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Growth and Decay Applications

Key Terms

Growth and decay problems use geometric sequences with r = 1 + rate (growth) or r = 1 − rate (decay)
Compound interest
A = P(1 + r)n, where r = annual rate, n = years
Depreciation
V = V0(1 − d)n, where d = annual depreciation rate
Population growth
Pn = P0 × growth factorn
To find the number of periods: use logarithms or systematic testing
Always check whether growth or decay applies to the context
Compound growth: A = P(1 + r)n

Depreciation: V = V0(1 − d)n

Growth rate r as a decimal:
6% growth → r = 0.06, multiply by 1.06
15% decay → d = 0.15, multiply by 0.85
Worked Example: A town’s population of 12 000 grows at 2.5% per year. Find the population after 10 years.

r = 1.025, a = 12 000, n = 10
P10 = 12 000 × (1.025)10 = 12 000 × 1.2801 ≈ 15 361
Hot Tip: When the question asks “after n years”, use tn+1 = arn if you define a as the initial value. Be careful: “after 3 years” means 3 multiplications, so use r3.

Modelling with Sequences

Many real-world situations involve quantities that change by a fixed percentage each period. These are modelled by geometric sequences:

  • Growth (r > 1): population growth, investment returns, inflation
  • Decay (0 < r < 1): depreciation, radioactive decay, drug concentration

Setting Up the Model

  1. Identify the initial value a (at time 0 or year 0)
  2. Determine the growth/decay factor r from the percentage change
  3. Choose the correct term number: value after n periods = a × rn

Finding When a Target is Reached

To find how many periods until a value is reached, either:

  • Test values systematically (acceptable in General Maths)
  • Use logarithms: n = log(target/a) / log(r)
Strategy: Check your answer by substituting back. If you find “n = 7.3” for a whole-number context (complete years), the answer is 8 (the next whole number after 7.3).

Mastery Practice

See Answers ➔
  1. Fluency A population of 8 000 grows at 3% per year. Find the population after 6 years.
  2. Fluency A machine is worth $50 000 and depreciates at 20% per year. Find its value after 4 years.
  3. Fluency $10 000 is invested at 5% compound interest per annum. How much is the investment worth after 8 years?
  4. Fluency A radioactive substance has a mass of 200 g. It decays at 10% per year. What mass remains after 5 years?
  5. Understanding A town has 5 000 residents. The population grows at 1.8% per year. In which year will the population first exceed 6 000?
  6. Understanding A laptop costs $2 400 new and depreciates at 25% per year. After how many complete years will it be worth less than $500?
  7. Understanding Two investments are compared. Investment A: $8 000 at 6% per annum for 5 years. Investment B: $8 000 at 5.5% per annum for 6 years. Which gives the higher final amount?
  8. Understanding A rumour spreads through a school. Each hour, the number of students who have heard it triples. If 4 students know at 8 am, how many know by 2 pm?
  9. Problem Solving A bank offers 3.5% interest per annum, compounded monthly. An investor deposits $15 000. Find the value after 10 years using the monthly rate r = 0.035/12.
  10. Problem Solving The population of a city was 250 000 in 2010 and has been growing at a constant annual rate. By 2020, it had reached 310 000.
    • (a) Find the annual growth rate (as a percentage, to 2 decimal places).
    • (b) Predict the population in 2030 at the same rate.
    • (c) In which year will the population exceed 400 000?

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