Iteration GCSE Questions, Worked Examples and Answers

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Hi, I’m Ari. We can follow one repeated calculation, unpack the subscripts and then use a calculator to estimate a solution.

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Start with a repeated instruction

What you need to know about iteration

Start with 2. Halve it and add 3: the answer is 4. Now apply the same instruction to 4: the answer is 5. Apply it to 5 and you get 5.5. Repeating a calculation in this way is called iteration. The important change each time is the input: always use the answer you just obtained.

Use the result as the next input

Name the current and next values

We call the starting value x₀ (‘x zero’). After one calculation we have x₁; after two we have x₂. The small lower number counts steps: x₂ does not mean x squared.

Startx₀ = 2before applying the rule

One calculationx₁ = 4halve 2, then add 3

Next calculationx₂ = 5halve 4, then add 3

The letter $n$ counts how many steps have been taken. $x_n$ is the current value and $x_{n+1}$ is the next value. Our instruction becomes:

x_{n+1}=\frac{x_n}{2}+3

How can repeating help solve an equation?

Here the values approach 6: $2,4,5,5.5,5.75,\ldots$ . If the input were exactly 6, the output would also be 6. Such an unchanged value is called a fixed point.

6=\frac62+3

So the limit solves $x=x/2+3$ . Each step halves the distance to 6, which explains why this particular rule approaches it. Other rules may move away or alternate forever; repetition alone does not guarantee an answer.

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Calculate the next few values

What the problem asks: Given x₀ and a rule for xₙ₊₁, find x₁, x₂ and x₃.

How to solve it: Substitute the start once to find x₁. Put that answer back in for x₂, then repeat. Record the step numbers so that the starting value is not counted as the first iteration.

Show how a rule comes from an equation

What the problem asks: Rearrange x² − x − 8 = 0 into a supplied square-root form for the positive solution.

How to solve it: Isolate x², then take the positive square root because the required solution is positive. Replace the left x with the next value and the right x with the current value.

Estimate a root

What the problem asks: Use the supplied rule to find a solution to a stated number of decimal places.

How to solve it: Continue with full calculator precision until the values settle to the requested rounding. Substitute back to check. A root means a value making the original expression zero.

Justify an interval or accuracy

What the problem asks: Show that a root is between two values.

How to solve it: Evaluate the original continuous expression at both endpoints. Opposite signs show a crossing of zero between them. To prove rounding, choose bounds inside the same rounding interval.

A reliable routine

Use a supplied iterative formula

Use this routine when a question provides a starting value and a rule for the next value. Each output becomes the next input by definition. If the task is root finding, the rule must also converge to a fixed point that satisfies the original equation.

Identify the starting value and its subscript.
Substitute it into the whole right-hand side, using brackets carefully.
Label the output with the next subscript.
Feed the full output back into the formula; a stored-answer key can help.
Stop after the requested number of steps, or establish the requested accuracy and check in the original equation.

Check: Repeatedly seeing the same rounded answer is evidence of settling, not a general proof of convergence. Use a sign-change bracket when the question asks you to justify accuracy.

Fully worked

Iteration GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Read the subscripts correctly

Higher only3 marks

Question

Given $x_0=1$ and $x_{n+1}=3x_n+2$ , find $x_1$ , $x_2$ and $x_3$ .

Each step triples the latest value and adds 2.

x_1=3(1)+2=5

x_2=3(5)+2=17

x_3=3(17)+2=53

\boxed{x_1=5,\quad x_2=17,\quad x_3=53}

This sequence does not settle down. The question asks for three outputs, not a root.

Example 2

Rearrange for a positive solution

Higher only2 marks

Question

Show how $x^2-x-8=0$ gives the iterative rule $x_{n+1}=\sqrt{8+x_n}$ for estimating its positive solution.

Add $x+8$ to both sides.

x^2=x+8

For the positive solution, take the positive square root.

x=\sqrt{x+8}

Use the current value on the right to generate a new value on the left:

\boxed{x_{n+1}=\sqrt{8+x_n}}

The original quadratic also has a negative root; this rule targets the positive one.

