What is A20: Iteration?

Higher tier only. Iteration rewrites an equation as x=g(x) and repeatedly applies x_n+1=g(x_n) from the stated starting value.

How many marks are A20 questions worth?

In verified past-paper appearances, A20 questions on Edexcel GCSE Maths 1MA1 have been worth 4–6 marks.

Which past exam papers has A20 appeared in?

A20 has appeared in 3 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A20

Find approximate solutions to equations numerically using iteration [Higher only]

Higher only

Iteration

Worked answers, methods and verified real exam appearances for A20 on Edexcel GCSE Maths 1MA1.

Explanation

Higher tier only. Iteration rewrites an equation as $x=g(x)$ and repeatedly applies $x_{n+1}=g(x_n)$ from the stated starting value.
Each new output becomes the next input. Keep the calculator's full stored value and round only values you are asked to record, because premature rounding can change later iterates.
Convergence is indicated when successive values agree to the required accuracy, although not every rearrangement converges.
The examiner expects the requested named iterate or a justified approximate root, with enough intermediate values to show the process.
Substitute the final approximation into the original equation as a reasonableness check.

Worked example

The iteration $x_{n+1}=\sqrt{10-x_n}$ starts with $x_0=3$ . Find $x_1$ and $x_2$ to three decimal places.

1. $x_1=\sqrt{10-3}=\sqrt7=2.64575\ldots$ .
2.Use the unrounded value: $x_2=\sqrt{10-2.64575\ldots}=2.71187\ldots$ .
3.Round each requested result to three decimal places.

Answer: $x_1=2.646$ and $x_2=2.712$ .

Common mistakes

Don't substitute $x_0$ again when calculating $x_2$ instead of using $x_1$ .
Don't round each iterate heavily before using it as the next input.

Exam tip

Use the calculator answer key to carry the full previous iterate into the next substitution.

Worked practice

Tier 1 · Easy

The iteration

x_{n+1}=\sqrt{10-x_n}

starts with

x_0=3

. Work out

x_1

and

x_2

, giving each to three decimal places.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$x_1=2.646$ $x_2=2.712$	2	$x_1=\sqrt{10-3}=\sqrt7=2.64575\ldots$ . Then $x_2=\sqrt{10-2.64575\ldots}=2.71187\ldots$ . To three decimal places these are $2.646$ and $2.712$ .

Tier 2 · Standard

Use

x_{n+1}=\sqrt{(18-x_n)/2}

with

x_0=3

to find

x_4

. Give the result to four decimal places.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$x_4=2.7604$	4	$x_1=2.7386\ldots$ , $x_2=2.7624\ldots$ , $x_3=2.7602\ldots$ and $x_4=2.7604\ldots$ . Retaining the unrounded values at each stage gives $x_4=2.7604$ to four decimal places.

Tier 3 · Hard

The equation

x^3+2x-7=0

is rearranged in two ways:

x_{n+1}=\sqrt[3]{7-2x_n}

and

x_{n+1}=\dfrac{7}{x_n^2+2}

. For each iteration, start with

x_0=1.6

and work out

x_1

x_4

, giving each value to four decimal places. State which iteration should be continued to estimate the root, explain your choice, and give the root correct to three decimal places.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	For $x_{n+1}=\sqrt[3]{7-2x_n}$ : $x_1=1.5605$ , $x_2=1.5712$ , $x_3=1.5683$ , $x_4=1.5691$ . For $x_{n+1}=7/(x_n^2+2)$ : $x_1=1.5351$ , $x_2=1.6068$ , $x_3=1.5278$ , $x_4=1.6151$ . Continue the cube-root iteration because its values settle to a common value; the second iteration oscillates with increasing changes. $x\approx1.569$ .	6	Both rearrangements have fixed points satisfying $x^3+2x-7=0$ . The cube-root iteration gives the unrounded values $1.560490751\ldots$ , $1.571233091\ldots$ , $1.568326861\ldots$ and $1.569114174\ldots$ ; continuing gives $1.568900965\ldots$ , so the values converge to $1.568946403\ldots$ . The rational iteration gives $1.535087719\ldots$ , $1.606796545\ldots$ , $1.527785462\ldots$ and $1.615088278\ldots$ ; its successive changes grow rather than settle. Therefore the cube-root iteration is suitable from $x_0=1.6$ , and the root is $1.569$ correct to three decimal places.

Tier 1 · Easy

The equation

x^2-3x+1=0

has a solution between

2

and

3

. Starting with

x_0=2

, use the iteration formula

x_{n+1}=3-\dfrac{1}{x_n}

. Work out

x_1

and

x_2

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$x_1=2.5$ $x_2=2.6$	2	$x_1=3-\dfrac12=2.5$ . Then $x_2=3-\dfrac{1}{2.5}=3-0.4=2.6$ .

