Hard GCSE Special sequences Questions

Challenging, exam-style GCSE Special sequences questions with worked solutions. Stretch yourself on the hardest triangular numbers, solving a quadratic, justifying, geometric sequences problems.

triangular numberssolving a quadraticjustifyinggeometric sequencesnegative common ratioterm rule
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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 20th20\mathrm{th} term.
Show worked solution

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 33, 55, 77, 99.

  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 00 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 22.

  5. Confirm it is quadratic

    constant 2nd difference⇒quadratic\text{constant }2\mathrm{nd}\text{ 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 −1-1 every time.

  10. Write the nnth term

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

    The sequence is the square numbers, each reduced by 11.

  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 5th5\mathrm{th} term.

  13. Substitute n=20n = 20

    t20=202−1t_{20} = 20^2 - 1

    Now use the rule for the 20th20\mathrm{th} 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 20th20\mathrm{th} term is 399399. Notice the structure: this named sequence is just the square numbers shifted down by 11.

Answer
399399
Question 2
6 markschallenging
Is 528528 a triangular number? If it is, state its position in the sequence.
Show worked solution

Worked solution

  1. Recall the formula

    Tn=n(n+1)2T_n = \frac{n(n+1)}{2}

    A number is triangular exactly when this equation has a positive whole-number solution.

  2. Set up the equation

    n(n+1)2=528\frac{n(n+1)}{2} = 528

    Test 528528 against the formula.

  3. Clear the fraction

    n(n+1)=1056n(n+1) = 1056

    Two consecutive whole numbers must multiply to 10561056.

  4. Rearrange into a quadratic

    n2+n−1056=0n^2 + n - 1056 = 0

    Bring everything to one side.

  5. Estimate nn

    1056≈32.5\sqrt{1056} \approx 32.5

    n should be around 32.

  6. Test the consecutive pair

    32×33=1056 ✓32 \times 33 = 1056 \ \checkmark

    3232 and 3333 are consecutive whole numbers with product exactly 10561056.

  7. Factorise

    (n−32)(n+33)=0(n - 32)(n + 33) = 0

    -32 x 33=−105633 = -1056 and −32+33=1-32 + 33 = 1.

  8. Solve

    n=32orn=−33n = 32 \quad \text{or} \quad n = -33

    Two roots appear.

  9. Reject the negative root

    n=−33 is not a valid positionn = -33 \text{ is not a valid position}

    Positions in a sequence are positive whole numbers.

  10. Verify by substitution

    T32=32×332=10562=528 ✓T_{32} = \frac{32 \times 33}{2} = \frac{1056}{2} = 528 \ \checkmark

    The 32nd32\mathrm{nd} triangular number really is 528528.

  11. Confirm with the discriminant test

    b2−4ac=1+4×1056=4225b^2 - 4ac = 1 + 4 \times 1056 = 4225

    For n to be whole, the discriminant must be a perfect square.

  12. Check the discriminant is square

    652=4225 ✓65^2 = 4225 \ \checkmark

    42254225 is a perfect square, which is exactly the integer test passing.

  13. Complete the formula

    n=−1+652=642=32 ✓n = \frac{-1 + 65}{2} = \frac{64}{2} = 32 \ \checkmark

    The quadratic formula gives the same position, 3232.

  14. Contrast with a non-triangular number

    for 530: 1+8×530=4241,  652=4225<4241<4356=662\text{for } 530:\ 1 + 8 \times 530 = 4241,\ \ 65^2 = 4225 < 4241 < 4356 = 66^2

    42414241 is not a perfect square, so 530530 is NOT triangular - the integer check is what decides it, not appearance.

  15. State the answer

    528=T32528 = T_{32}

    Yes: 528528 is triangular, and it is the 32nd32\mathrm{nd} triangular number.

Answer
3232
Question 3
5 markschallenging
A geometric sequence has first term 22 and common ratio −12-\frac{1}{2}. Work out the 7th7\mathrm{th} term. Give your answer as a fraction.
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Worked solution

  1. Write down what is given

    a=2,r=−12a = 2,\qquad r = -\frac{1}{2}

    The ratio is both negative and fractional, so the terms shrink AND alternate in sign.