Example 3

Use the stored calculator answer

Higher only3 marks

Question

Starting with $x_0=3$ , use $x_{n+1}=\sqrt{8+x_n}$ to find $x_1$ , $x_2$ and $x_3$ . Show each to 4 decimal places.

x_1=\sqrt{8+3}=3.316624790\ldots

x_2=\sqrt{8+3.316624790\ldots}=3.364019142\ldots

x_3=\sqrt{8+3.364019142\ldots}=3.371056087\ldots

\boxed{x_1=3.3166,\ x_2=3.3640,\ x_3=3.3711}

On a suitable calculator, enter the start, then evaluate $\sqrt{8+\mathrm{Ans}}$ repeatedly. Keep the whole sum under the square root; button sequences vary by model.

Example 4

Continue to an estimate

Higher only3 marks

Question

Use the same rule and start to estimate the positive root of $x^2-x-8=0$ to 3 decimal places.

Add 8 to the previous value, then take the square root of the whole sum
Step n	Value xₙ (display rounded)
0	3 (start)
1	3.316624790
2	3.364019142
3	3.371056087
4	3.372099656
5	3.372254388
6	3.372277330

The table displays nine decimal places. Calculations use the full previous calculator value, not the rounded display typed in again.

From $x_4$ onwards the displayed values round to the same 3-decimal-place result.

\boxed{x\approx3.372}

Substitution gives $3.372^2-3.372-8=-0.001616$ , close to zero. This checks the estimate; the next example gives an actual bracket to justify the rounding.

Example 5

Justify the rounding

Higher only3 marks

Question

Let $f(x)=x^2-x-8$ . Show that the positive root rounds to 3.372 to 3 decimal places using 3.3722 and 3.3723.

$f(x)$ means evaluate the expression at x.

f(3.3722)=-0.00046716

f(3.3723)=0.00010729

This polynomial is continuous: its graph has no jump between these inputs. The sign change places a root between them. It is the unique positive root, since the quadratic has one positive and one negative root.

Both endpoints lie between $3.3715$ and $3.3725$ , so every number in this bracket rounds to $\boxed{3.372}$ to 3 decimal places.

Example 6

Recognise a non-converging rule

Higher only3 marks

Question

For $x_{n+1}=5-x_n$ and $x_0=1$ , calculate three values and explain why this start does not lead to an approximate fixed point.

x_1=5-1=4

x_2=5-4=1

x_3=5-1=4

The values repeat $1,4,1,4,\ldots$ and do not get closer to one number. They oscillate. Although $x=2.5$ solves $x=5-x$ , this starting value and rule do not approach it.

10 original questions · total 23 marks

Iteration GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 35 minutes · Use a calculator. Label the iteration number, carry full precision forward and distinguish a rounded display from the value used in the next step. · answers start collapsed

1

Identify current and next

Higher only1 mark

What does $x_{n+1}$ mean in an iterative formula?

Show worked answer

It means the next value, after applying the rule to the current value. The subscript is a step label, not a power.

2

Apply two iterations

Higher only2 marks

$x_0=2$ and $x_{n+1}=2x_n+3$ . Find $x_1$ and $x_2$ .

Show worked answer

x_1=2(2)+3=7

x_2=2(7)+3=17

\boxed{x_1=7,\ x_2=17}

3

Start with a different subscript

Higher only2 marks

$u_1=10$ and $u_{n+1}=u_n/2+1$ . Find $u_2$ and $u_3$ .

Show worked answer

The supplied starting subscript is 1, so the first new value is labelled 2.

u_2=10/2+1=6

u_3=6/2+1=4

\boxed{u_2=6,\ u_3=4}

4

Use a square-root rule

Higher only3 marks

$x_0=2$ , $x_{n+1}=\sqrt{6+x_n}$ . Find $x_1$ and $x_2$ to 3 decimal places.

Show worked answer

x_1=\sqrt{6+2}=\sqrt8=2.828427124\ldots

x_2=\sqrt{6+2.828427124\ldots}=2.971266922\ldots

\boxed{x_1=2.828,\ x_2=2.971}

Use full precision from the first step.

