Hard GCSE nth term of quadratic sequences Questions

Challenging, exam-style GCSE nth term of quadratic sequences questions with worked solutions. Stretch yourself on the hardest second differences, nth term of a quadratic sequence, negative n squared coefficient, fractional n squared coefficient problems.

second differencesnth term of a quadratic sequencenegative n squared coefficientfractional n squared coefficientfar termssolving quadratic equations
GCSE Higher34 questionsStep-by-step solutions
Question 1
5 markschallenging
A quadratic sequence has nnth term an2+bn+can^2 + bn + c. Its first three terms are T1T_1, T2T_2 and T3T_3. Which one of these expressions always gives the value of aa?
Show worked solution

Worked solution

  1. Write the first three terms in terms of a, b and c

    T1=a+b+c,T2=4a+2b+c,T3=9a+3b+cT_1 = a + b + c, \quad T_2 = 4a + 2b + c, \quad T_3 = 9a + 3b + c

    Substitute n=1,2,3n = 1, 2, 3 into an2+bn+can^2 + bn + c.

  2. Work out the first differences

    T2−T1=3a+b,T3−T2=5a+bT_2 - T_1 = 3a + b, \qquad T_3 - T_2 = 5a + b

    Subtract consecutive terms.

  3. Work out the second difference

    (T3−T2)−(T2−T1)=(5a+b)−(3a+b)=2a(T_3 - T_2) - (T_2 - T_1) = (5a + b) - (3a + b) = 2a

    The bb terms cancel.

  4. Expand the second difference in terms of the T values

    T3−2T2+T1=2aT_3 - 2T_2 + T_1 = 2a

    Rearranging (T3−T2)−(T2−T1)(T_3 - T_2) - (T_2 - T_1) gives T1−2T2+T3T_1 - 2T_2 + T_3.

  5. Solve for a

    a=T1−2T2+T32a = \frac{T_1 - 2T_2 + T_3}{2}

    Divide the second difference by 22 -- the halving step.

  6. Check the coefficients directly

    T1−2T2+T3=(1−8+9)a+(1−4+3)b+(1−2+1)c=2aT_1 - 2T_2 + T_3 = (1 - 8 + 9)a + (1 - 4 + 3)b + (1 - 2 + 1)c = 2a

    The bb and cc coefficients are both zero, so only aa survives.

  7. Test with a known example

    un=n2:T1=1,T2=4,T3=9u_n = n^2: \quad T_1 = 1, T_2 = 4, T_3 = 9

    Here a=1a = 1, so the formula should give 11.

  8. Apply the formula to the example

    1−8+92=22=1✓\frac{1 - 8 + 9}{2} = \frac{2}{2} = 1 \quad \checkmark

    The formula returns a=1a = 1, as it should.

  9. Show the unhalved version fails

    T1−2T2+T3=2≠1T_1 - 2T_2 + T_3 = 2 \ne 1

    Forgetting to halve gives 22, which is 2a2a, not aa. This is the classic error.

  10. Show a difference-based distractor fails

    T3−T12=9−12=4≠1\frac{T_3 - T_1}{2} = \frac{9 - 1}{2} = 4 \ne 1

    This is not aa either.

  11. Show the first-difference distractor fails

    T2−T1=4−1=3≠1T_2 - T_1 = 4 - 1 = 3 \ne 1

    The first difference is 3a+b3a + b, not aa.

  12. Show the halved first difference fails

    T2−T12=32≠1\frac{T_2 - T_1}{2} = \frac{3}{2} \ne 1

    Halving the first difference is not the same as halving the second difference.

  13. Test the correct formula on a second example

    un=−2n2+n:T1=−1,T2=−6,T3=−15u_n = -2n^2 + n: \quad T_1 = -1, T_2 = -6, T_3 = -15

    Here a=−2a = -2.

  14. Apply the formula again

    −1−2(−6)+(−15)2=−1+12−152=−42=−2✓\frac{-1 - 2(-6) + (-15)}{2} = \frac{-1 + 12 - 15}{2} = \frac{-4}{2} = -2 \quad \checkmark

    It works for a negative aa too.

  15. State the correct expression

    a=T1−2T2+T32a = \frac{T_1 - 2T_2 + T_3}{2}

    This is half the second difference, written directly in terms of the first three terms.

