GCSE Special sequences Practice Questions

Free GCSE Special sequences practice questions with full step-by-step worked solutions. Covers square numbers, recognising sequences, cube numbers, triangular numbers. Practise exam-style problems and check your method.

square numbersrecognising sequencescube numberstriangular numberscontinuing a sequencearithmetic sequences
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
The square numbers begin 1,4,9,16,…1, 4, 9, 16, \ldots Write down the 6th square number.
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Worked solution

  1. Recall what a square number is

    nth square number=n2\text{nth square number} = n^2

    A square number is a whole number multiplied by itself.

  2. Substitute n=6n = 6

    62=6×66^2 = 6 \times 6

    The 6th square number comes from squaring 6.

  3. State the answer

    3636

    A 6 by 6 array of dots holds 36 dots, so the 6th square number is 36.

Answer
3636
Question 2
1 markeasy
A Fibonacci-type sequence begins 2,5,…2, 5, \ldots Work out the 3rd term.
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Worked solution

  1. State the rule

    t3=t1+t2t_3 = t_1 + t_2

    In a Fibonacci-type sequence each term is the sum of the two before it.

  2. Add the first two terms

    2+5=72 + 5 = 7

    Add 2 and 5.

  3. State the answer

    77

    The sequence continues 2, 5, 7, 12, 19, ...

Answer
77
Question 3
2 marksintermediate
The first term of a geometric sequence is 55 and the common ratio is −2-2. Work out the 4th term.
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Worked solution

  1. Write down what is given

    a=5,r=−2a = 5,\quad r = -2

    The first term is 5 and each term is multiplied by -2.

  2. Find the 2nd term

    5×(−2)=−105 \times (-2) = -10

    Multiplying a positive by a negative gives a negative.

  3. Find the 3rd term

    −10×(−2)=20-10 \times (-2) = 20

    Multiplying a negative by a negative gives a positive.

  4. Find the 4th term

    20×(−2)=−4020 \times (-2) = -40

    The signs alternate because the ratio is negative.

  5. Check with the term rule

    5×(−2)3=5×(−8)=−405 \times (-2)^3 = 5 \times (-8) = -40

    The power 3 is odd, so the result is negative.

  6. State the answer

    −40-40

    The sequence is 5, -10, 20, -40.

Answer
−40-40
Question 4
4 markshard
A ball is dropped from a height of 6464 cm. After each bounce it reaches 34\frac{3}{4} of its previous height. Work out the exact height reached after the 3rd bounce.
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Worked solution

  1. Recognise the sequence

    each height=34×previous height\text{each height} = \frac{3}{4} \times \text{previous height}

    A constant multiplier means the heights form a geometric sequence with r=3/4r = 3/4.

  2. Keep the ratio exact

    r=34r = \frac{3}{4}

    Work in fractions throughout so that nothing is rounded.

  3. Height after the 1st bounce

    64×34=192464 \times \frac{3}{4} = \frac{192}{4}

    Multiply the drop height by three quarters.

  4. Evaluate

    =48 cm= 48 \text{ cm}

    After one bounce the ball reaches 48 cm.

  5. Height after the 2nd bounce

    48×34=36 cm48 \times \frac{3}{4} = 36 \text{ cm}

    Three quarters of 48 is 36.

  6. Height after the 3rd bounce

    36×34=27 cm36 \times \frac{3}{4} = 27 \text{ cm}

    Three quarters of 36 is 27.

  7. Do it in one go as a check

    64×(34)364 \times \left(\frac{3}{4}\right)^{3}

    Three bounces means multiplying by 3/4 three times.

  8. Cube the fraction

    (34)3=2764\left(\frac{3}{4}\right)^{3} = \frac{27}{64}

    Cube the top and cube the bottom.

  9. Multiply and cancel

    64×2764=2764 \times \frac{27}{64} = 27

    The 64s cancel exactly, giving 27.

  10. State the answer

    27 cm27 \text{ cm}

    Rounding 3/4 to 0.8 first would give 32.8 cm - wrong. Keeping the fraction exact is essential.

Answer
2727
Question 5
6 markschallenging
A quadratic sequence begins 0,3,8,15,24,…0, 3, 8, 15, 24, \ldots Find the nnth term and hence work out the 20th term.
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Worked solution

  1. Find the first differences

    3−0=3, 8−3=5, 15−8=7, 24−15=93-0 = 3,\ 8-3 = 5,\ 15-8 = 7,\ 24-15 = 9

    The first differences are 3, 5, 7, 9.

  2. Rule out arithmetic

    3≠53 \neq 5

    The first differences change, so there is no common difference.

  3. Rule out geometric

    83≈2.67,158≈1.88\frac{8}{3} \approx 2.67,\quad \frac{15}{8} \approx 1.88

    The ratios are not equal, so it is not geometric. (Also, dividing by the first term 0 is impossible.)

  4. Find the second differences

    5−3=2, 7−5=2, 9−7=25-3 = 2,\ 7-5 = 2,\ 9-7 = 2

    The second differences are a constant 2.

  5. Confirm it is quadratic

    constant 2nd difference⇒quadratic\text{constant 2nd difference} \Rightarrow \text{quadratic}

    This is the defining test for a quadratic sequence.

  6. Find the coefficient of n2n^2

    22=1\frac{2}{2} = 1

    The coefficient of n squared is half the second difference.

  7. Write down n2n^2

    n2: 1, 4, 9, 16, 25n^2:\ 1,\ 4,\ 9,\ 16,\ 25

    These are the square numbers.

  8. Subtract term by term

    0−1=−1,3−4=−10-1 = -1,\qquad 3-4 = -1

    Take the square numbers away from the sequence.

  9. Continue subtracting

    8−9=−1,15−16=−1,24−25=−18-9 = -1,\quad 15-16 = -1,\quad 24-25 = -1

    The leftover is a constant -1 every time.

  10. Write the nnth term

    nth term=n2−1\text{nth term} = n^2 - 1

    The sequence is the square numbers, each reduced by 1.

  11. Check at n=1n = 1

    12−1=0 ✓1^2 - 1 = 0 \ \checkmark

    The rule reproduces the first term.

  12. Check at n=5n = 5

    52−1=25−1=24 ✓5^2 - 1 = 25 - 1 = 24 \ \checkmark

    The rule reproduces the 5th term.

  13. Substitute n=20n = 20

    t20=202−1t_{20} = 20^2 - 1

    Now use the rule for the 20th term.

  14. Work out the square first

    202=40020^2 = 400

    Square before subtracting - BIDMAS.

  15. Finish and state

    400−1=399400 - 1 = 399

    The 20th term is 399. Notice the structure: this named sequence is just the square numbers shifted down by 1.

Answer
399399

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