5

Derive the given rearrangement

Higher only2 marks

Rearrange $x^2-2x-7=0$ into a square-root iterative rule for a positive solution.

Show worked answer

x^2=2x+7

x=\sqrt{2x+7}

\boxed{x_{n+1}=\sqrt{2x_n+7}}

The positive square root matches the requested positive solution.

6

Recover the equation

Higher only3 marks

A convergent rule is $x_{n+1}=\sqrt{12-x_n}$ . Write the quadratic equation satisfied by its fixed point.

Show worked answer

At a fixed point, current and next values are equal.

x=\sqrt{12-x}

x^2=12-x

\boxed{x^2+x-12=0}

The square-root form requires a non-negative fixed point, so it selects 3 rather than the other quadratic root, −4.

7

Locate a root by a sign change

Higher only3 marks

Show that $x^3+x-5=0$ has a root between 1 and 2.

Show worked answer

Let $f(x)=x^3+x-5$ .

f(1)=1+1-5=-3

f(2)=8+2-5=5

The continuous polynomial changes sign, so it crosses zero somewhere between 1 and 2. Thus a root lies in that interval.

8

Spot an input error

Higher only2 marks

For $x_{n+1}=\sqrt{8+x_n}$ with $x_0=3$ , a student enters $\sqrt{8+3}$ three times and reports $x_1=x_2=x_3=3.3166$ to 4 decimal places. Explain the mistake.

Show worked answer

They restarted from $3$ every time. The first value is correct, but the second calculation must use the full $x_1$ value as input:

x_2=\sqrt{8+3.316624790\ldots}=3.364019142\ldots

Then use $x_2$ for the third calculation. This rule does not leave the starting value unchanged, so the repeated answers are not successive iterations.

9

Find a fixed point directly

Higher only3 marks

Find the fixed point of $x_{n+1}=0.75x_n+2$ .

Show worked answer

Set next equal to current.

x=0.75x+2

0.25x=2

\boxed{x=8}

Check: $0.75(8)+2=8$ .

10

Distinguish exact from rounded

Higher only2 marks

An iteration table approaches 2.7182818… . A student writes that the exact solution equals 2.718. What should they write if 3 decimal places are requested?

Show worked answer

They should write $\boxed{x\approx2.718}$ , to 3 decimal places. A rounded decimal is an approximation, not an exact equality. Settling digits alone are not a proof that the original equation has that exact decimal root.

Examiner-style feedback

Common iteration mistakes

Reading a subscript as a power

x₃ names the third output after x₀; x³ cubes x. They mean different things.

Typing rounded values back in

Keep the stored result for the next calculation. Round only the values displayed in your answer.

Using a rule without checking brackets

For a square root of 8 plus x, both 8 and x must sit inside the root.

Assuming every rearrangement converges

A rule can diverge or oscillate. Use the supplied rule and investigate unexpected values rather than declaring a root.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

Iteration reuses the latest answer.
Subscripts label steps.
A convergent fixed point must satisfy the original equation.
Check accuracy using the original expression.

Quick answers

Iteration FAQ

Is iteration Higher tier?

Numerically approximating equation solutions using iteration is Higher-tier GCSE content.

Does x₀ count as the first iteration?

No. It is the supplied start. Applying the rule once gives x₁, unless the question starts with another subscript.

When should I stop?

Follow the question: calculate the specified number of terms or continue to the required accuracy. If asked to justify rounding, use a suitable root bracket.

Does iteration always work?

No. Some rules and starting values move away from a root or cycle between values.

Build connected skills

What to revise next

Prerequisite

Functions

Understand inputs, outputs and substitution.

Revise functions

Sequences

Read term labels and repeated rules.

Revise sequences

Exact alternative

Quadratic formula

Compare numerical estimates with exact quadratic roots.

Revise quadratic formula

Content standards

Curriculum and rights review

Reviewed 7 September 2026 against GCSE A20 and Pearson teaching guidance. All examples, numerical sequences and practice are original. Suggested marks are Pass an Exam estimates.

Official specification references