Compound Growth, Decay and Iteration Revision Notes | GCSE AQA Higher Maths

1. Why Growth and Decay Problems Get Harder

You already know how to find a single percentage change. The challenge at this level is chaining those changes together — different rates in different years, quantities that feed back into themselves, or sequences where each term depends on the one before. That is where exponential models and iterative processes come in.

Both ideas share the same core logic: the output of one step becomes the input of the next. Once you see that pattern, the two topics click together naturally.

2. Exponential Growth and Decay — The Multiplier Method

The cleanest way to handle repeated percentage change is with a percentage multiplier. Instead of finding the percentage and adding or subtracting it each time, you multiply by a single decimal that does both jobs at once.

The multiplier $m$ is the engine of the formula. For growth, $m > 1$; for decay, $0 < m < 1$. Raising it to the power $n$ stacks the repeated change automatically — no need to recalculate year by year.

Worked example: Exponential Decay — Depreciation Problem

A car is bought for £18 000. Its value depreciates by 15% each year. Calculate its value after 4 years. Give your answer to the nearest pound. [3 marks]

1. $0.85^2 = 0.85 \times 0.85 = 0.7225$\n$0.85^4 = 0.7225 \times 0.7225 = 0.522006$

Answer: £9396

Notice how the multiplier method avoids recalculating the value after each individual year. One substitution handles all four years simultaneously — that efficiency is exactly why examiners favour this approach at Higher tier.

3. Mixed-Rate and Reverse Problems

Exam questions at this level often combine different rates across different periods, or ask you to work backwards from a final value. Both situations use the same multiplier logic — you need to set up the chain carefully.

For a reverse problem, you know the final amount $A$ and need to find the starting amount $P$. Rearranging $A = P \times m^n$ gives $P = \frac{A}{m^n}$. Dividing by a decimal multiplier less than 1 will give a larger result — that is correct, because you are recovering the original larger value before decay occurred.

Worked example: Reverse Exponential — Finding the Original Value

After 3 years of 8% annual decay, a machine is worth £10 450. Work out its original value. Give your answer to the nearest pound. [3 marks]

1. $m = 1 - 0.08 = 0.92$ — 8% decay means the multiplier is 0.92 each year. [M1]
2. $10450 = P \times 0.92^3$ — A = 10 450, m = 0.92, n = 3. We need to find P.
3. $0.92^2 = 0.92 \times 0.92 = 0.8464$ $0.92^3 = 0.8464 \times 0.92 = 0.778688$ — Expand each power step by step to avoid errors.
4. $P = \frac{10450}{0.778688} \approx 13419.8...$ — Dividing by 0.778688 gives a larger result than 10 450 — correct, since the machine was worth more originally. [M1 — correct rearrangement]
5. $P \approx £13420$ — [A1 — correct final answer to the nearest pound]

Answer: £13 420

4. Iterative Processes — Generating Sequences Step by Step

An iterative process is a rule that generates a sequence by feeding each output back in as the next input. The rule is written in the form $x_{n+1} = f(x_n)$, which means: to get the next term, apply the function $f$ to the current term.

The subscript notation can look unfamiliar at first. Think of $x_n$ as 'the current value' and $x_{n+1}$ as 'the next value'. The formula is the instruction for moving from one to the other.

Worked example: Generating Terms from an Iterative Formula

An iterative formula is given as $x_{n+1} = \frac{1}{2}x_n + 6$, with $x_1 = 4$. Find $x_2$, $x_3$, and $x_4$, giving each answer to 3 decimal places where necessary. [3 marks]

1. $x_2 = \frac{1}{2}(4) + 6 = 2 + 6 = 8$ — Substitute $x_1 = 4$ into the formula. [M1 — correct application of iterative rule]
2. $x_3 = \frac{1}{2}(8) + 6 = 4 + 6 = 10$ — Now use $x_2 = 8$ as the input.
3. $x_4 = \frac{1}{2}(10) + 6 = 5 + 6 = 11$ — Use $x_3 = 10$ as the input. [A1 — correct sequence of values]
4. $x_1 = 4,\; x_2 = 8,\; x_3 = 10,\; x_4 = 11$ — The terms are increasing but the gaps are shrinking: 4, 8, 10, 11... The sequence appears to be converging towards a fixed value. [A1]

Answer: $x_2 = 8$, $x_3 = 10$, $x_4 = 11$

The sequence above is converging — the terms are getting closer and closer to a fixed value. When a sequence converges, the iterative process is finding the solution to an equation. That fixed value is called the limit of the sequence.

