Printable · GCSE Higher · ages 14-16
The nth term of linear and quadratic sequences worksheet — GCSE Higher
Fifteen questions on "the nth term of linear and quadratic sequences" — DfE statement A25. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
The nth term of linear and quadratic sequences worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.The first five terms of a quadratic sequence are 4, 7, 12, 19, 28. Work out an expression, in terms of n, for the nth term.
- 2.A square patio of side length n slabs is surrounded by a single border of square paving slabs of the same size. For a patio with side length n, the total number of slabs used for the patio and its border together is 9 when n = 1, 16 when n = 2, 25 when n = 3, and 36 when n = 4. Work out an expression, in terms of n, for the total number of slabs.
- 3.The nth term of a sequence is n² + 2n − 4. Work out the 7th term.
- 4.A theatre has rows of seats arranged so that row 1 has 12 seats, row 2 has 17 seats, row 3 has 22 seats and row 4 has 27 seats, with each row having 5 more seats than the row before. Work out an expression, in terms of n, for the number of seats in row n.
- 5.The nth term of a sequence is 3n² + 2n − 1. Work out the 10th term.
- 6.A market trader builds a display of oranges in a rectangular block. The block is n oranges wide and 4 oranges longer than it is wide. The number of oranges needed is 5 when n = 1, 12 when n = 2, 21 when n = 3, 32 when n = 4, and 45 when n = 5. Work out an expression, in terms of n, for the number of oranges needed for a display of width n.
- 7.A gym charges a reduced price for the first month of membership and a standard price for every month after that. The total cost is £5 for 1 month, £12 for 2 months, £19 for 3 months and £26 for 4 months. Work out an expression, in terms of n, for the total cost, in pounds, of n months of membership.
- 8.The nth term of a sequence is 4n + 2. A term in the sequence has value 30. Work out its position, n, in the sequence.
- 9.A hall is set out in rows of chairs for an assembly. Row 1 has 50 chairs, row 2 has 45 chairs, row 3 has 40 chairs, and row 4 has 35 chairs, with each row after the first having 5 fewer chairs than the one before. Work out an expression, in terms of n, for the number of chairs in row n.
- 10.The first five terms of a quadratic sequence are 6, 11, 18, 27, 38. Work out an expression, in terms of n, for the nth term.
- 11.The first five terms of a quadratic sequence are −1, 4, 13, 26, 43. Work out an expression, in terms of n, for the nth term.
- 12.The first four terms of a sequence are 4, 9, 14, 19. Work out an expression, in terms of n, for the nth term.
- 13.The first five terms of a sequence are 7, 9, 13, 19, 27. By finding the second difference, work out the coefficient of n² in the nth term.
- 14.A company charges for hiring chairs for a day. Hiring 1 chair costs £12, hiring 2 chairs costs £19, hiring 3 chairs costs £26, and hiring 4 chairs costs £33. Work out an expression, in terms of n, for the cost in pounds of hiring n chairs for a day.
- 15.The first four terms of a sequence are 9, 14, 19, 24. Work out an expression, in terms of n, for the nth term.
Answer key
- (d) n² + 3 — First differences: 3, 5, 7, 9. Second differences: 2, 2, 2, so the sequence is quadratic and the coefficient of n² is half the second difference: a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the terms (4, 7, 12, 19, 28) leaves 3, 3, 3, 3, 3, a constant, so the nth term is n² + 3. Using the second difference itself as a, without halving it, gives 2n² + 3. Finding a = 1 correctly but then dropping the constant remainder gives n². Treating the first first difference (3) as a common difference and building a linear formula a + (n − 1)d = 4 + 3(n − 1) gives 3n + 1, which fits only the first term.
- (a) n² + 4n + 4 — The whole patio-plus-border square has side length (n + 2), so the total is (n + 2)². Expanding this bracket correctly gives n² + 4n + 4, which matches 9, 16, 25, 36 for n = 1, 2, 3, 4. Expanding (n + 2)² by squaring each term separately, as if (a + b)² = a² + b², gives n² + 4, which is already wrong at n = 1 (it gives 5, not 9). Multiplying out (n + 2)(n + 2) as n² + 2n + 2n but forgetting the final 2 × 2 gives n² + 4n, which is 5 short at every value of n. Counting the patio twice — once as the n² inner square and again inside the (n + 2)² total — gives n² + (n² + 4n + 4) = 2n² + 4n + 4.
- (c) 59 — Substitute n = 7: 7² + 2 × 7 − 4 = 49 + 14 − 4 = 59. A sign error on the +2n term, treating it as −2n, gives 49 − 14 − 4 = 31. Working out 7² + 2 × 7 but forgetting to subtract the final 4 gives 49 + 14 = 63. Using n = 6 instead of n = 7 gives 36 + 12 − 4 = 44.
- (c) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (b) 319 — Substitute n = 10: 3 × 10² + 2 × 10 − 1 = 3 × 100 + 20 − 1 = 300 + 20 − 1 = 319. A sign error on the +2n term, treating it as −2n, gives 300 − 20 − 1 = 279. Working out 3 × 10² + 2 × 10 but forgetting to subtract the final 1 gives 300 + 20 = 320. Using n = 9 instead of n = 10 gives 3 × 81 + 18 − 1 = 243 + 18 − 1 = 260.