  2. Keep the ratio exact

    r=−12 (never rounded)r = -\frac{1}{2} \text{ (never rounded)}

    Working in fractions guarantees an exact final answer.

  3. Find the 2nd2\mathrm{nd} term

    2×(−12)=−12 \times \left(-\frac{1}{2}\right) = -1

    Half of 22 is 11, and the sign flips.

  4. Find the 3rd3\mathrm{rd} term

    −1×(−12)=12-1 \times \left(-\frac{1}{2}\right) = \frac{1}{2}

    Negative times negative is positive.

  5. Find the 4th4\mathrm{th} term

    12×(−12)=−14\frac{1}{2} \times \left(-\frac{1}{2}\right) = -\frac{1}{4}

    The sign flips again.

  6. Find the 5th5\mathrm{th} term

    −14×(−12)=18-\frac{1}{4} \times \left(-\frac{1}{2}\right) = \frac{1}{8}

    Back to positive.

  7. Find the 6th6\mathrm{th} term

    18×(−12)=−116\frac{1}{8} \times \left(-\frac{1}{2}\right) = -\frac{1}{16}

    Negative again.

  8. Find the 7th7\mathrm{th} term

    −116×(−12)=132-\frac{1}{16} \times \left(-\frac{1}{2}\right) = \frac{1}{32}

    Positive, and one thirty-second in size.

  9. Now check with the term rule

    tn=arn−1⇒t7=2×(−12)6t_n = ar^{n-1} \Rightarrow t_7 = 2 \times \left(-\frac{1}{2}\right)^{6}

    The power is 7−1=67 - 1 = 6.

  10. Decide the sign of the power

    6 is even⇒the result is positive6 \text{ is even} \Rightarrow \text{the result is positive}

    An even power of a negative number is positive.

  11. Work out the size

    (12)6=164\left(\frac{1}{2}\right)^{6} = \frac{1}{64}

    Two to the sixth is 6464.

  12. Multiply

    t7=2×164=264t_7 = 2 \times \frac{1}{64} = \frac{2}{64}

    Multiply the first term by the power.

  13. Cancel

    264=132 ✓\frac{2}{64} = \frac{1}{32} \ \checkmark

    Both methods agree: the 7th7\mathrm{th} term is one thirty-second.

  14. Summarise the sign pattern

    +, −, +, −, +, −, ++,\ -,\ +,\ -,\ +,\ -,\ +

    Odd-numbered terms are positive and even-numbered terms are negative, so a positive 7th7\mathrm{th} term is exactly what we expect.

  15. State the exact answer

    132\frac{1}{32}

    The full sequence is 22, −1-1, 1/21/2, −1/4-1/4, 1/81/8, −1/16-1/16, 1/321/32.

Answer
132\frac{1}{32}
Question 4
6 markschallenging
Prove that the sum of any two consecutive triangular numbers is always a square number.
Show worked solution

Worked solution

  1. Test the claim numerically

    T1+T2=1+3=4=22T_1 + T_2 = 1 + 3 = 4 = 2^2

    Start by checking whether the claim even looks true.

  2. Test it again

    T2+T3=3+6=9=32T_2 + T_3 = 3 + 6 = 9 = 3^2

    Another square number.

  3. And again

    T3+T4=6+10=16=42T_3 + T_4 = 6 + 10 = 16 = 4^2

    The pattern is holding.

  4. Form the conjecture

    Tn+Tn+1=(n+1)2T_n + T_{n+1} = (n+1)^2

    The square root seems to be one more than the first position.

  5. Note that examples are not a proof

    three cases≠all cases\text{three cases} \neq \text{all cases}

    Checking examples can never prove a statement about EVERY pair - algebra is needed.

  6. Write the first triangular number

    Tn=n(n+1)2T_n = \frac{n(n+1)}{2}

    The standard formula.

  7. Write the next triangular number

    Tn+1=(n+1)(n+2)2T_{n+1} = \frac{(n+1)(n+2)}{2}

    Replace n by n + 1 throughout the formula.

  8. Add them

    Tn+Tn+1=n(n+1)2+(n+1)(n+2)2T_n + T_{n+1} = \frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2}

    Both fractions already have denominator 22.