Answer
a=T1−2T2+T32a = \frac{T_1 - 2T_2 + T_3}{2}
Question 2
5 markschallenging
A geometric sequence has all positive terms. Its second term is 232\sqrt{3} and its fourth term is 636\sqrt{3}. Work out the 66th term. Give your answer in surd form.
Show worked solution

Worked solution

  1. Write down what you know

    u2=23,u4=63u_2 = 2\sqrt{3}, \qquad u_4 = 6\sqrt{3}

    All the terms are positive.

  2. Relate the two terms

    u4=u2×r2u_4 = u_2 \times r^2

    Two applications of the ratio take you from the 2nd term to the 4th.

  3. Form an equation

    23×r2=632\sqrt{3} \times r^2 = 6\sqrt{3}

    Substitute the given terms.

  4. Divide both sides

    r2=6323=3r^2 = \frac{6\sqrt{3}}{2\sqrt{3}} = 3

    The 3\sqrt{3} factors cancel exactly, leaving 62=3\frac{6}{2} = 3.

  5. Take the square root

    r=±3r = \pm\sqrt{3}

    Both signs square to 33.

  6. Use the positive-terms condition

    r=3r = \sqrt{3}

    A negative ratio would make the terms alternate in sign.

  7. Find the 6th term from the 4th

    u6=u4×r2=63×3u_6 = u_4 \times r^2 = 6\sqrt{3} \times 3

    Two more applications of the ratio, and r2=3r^2 = 3.

  8. Multiply

    63×3=1836\sqrt{3} \times 3 = 18\sqrt{3}

    Multiply the rational part; the surd part is unchanged.

  9. Find the first term as a check

    u1=u2r=233=2u_1 = \frac{u_2}{r} = \frac{2\sqrt{3}}{\sqrt{3}} = 2

    The first term is 22.

  10. Write the whole sequence out

    2, 23, 6, 63, 18, 1832,\ 2\sqrt{3},\ 6,\ 6\sqrt{3},\ 18,\ 18\sqrt{3}

    Each term is 3\sqrt{3} times the one before it.

  11. Check the 4th term

    u4=2×(3)3=2×33=63u_4 = 2 \times (\sqrt{3})^{3} = 2 \times 3\sqrt{3} = 6\sqrt{3}

    Matches the given 4th term.

  12. Check the 6th term directly

    u6=2×(3)5=2×93=183u_6 = 2 \times (\sqrt{3})^{5} = 2 \times 9\sqrt{3} = 18\sqrt{3}

    Since (3)4=9(\sqrt{3})^4 = 9, (3)5=93(\sqrt{3})^5 = 9\sqrt{3}. This agrees with the earlier working.

  13. Note why exactness matters

    (3)2=3 exactly(\sqrt{3})^2 = 3 \text{ exactly}

    Using 1.7321.732 throughout would give 17.99…17.99\ldots instead of exactly 18318\sqrt{3}.

  14. Note the alternating structure

    odd positions are integers, even positions carry 3\text{odd positions are integers, even positions carry } \sqrt{3}

    This is a useful check on any answer.

  15. State the 6th term

    18318\sqrt{3}

    The 6th term is 18318\sqrt{3}.

Answer
18318\sqrt{3}
Question 3
5 markschallenging
The nnth term of a quadratic sequence is n2+kn+5n^2 + kn + 5, where kk is a whole number. The 44th term of the sequence is 3333. Work out the 1010th term.
Show worked solution

Worked solution

  1. Write down the rule

    un=n2+kn+5u_n = n^2 + kn + 5

    Only kk is unknown.

  2. Substitute n equals 4

    u4=42+4k+5u_4 = 4^2 + 4k + 5

    The 4th term uses n=4n = 4.

  3. Simplify

    u4=16+4k+5=21+4ku_4 = 16 + 4k + 5 = 21 + 4k

    Collect the numbers.

  4. Form an equation

    21+4k=3321 + 4k = 33

    The 4th term is given as 3333.

  5. Solve for k

    4k=12⇒k=34k = 12 \quad \Rightarrow \quad k = 3

    Subtract 2121, then divide by 44.

  6. Write the complete rule

    un=n2+3n+5u_n = n^2 + 3n + 5

    Now every term can be found.

