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 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
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 528 against the formula.

  3. Clear the fraction

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

    Two consecutive whole numbers must multiply to 1056.

  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

    32 and 33 are consecutive whole numbers with product exactly 1056.

  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 32nd triangular number really is 528.

  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

    4225 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, 32.

  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

    4241 is not a perfect square, so 530 is NOT triangular - the integer check is what decides it, not appearance.

  15. State the answer

    528=T32528 = T_{32}

    Yes: 528 is triangular, and it is the 32nd 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 7th 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 2nd term

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

    Half of 2 is 1, and the sign flips.

  4. Find the 3rd term

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

    Negative times negative is positive.

  5. Find the 4th term

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

    The sign flips again.

  6. Find the 5th term

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

    Back to positive.

  7. Find the 6th term

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

    Negative again.

  8. Find the 7th 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 64.

  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 7th term is one thirty-second.

  14. Summarise the sign pattern

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

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

  15. State the exact answer

    132\frac{1}{32}

    The full sequence is 2, -1, 1/2, -1/4, 1/8, -1/16, 1/32.

Answer
132\frac{1}{32}
Question 4
6 markschallenging
Prove that the sum of any two consecutive triangular numbers is always a square number.
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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 2.

  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 2

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

    The 2 on the top cancels with the 2 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 9 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 1 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 9th 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 10 people.

  8. Evaluate

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

    There are 45 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 5 is left over, giving 45 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 9th triangular number is 45.

  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 45.

  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 45 handshakes in total.

Answer
4545

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