Tier 2 · Standard

The equation

x^2+x-5=0

has a solution between

1

and

2

. Starting with

x_0=2

, use the iteration formula

x_{n+1}=\dfrac{5}{x_n+1}

to find

x_3

. Give the result to three decimal places.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$x_3=1.739$	3	$x_1=5/(2+1)=5/3$ , $x_2=5/(5/3+1)=15/8$ and $x_3=5/(15/8+1)=40/23=1.73913\ldots$ . Therefore $x_3=1.739$ to three decimal places.

Tier 3 · Hard

The equation

x^3=15

has a solution between

2

and

3

. Starting with

x_0=2.5

, use the iteration formula

x_{n+1}=\dfrac13\left(2x_n+\dfrac{15}{x_n^2}\right)

. Write down the next three iterates and hence give the solution correct to

4

decimal places.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$x_1=2.466666\ldots$ $x_2=2.466212158\ldots$ $x_3=2.466212074\ldots$ $x=2.4662$	3	$x_1=2.466666\ldots$ , $x_2=2.466212158\ldots$ and $x_3=2.466212074\ldots$ . The next iterate remains $2.466212074\ldots$ , so the solution is $x=2.4662$ to four decimal places.

Tier 2 · Standard

The equation

x^2+20x-50=0

has a positive solution. It can be rearranged to give

x=\dfrac{50-x^2}{20}

. Starting with

x_0=2.2

, use the iteration formula

x_{n+1}=\dfrac{50-x_n^2}{20}

. Work out

x_1

x_2

and

x_3

, giving each value to

3

decimal places. Hence estimate the positive solution correct to

2

decimal places.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x_1=2.258$ , $x_2=2.245$ , $x_3=2.248$ $x\approx2.25$	3	Using unrounded values, $x_1=2.258$ , $x_2=2.2450718$ and $x_3=2.247982631\ldots$ . Therefore the iterates to three decimal places are $2.258$ , $2.245$ and $2.248$ , and the positive solution is $2.25$ to two decimal places. The smallest iterate rounding margin is $0.428$ final-digit units. The root is $-10+5\sqrt6=2.247448714\ldots$ , which is $0.002448714\ldots$ , or $0.245$ final-digit units, from the nearest two-decimal-place rounding boundary.

Tier 3 · Hard

The positive solution of

x^2-2x-5=0

can be found using

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

. Starting with

x_0=3.5

, work out

x_1

x_5

. Hence give the solution correct to

4

decimal places.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$x_1=3.444444\ldots$ , $x_2=3.45$ , $x_3=3.449438\ldots$ , $x_4=3.449494\ldots$ , $x_5=3.449489\ldots$ $x=3.4495$	4	Repeated substitution gives $x_1=3.444444\ldots$ , $x_2=3.45$ , $x_3=3.449438\ldots$ , $x_4=3.449494\ldots$ and $x_5=3.449489\ldots$ . The successive values agree to four decimal places, so the positive solution is $3.4495$ .

Tier 3 · Hard

Starting with

x_0=2

, use

x_{n+1}=\sqrt{\dfrac{28}{x_n+3}}

to work out

x_1

x_6

. Give each iterate to at least

5

decimal places. Hence write down the positive solution approached by the iteration, correct to

3

decimal places.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$x_1=2.36643$ , $x_2=2.28421$ , $x_3=2.30191$ , $x_4=2.29807$ , $x_5=2.29890$ , $x_6=2.29872$ $x=2.299$	5	Repeated substitution gives $2.3664319\ldots$ , $2.2842109\ldots$ , $2.3019132\ldots$ , $2.2980671\ldots$ , $2.2989011\ldots$ and $2.2987202\ldots$ . The smallest five-decimal-place rounding margin among these iterates is $0.180$ final-digit units. The values converge to $2.2987524\ldots$ , so the positive solution is $2.299$ to three decimal places. It is $0.248$ final-digit units from the nearest rounding boundary.

Q10

Tier 3 · Hard

The equation

x^2+4x-24=0

has one positive solution. Use

x_{n+1}=\dfrac{24+x_n}{x_n+5}

with

x_0=3.5

. Continue the iteration until, for the first time, two consecutive values agree when rounded to

4

decimal places. Write down every iterate used and hence give the positive solution correct to

4

decimal places.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$x_1=3.235294\ldots$ , $x_2=3.307143\ldots$ , $x_3=3.287188\ldots$ , $x_4=3.292696\ldots$ , $x_5=3.291173\ldots$ , $x_6=3.291594\ldots$ , $x_7=3.291477\ldots$ , $x_8=3.291510\ldots$ $x=3.2915$	5	Successive substitution gives the eight values shown. To four decimal places, $x_6=3.2916$ , while $x_7=3.2915$ and $x_8=3.2915$ , so $x_7$ and $x_8$ are the first consecutive pair that agree at the requested precision. The limiting value is $-2+2\sqrt7=3.2915026\ldots$ , so the answer is $3.2915$ . It is $0.474$ final-digit units from the nearest four-decimal-place rounding boundary.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-11	2H	Q17	6	Allowed	Higher	QP MS
2021-11	2H	Q16	4	Allowed	Higher	QP MS
2023-06	2H	Q17	4	Allowed	Higher	QP MS

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