  7. Check the 4th term

    42+3×4+5=16+12+5=334^2 + 3 \times 4 + 5 = 16 + 12 + 5 = 33

    The rule reproduces the given term, so k=3k = 3 is right.

  8. List the first few terms as a sense check

    9, 15, 23, 33, 459,\ 15,\ 23,\ 33,\ 45

    Substituting n=1n = 1 to 55 gives these values.

  9. Check the second difference

    1st:6, 8, 10, 122nd:2, 2, 2\text{1st}: 6,\ 8,\ 10,\ 12 \qquad \text{2nd}: 2,\ 2,\ 2

    Constant at 2=2a2 = 2a with a=1a = 1, exactly as the rule predicts.

  10. Substitute n equals 10

    u10=102+3×10+5u_{10} = 10^2 + 3 \times 10 + 5

    Now work out the 10th term.

  11. Square first

    100+3×10+5100 + 3 \times 10 + 5

    Indices come before multiplication.

  12. Multiply

    100+30+5100 + 30 + 5

    Then 3×10=303 \times 10 = 30.

  13. Add

    135135

    The 10th term is 135135.

  14. Confirm k was needed

    k=3k = 3

    Without finding kk first, the 10th term could not be worked out.

  15. State the 10th term

    135135

    The 10th term of the sequence is 135135.

Answer
135135
Question 4
6 markschallenging
Here are the first five terms of a quadratic sequence. 4, 9, 15, 22, 30, …4,\ 9,\ 15,\ 22,\ 30,\ \ldots Find the nnth term of the sequence and use it to work out the 2020th term.
Show worked solution

Worked solution

  1. Write out the terms

    4, 9, 15, 22, 304,\ 9,\ 15,\ 22,\ 30

    These are the first 5 terms of the sequence.

  2. Work out the first differences

    5, 6, 7, 85,\ 6,\ 7,\ 8

    Subtract each term from the one after it. The first differences are not constant, so the sequence is not linear.

  3. Work out the second differences

    1, 1, 11,\ 1,\ 1

    Now take the differences of the first differences.

  4. Confirm the sequence is quadratic

    second difference=1\text{second difference} = 1

    The second difference is constant and not zero, so the sequence is quadratic: its nnth term has the form an2+bn+can^2 + bn + c.

  5. Halve the second difference to get the coefficient of n squared

    a=12=12a = \frac{1}{2} = \frac{1}{2}

    The second difference of an2+bn+can^2+bn+c is always 2a2a, so aa is HALF the second difference. Forgetting to halve is the classic error here.

  6. Write out a times n squared for each position

    12, 2, 92, 8, 252\frac{1}{2},\ 2,\ \frac{9}{2},\ 8,\ \frac{25}{2}

    These are the values of n22\frac{n^{2}}{2} for n=1,2,3,…n = 1, 2, 3, \ldots

  7. Subtract a times n squared from each term

    72, 7, 212, 14, 352\frac{7}{2},\ 7,\ \frac{21}{2},\ 14,\ \frac{35}{2}

    What is left is the linear part bn+cbn + c of the rule.

  8. Find the linear part

    bn+c:b=72, c=0bn + c: \quad b = \frac{7}{2},\ c = 0

    The leftover sequence goes up by 72\frac{7}{2} each time, so b=72b = \frac{7}{2}. Its first value is 72\frac{7}{2}, and 72+c=72\frac{7}{2} + c = \frac{7}{2} gives c=0c = 0.

  9. Write down the nth term

    n2+7n2\frac{n^{2} + 7 n}{2}

    Putting the quadratic part and the linear part together gives the rule for the nnth term.

  10. Check the rule against a given term

    n=5:52+7×52=30n = 5: \quad \frac{5^2 + 7 \times 5}{2} = 30

    Substituting n=5n = 5 regenerates the given term 3030, so the rule is right.

  11. Rewrite the rule as a single fraction

    n2+7n2\frac{n^2 + 7n}{2}

    Both coefficients have denominator 22, so combine them over one fraction.

  12. Factorise the numerator as a check

    n(n+7)2\frac{n(n + 7)}{2}

    This form makes the substitution quick and shows the rule is a whole number whenever nn and n+7n+7 are of opposite parity.

  13. Substitute n equals 20

    20×272\frac{20 \times 27}{2}

    Replace nn with 2020: 20+7=2720 + 7 = 27.

