}{100}\right)^2

$\displaystyle R=1+\frac{8}{100}=1.08$An 8% yearly increase gives a multiplier of 1.08
$\displaystyle R^2=1.08^2=1.1664$Two years means that the yearly multiplier is used twice
$\displaystyle V_2=12000\times1.1664=13996.80$Multiply the starting loan by the two-year multiplier to get $13,996.80

Common mistakes

Using the rate alone

✕$\displaystyle V_3=V_0\left(0.05\right)^3$✓$\displaystyle V_3=V_0\left(1.05\right)^3$

The rate is part of the multiplier, but the multiplier also needs the starting 1.

Multiplying by the number of periods

✕$\displaystyle V_4=V_0\left(1.06\times4\right)$✓$\displaystyle V_4=V_0\left(1.06\right)^4$

A geometric model applies the same multiplier repeatedly, so the multiplier is raised to a power

Adding a depreciation rate

✕$\displaystyle V_n=V_0\left(1+\frac{12}{100}\right)^n$✓$\displaystyle V_n=V_0\left(1-\frac{12}{100}\right)^n$

Reducing-balance depreciation makes the value smaller, so its multiplier is below 1.

Difficulty levels

introFind a compound-interest value after a small number of yearly periods using an exact multiplier

standardUse growth or decay and find both a later value and the total interest or value lost

challengeUse longer spans, reducing-balance values, or a changed compounding period

examSelect the correct value or total from options based on a geometric finance rule

Questions about this topic

Why is the multiplier raised to a power?+

The balance is multiplied once for each period. Applying the same multiplier n times means R raised to the power of n

How is compound interest different from simple interest?+

Compound interest applies the rate to the changing balance, so it uses a multiplier. Simple interest adds the same amount each period based on the original principal

Why does reducing-balance depreciation use 1 minus the rate?+

The value loses a percentage each period. Subtracting that percentage from 1 gives a multiplier between 0 and 1, so the value decreases

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, General Mathematics, chapter 3: Sequences and finance

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