  9. Combine over one denominator

    =n(n+1)+(n+1)(n+2)2= \frac{n(n+1) + (n+1)(n+2)}{2}

    Add the numerators.

  10. Take out the common factor

    =(n+1)[n+(n+2)]2= \frac{(n+1)\big[n + (n+2)\big]}{2}

    Both terms in the numerator contain (n + 1).

  11. Simplify inside the bracket

    =(n+1)(2n+2)2= \frac{(n+1)(2n+2)}{2}

    n+n+2=2n+2n + n + 2 = 2n + 2.

  12. Factorise the bracket

    =(n+1)⋅2(n+1)2= \frac{(n+1) \cdot 2(n+1)}{2}

    2n+2=2(n+1)2n + 2 = 2(n + 1).

  13. Cancel the 22

    =(n+1)(n+1)=(n+1)2= (n+1)(n+1) = (n+1)^2

    The 22 on the top cancels with the 22 on the bottom.

  14. Interpret the result

    (n+1)2 is a whole number squared(n+1)^2 \text{ is a whole number squared}

    Since n is a positive whole number, n + 1 is too, so the sum is always a square number. Proved.

  15. Check the proof on a new case

    T7+T8=28+36=64=(7+1)2 ✓T_7 + T_8 = 28 + 36 = 64 = (7+1)^2 \ \checkmark

    The general result correctly predicts this case as well.

Answer
Tn+Tn+1=(n+1)2T_n + T_{n+1} = (n+1)^2
Question 5
5 markschallenging
There are 1010 people at a meeting. Each person shakes hands exactly once with every other person. Work out the total number of handshakes.
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Worked solution

  1. Count the first person's handshakes

    person 1 shakes 9 hands\text{person 1 shakes } 9 \text{ hands}

    There are 99 other people.

  2. Count the second person's NEW handshakes

    person 2 adds 8 new\text{person 2 adds } 8 \text{ new}

    The handshake with person 11 has already been counted.

  3. Continue the count

    person 3 adds 7, person 4 adds 6, …\text{person 3 adds } 7,\ \text{person 4 adds } 6,\ \ldots

    Each new person adds one fewer new handshake.

  4. Write the total as a sum

    9+8+7+6+5+4+3+2+19 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

    The last person adds no new handshakes at all.

  5. Recognise the sequence

    1+2+⋯+9=T91 + 2 + \cdots + 9 = T_9

    This is exactly the 9th9\mathrm{th} triangular number.

  6. Write the formula

    Tn=n(n+1)2T_n = \frac{n(n+1)}{2}

    The standard triangular number formula.

  7. Substitute n=9n = 9

    T9=9×102T_9 = \frac{9 \times 10}{2}

    Careful: it is T9T_9, not T10T_10, even though there are 1010 people.

  8. Evaluate

    T9=902=45T_9 = \frac{90}{2} = 45

    There are 4545 handshakes.

  9. Check by pairing the sum

    (9+1)+(8+2)+(7+3)+(6+4)=4×10=40(9+1) + (8+2) + (7+3) + (6+4) = 4 \times 10 = 40

    Pair the numbers from the outside in.

  10. Finish the pairing check

    40+5=45 ✓40 + 5 = 45 \ \checkmark

    The middle number 55 is left over, giving 4545 in total.

  11. Check against the triangular list

    1,3,6,10,15,21,28,36,451, 3, 6, 10, 15, 21, 28, 36, 45

    The 9th9\mathrm{th} triangular number is 4545.

  12. Explain the shift

    for n people the answer is Tn−1\text{for } n \text{ people the answer is } T_{n-1}

    The last person has nobody new left to shake hands with, which is why the position drops by one.

  13. Check the general rule

    10×92=45 ✓\frac{10 \times 9}{2} = 45 \ \checkmark

    n(n - 1) / 2 with n=10n = 10 gives the same 4545.

  14. Beware the double count

    10×9=90 counts A-B and B-A separately10 \times 9 = 90 \text{ counts A-B and B-A separately}

    Forgetting to halve is the classic error here.

  15. State the answer

    4545

    There are 4545 handshakes in total.

Answer
4545

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