  14. Multiply out

    5402\frac{540}{2}

    20×27=54020 \times 27 = 540.

  15. Divide

    5402=270\frac{540}{2} = 270

    The 20th term is 270270. Working with exact halves throughout avoids any rounding error.

Answer
270270
Question 5
6 markschallenging
A sequence of dot patterns is made from rectangular blocks of dots. Pattern 11 has 55 dots, pattern 22 has 1212 dots, pattern 33 has 2121 dots and pattern 44 has 3232 dots. Find an expression for the number of dots in pattern nn, work out how many dots pattern 1212 has, and find which pattern has exactly 165165 dots.
Show worked solution

Worked solution

  1. Write out the number of dots in each pattern

    5, 12, 21, 325,\ 12,\ 21,\ 32

    Pattern nn is a rectangle of dots with nn columns and n+4n + 4 rows, so pattern 11 is 11 by 55, pattern 22 is 22 by 66, and so on.

  2. Work out the first differences

    7, 9, 117,\ 9,\ 11

    Subtract each term from the one after it. The first differences are not constant, so the sequence is not linear.

  3. Work out the second differences

    2, 22,\ 2

    Now take the differences of the first differences.

  4. Confirm the sequence is quadratic

    second difference=2\text{second difference} = 2

    The second difference is constant and not zero, so the sequence is quadratic: its nnth term has the form an2+bn+can^2 + bn + c.

  5. Halve the second difference to get the coefficient of n squared

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

    The second difference of an2+bn+can^2+bn+c is always 2a2a, so aa is HALF the second difference. Forgetting to halve is the classic error here.

  6. Write out a times n squared for each position

    1, 4, 9, 161,\ 4,\ 9,\ 16

    These are the values of n2n^{2} for n=1,2,3,…n = 1, 2, 3, \ldots

  7. Subtract a times n squared from each term

    4, 8, 12, 164,\ 8,\ 12,\ 16

    What is left is the linear part bn+cbn + c of the rule.

  8. Find the linear part

    bn+c:b=4, c=0bn + c: \quad b = 4,\ c = 0

    The leftover sequence goes up by 44 each time, so b=4b = 4. Its first value is 44, and 4+c=44 + c = 4 gives c=0c = 0.

  9. Write down the nth term

    n2+4nn^{2} + 4 n

    Putting the quadratic part and the linear part together gives the rule for the nnth term.

  10. Check the rule against pattern 4

    n=4:42+4×4=16+16=32n = 4: \quad 4^2 + 4 \times 4 = 16 + 16 = 32

    Pattern 44 has 3232 dots, matching the given information.

  11. Work out the number of dots in pattern 12

    n=12:122+4×12=144+48=192n = 12: \quad 12^2 + 4 \times 12 = 144 + 48 = 192

    Pattern 1212 has 192192 dots.

  12. Check the rule against the picture

    n(n+4)=n2+4nn(n + 4) = n^2 + 4n

    Pattern nn is an nn by (n+4)(n+4) rectangle, so the number of dots is n(n+4)n(n+4), which expands to the rule found from the differences.

  13. Set the rule equal to 165

    n2+4n=165n^2 + 4n = 165

    Now find which pattern uses 165165 dots.

  14. Rearrange and factorise

    n2+4n−165=0⇒(n+15)(n−11)=0n^2 + 4n - 165 = 0 \quad \Rightarrow \quad (n + 15)(n - 11) = 0

    Two numbers multiplying to −165-165 and adding to 44 are 1515 and −11-11.

  15. Solve and check

    n=11:112+4×11=121+44=165n = 11: \quad 11^2 + 4 \times 11 = 121 + 44 = 165

    The pattern number n=−15n = -15 is rejected, so pattern 1111 uses 165165 dots. It is an 1111 by 1515 rectangle of dots, and 11×15=16511 \times 15 = 165.

Answer
n=11n = 11

Unlock 29 more nth term of quadratic sequences questions

Create a free account to work through every GCSE nth term of quadratic sequences question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More nth term of quadratic sequences practice

Practice QuestionsWorked Solutions

Related Algebra topics

Algebraic fractionsEquation of a circleCoordinatesCubic and reciprocal graphsExponential and trig graphsExpanding binomials