Resources For Teachers For Tutors For Students & Parents Pricing
Year 12 Maths Advanced (2027) Sequences and series

Growth & Decay with Sequences

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Sequences and series

Growth and decay with sequences models repeated percentage change — savings, depreciation and populations — as a geometric sequence with ratio \(r=1\pm\dfrac{p}{100}\).

Part of the NSW Year 12 Mathematics Advanced course, in the Sequences and series focus area of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

Growth and decay problems use sequences: a fixed amount each period is arithmetic (linear), a fixed percentage is geometric (exponential). This Year 12 Mathematics Advanced topic (MAV-12-03) covers compound interest and depreciation.

A quantity that changes by the same amount each period forms an arithmetic sequence \(a_n=a+(n-1)d\) (a straight line). A quantity that changes by the same percentage forms a geometric sequence \(a_n=ar^{\,n-1}\) (a curve).

Compound growth: value after \(n\) periods \(=P(1+r)^{n}\). Depreciation: value \(=P(1-r)^{n}\).

Linear versus percentage growthA straight line for a fixed amount each period and an upward curve for a fixed percentage each period.npercentage (GP)linear (AP)
A fixed amount gives a line; a fixed percentage curves upward.
\[\text{growth: }P(1+r)^{n}\qquad \text{decay: }P(1-r)^{n}\]
compound growth value equals P times (1 plus r) to the n; depreciation equals P times (1 minus r) to the n

Method

  1. Decide linear (fixed amount \(\to\) AP) or percentage (fixed rate \(\to\) GP).
  2. Use \(P(1+r)^n\) for growth, \(P(1-r)^n\) for depreciation.
  3. Substitute the number of periods \(n\).
Example 1 — Compound interest
\(\$1000\) at \(5\%\) p.a. compound. Value after \(3\) years?
Solution
\(A\)\(=\)\(1000(1.05)^3\)
\(\approx\)\(\$1157.63\)
Example 2 — Depreciation
A \(\$20\,000\) car loses \(15\%\) each year. Value after \(2\) years?
Solution
\(A\)\(=\)\(20000(0.85)^2\)
\(=\)\(\$14\,450\)
Example 3 — Linear vs percentage
Pay A: \(\$40\,000\) \(+\$3000\)/yr; Pay B: \(\$40\,000\) \(+5\%\)/yr. Year 3 each?
Solution
\(\text{A}\)\(=\)\(40000+2(3000)=46000\)
\(\text{B}\)\(=\)\(40000(1.05)^2=44100\)
Example 4 — Investment
\(\$5000\) at \(8\%\) p.a. compound: value after \(1\) and \(10\) years?
Solution
\(\text{yr }1\)\(=\)\(5000(1.08)=\$5400\)
\(\text{yr }10\)\(\approx\)\(\$10\,794.62\)

Common pitfalls

Amount vs percentage. \(\$2000\) a year is arithmetic; \(5\%\) a year is geometric.
Depreciation multiplies by \(1-r\). A \(15\%\) loss means \(\times0.85\), not \(-15\).
Mind the period count. After \(n\) years the factor applies \(n\) times.

Frequently asked questions

What is the difference between linear and percentage growth?

Linear growth adds the same amount each period and forms an arithmetic sequence (a straight line). Percentage growth multiplies by the same factor each period and forms a geometric sequence (a curve).

What is the compound interest formula?

The value after n periods is P times (1 plus r) to the power n, where P is the principal and r is the interest rate per period as a decimal.

How do you calculate depreciation?

Multiply by (1 minus r) for each year. A 15 percent yearly loss means multiplying by 0.85 each year, so after 2 years the value is P times 0.85 squared.

Is compound interest arithmetic or geometric?

It is geometric, because the balance is multiplied by the same factor (1 plus r) each period.