[discia] GCSE Maths › Algebra › Iteration

Iteration

discia.org 10 questions · 26 marks
Name Date
  1. Either paper 2 marks

    A sequence of numbers is formed by the iterative process

    \(u_{n+1} = 8 - \frac{ 3 }{ u_n }\) with \(u_1 = 6\)

    Work out the values of \(u_2\) and \(u_3\).

    (a)

    The value of \(u_2\)

    (b)

    The value of \(u_3\)

    Show answer
    (a) \(7.5\)
    (b) \(7.6\)
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  2. Calculator 3 marks

    Dev borrows £1800 from a bank.

    Each month the amount he owes is multiplied by 1.01 (interest) and then Dev pays back £230.

    \(A_n\) is the amount Dev owes, in pounds, at the end of month \(n\).

    \(A_{n+1} = 1.01A_n - 230\) with \(A_0 = 1800\)

    Calculate the amount Dev still owes at the end of the 2nd month.

    Show answer
    \(\text{£}1373.88\)
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  3. Either paper 2 marks

    \(f(x) = x^3 + 4x - 37\)

    Show that the equation \(f(x) = 0\) has a solution in the interval \(2 < x < 3\).

    Show answer
    \(f(2) = -21\) and \(f(3) = 2\). One is negative and the other is positive, so \(f(x) = 0\) has a solution between \(x = 2\) and \(x = 3\).
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  4. Either paper 2 marks

    Rearrange the equation \(x^3 + 9x - 49 = 0\) into the form \(x = \sqrt[3]{p + qx}\), where \(p\) and \(q\) are integers.

    Show answer
    \(x = \sqrt[3]{ 49 - 9x }\)
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  5. Either paper 4 marks

    At the start of year \(n\) the population of a species is \(p_n\) million.

    At the start of the following year the population of the species is given by

    \(p_{n+1} = kp_n\)

    where \(k\) is a positive constant.

    The population of the species at the start of year 1 is 32 million.

    The population of the species at the start of year 2 is 30.4 million.

    (a)

    Work out the population of the species at the start of year 3.

    (b)

    At the start of year 5 the value of \(k\) is increased by 0.45 to a new constant value.

    Aisha thinks that from the start of year 5 the population of the species would decrease year on year.

    Is Aisha correct?

    You must show how you get your answer.

    Show answer
    (a) \(28.88\ \text{million}\)
    (b) No: the new value of \(k\) is \(1.4\), which is greater than 1, so the population would increase year on year.
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  6. Either paper 4 marks

    A hot air balloon is descending.

    The height of the balloon \(n\) minutes after it starts to descend is \(h_n\) metres.

    The height of the balloon \((n + 1)\) minutes after it starts to descend, \(h_{n+1}\) metres, is given by

    \(h_{n+1} = k h_n - 20\)

    where \(k\) is a constant.

    The balloon starts to descend from a height of 1600 metres at 09:31.

    At 09:32 the height of the balloon is 1260 metres.

    Work out the height of the balloon at 09:34.

    Show answer
    \(770.4\ \text{metres}\)
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  7. Calculator 2 marks

    \(f(x) = x^3 - x - 40\)

    Show that the equation \(f(x) = 0\) has a solution between \(x = 3.5\) and \(x = 3.6\).

    Show answer
    \(f(3.5) = -0.625\) and \(f(3.6) = 3.056\). The sign changes between them, so \(f(x) = 0\) has a solution between \(x = 3.5\) and \(x = 3.6\).
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  8. Either paper 2 marks

    A student uses the iteration formula

    \(x_{n+1} = \sqrt[3]{ 4x_n + 8 }\)

    The values it produces get closer and closer to one number.

    Write down the equation, in the form \(x^3 + px + q = 0\), that this number is a solution of.

    Show answer
    \(x^3 - 4x - 8 = 0\)
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  9. Calculator 3 marks

    \(x_{n+1} = \sqrt[3]{ 3x_n + 10 }\)

    Starting with \(x_0 = 2\), work out the values of \(x_1\), \(x_2\) and \(x_3\).

    Give your answers correct to 4 decimal places.

    Show answer
    \(x_1 \approx 2.5198\), \(x_2 \approx 2.5992\), \(x_3 \approx 2.6109\)
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  10. Calculator 2 marks

    A student uses the iteration formula

    \(x_{n+1} = \sqrt[3]{ 4x_n + b }\)

    where \(b\) is an integer.

    Starting with \(x_0 = 3\), the first iterate is \(x_1 = 4.0412\), correct to 4 decimal places.

    Find the value of \(b\).

    Show answer
    \(54\)
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discia.org Iteration · sheet 1
Answers Iteration · sheet 1
  1. (a) \(7.5\)
    (b) \(7.6\)
  2. \(\text{£}1373.88\)
  3. \(f(2) = -21\) and \(f(3) = 2\). One is negative and the other is positive, so \(f(x) = 0\) has a solution between \(x = 2\) and \(x = 3\).
  4. \(x = \sqrt[3]{ 49 - 9x }\)
  5. (a) \(28.88\ \text{million}\)
    (b) No: the new value of \(k\) is \(1.4\), which is greater than 1, so the population would increase year on year.
  6. \(770.4\ \text{metres}\)
  7. \(f(3.5) = -0.625\) and \(f(3.6) = 3.056\). The sign changes between them, so \(f(x) = 0\) has a solution between \(x = 3.5\) and \(x = 3.6\).
  8. \(x^3 - 4x - 8 = 0\)
  9. \(x_1 \approx 2.5198\), \(x_2 \approx 2.5992\), \(x_3 \approx 2.6109\)
  10. \(54\)
discia.org Iteration · sheet 1