- (b) n² + 4n — First differences of 5, 12, 21, 32, 45 are 7, 9, 11, 13. Second differences are 2, 2, 2, so a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the totals (5, 12, 21, 32, 45) leaves 4, 8, 12, 16, 20, the linear expression 4n. So the number of oranges is n² + 4n. Using the second difference itself as the coefficient of n², without halving it, gives 2n² + 4n. Finding a = 1 correctly but dropping the linear remainder 4n entirely leaves n². Treating the first first difference (7) as a constant common difference and building a linear formula 5 + 7(n − 1) = 7n − 2 fits only the first two totals, and gives 19 for n = 3 instead of 21.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (c) 7 — To reverse the rule, subtract the constant then divide by the coefficient: 30−2=28, then 28÷4=7, so n=7. A candidate who adds the constant instead of subtracting it, a sign error when rearranging, would compute (30+2)÷4=32÷4=8. A candidate who subtracts the constant correctly but then forgets to divide by the coefficient would stop at 30−2=28. A candidate who treats 4n+2 as a single term 6n, adding the coefficient and constant together, would compute 30÷6=5.
- (a) 55 − 5n — The number of chairs decreases by 5 in each row after the first, so the common difference is d=−5, and the first term is a=50. The nth term is a+(n−1)d = 50+(n−1)(−5) = 50−5n+5 = 55−5n. A candidate who uses the common difference as the constant term instead of correctly finding 55, giving the constant as −5 instead, would write −5n−5. A candidate who uses the first term, 50, as the coefficient of n instead of the common difference, would write 50n−5. A candidate who does not multiply the common difference by n at all, treating the nth term as n+d instead of dn+c, would write n−5.
- (b) n² + 2n + 3 — First differences: 5, 7, 9, 11. Second differences: 2, 2, 2, so a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the terms (6, 11, 18, 27, 38) leaves 5, 7, 9, 11, 13, which is the linear expression 2n + 3. So the nth term is n² + 2n + 3. Using the second difference itself as a, without halving it, gives 2n² + 2n + 3. Finding a = 1 correctly but then dropping the linear part 2n, keeping only the constant, gives n² + 3. Finding a = 1 correctly but dropping the constant +3 gives n² + 2n.
- (d) 2n² − n − 2 — First differences: 5, 9, 13, 17. Second differences: 4, 4, 4, so a = 4 ÷ 2 = 2. Subtracting 2n² (2, 8, 18, 32, 50) from the terms (−1, 4, 13, 26, 43) leaves −3, −4, −5, −6, −7, which is the linear expression −n − 2. So the nth term is 2n² − n − 2. Using the second difference itself as a, without halving it, gives 4n² − n − 2. Finding a = 2 correctly but dropping the linear remainder −n − 2 entirely leaves 2n². Treating the first first difference (5) as a common difference and building a + (n − 1)d = −1 + 5(n − 1) gives 5n − 6, which only matches the first two terms.
- (d) 5n − 1 — The common difference is 5 (9−4=5), so the expression starts 5n. To match the first term when n=1, 5×1+c=4, so c=−1: the nth term is 5n−1. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 5n+4 (giving 9, 14, 19, 24 — one term too high throughout). A candidate who omits the constant term altogether would write just 5n (giving 5, 10, 15, 20, not matching the sequence). A candidate who adds the common difference to n instead of multiplying would write n+5 (giving 6, 7, 8, 9, far too small).
- (b) 1 — First differences: 2, 4, 6, 8. Second differences: 2, 2, 2, a constant, confirming the sequence is quadratic. The coefficient of n² is half the constant second difference: 2 ÷ 2 = 1. Using the second difference itself as the coefficient, without halving it, gives 2. Halving twice, dividing the second difference by 4 instead of by 2, gives 0.5. Doubling the second difference instead of halving it, on the mistaken rule that the n² coefficient is twice the second difference, gives 2 × 2 = 4.
- (d) 7n + 5 — Method: find the rate charged per extra chair, then find the fixed part of the cost that fits hiring 1 chair. Working: the cost rises by £7 for each extra chair (19 − 12 = 7, 26 − 19 = 7, 33 − 26 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 12, so c = 5. Answer: the cost in pounds is 7n + 5. The value 7n comes from ignoring the fixed part of the charge entirely. The value 7n + 12 comes from using the cost of 1 chair as the fixed part directly, without subtracting the per-chair rate first. The value 12n + 7 comes from swapping the roles of the cost of hiring 1 chair, £12, and the rate per extra chair, £7 — using the total for one chair as the coefficient of n and the rate as the fixed part.
- (d) 5n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 14 − 9 = 5, 19 − 14 = 5, 24 − 19 = 5, so the terms increase by 5 each time and the nth term has the form 5n + c. Substituting n = 1: 5(1) + c = 9, so c = 4. Answer: the nth term is 5n + 4. The value 5n comes from leaving out the constant. The value 5n + 9 comes from using the first term as the constant directly, without subtracting the common difference first. The value 9n + 5 comes from swapping the roles of the first term and the common difference — using the first term, 9, as the coefficient of n and the difference, 5, as the constant.
Build your own mix at the worksheet builder.