4a. Finding the Limit of a Convergent Sequence

If a sequence converges to a limit $L$, then eventually $x_{n+1} \approx x_n \approx L$. Substituting $L$ for both $x_{n+1}$ and $x_n$ in the iterative formula gives an equation you can solve for $L$.

Worked example: Finding the Limit of a Convergent Iterative Sequence

The iterative formula $x_{n+1} = \frac{1}{2}x_n + 6$ produces a convergent sequence. Find the exact value of the limit $L$. [2 marks]

1. $L = \frac{1}{2}L + 6$ — At the limit, the input and output are equal. Substituting $L$ for both sides gives an equation to solve. [M1]
2. $L - \frac{1}{2}L = 6$ $\frac{1}{2}L = 6$ — Subtract $\frac{1}{2}L$ from both sides.
3. $L = 6 \times 2 = 12$ — Multiply both sides by 2. [A1 — correct limit]
4. $x_{n+1} = \frac{1}{2}(12) + 6 = 6 + 6 = 12$ ✓ — Substituting $L = 12$ back into the formula returns 12, confirming convergence to this value.

Answer: The limit is $L = 12$

The graph above shows the iterative formula $y = \frac{1}{2}x + 6$ alongside the line $y = x$. The intersection of these two lines is the fixed point — the limit $L = 12$. This is a useful visual check: wherever the formula line crosses $y = x$, that is where $x_{n+1} = x_n$, confirming the limit.

5. Iterative Processes as Equation Solvers

Iterative formulae are not only used to generate sequences — they are a powerful method for approximating solutions to equations that are difficult or impossible to solve algebraically. The higher-tier specification requires you to work with general iterative processes in this context.

The idea is to rearrange an equation into the form $x = f(x)$, then use $x_{n+1} = f(x_n)$ as the iterative formula. Starting from an approximate value, repeated application homes in on the exact solution.

Worked example: Using Iteration to Approximate a Solution

Show that the equation $x^3 - 2x - 5 = 0$ can be rearranged to give the iterative formula $x_{n+1} = \sqrt[3]{2x_n + 5}$. Then use $x_1 = 2$ to find $x_2$, $x_3$, and $x_4$, giving your answers to 4 decimal places. [4 marks]

1. $x^3 - 2x - 5 = 0$ $x^3 = 2x + 5$ $x = \sqrt[3]{2x + 5}$ — Add $2x + 5$ to both sides, then take the cube root. This confirms the iterative formula. [M1 — valid rearrangement shown]
2. $x_2 = \sqrt[3]{2(2) + 5} = \sqrt[3]{4 + 5} = \sqrt[3]{9} \approx 2.0801$ — $\sqrt[3]{9} \approx 2.0801$ (to 4 d.p.). [M1 — correct application of formula]
3. $x_3 = \sqrt[3]{2(2.0801) + 5} = \sqrt[3]{4.1602 + 5} = \sqrt[3]{9.1602} \approx 2.0921$ — Use the full value of $x_2$ without rounding mid-calculation.
4. $x_4 = \sqrt[3]{2(2.0921) + 5} = \sqrt[3]{4.1842 + 5} = \sqrt[3]{9.1842} \approx 2.0934$ — [A1 — correct values to 4 d.p. for $x_3$ and $x_4$]
5. $x_1 = 2,\; x_2 \approx 2.0801,\; x_3 \approx 2.0921,\; x_4 \approx 2.0934$ — The sequence is converging towards approximately 2.0946, which is the root of $x^3 - 2x - 5 = 0$. [A1 — convergence observed]

Answer: $x_2 \approx 2.0801$, $x_3 \approx 2.0921$, $x_4 \approx 2.0934$

7. Exam Technique and Common Mistakes

Growth and decay questions reward careful setup. The most common source of lost marks is not arithmetic — it is choosing the wrong multiplier or rounding too early. The guidance below covers the traps that appear most frequently in Higher-tier papers.

Across this lesson, every technique — from the multiplier formula to iterative sequences and their limits — rests on the same principle: the output of one step feeds directly into the next. Recognising that structure in any question is the key to selecting the right method and setting up the working that earns marks.