Printable · GCSE Foundation · ages 14-16
The nth term of linear and quadratic sequences worksheet — GCSE Foundation
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.
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The nth term of linear and quadratic sequences worksheet — GCSE Foundation
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- 1.The first five terms of a sequence are 6, 10, 14, 18, 22. Work out an expression, in terms of n, for the nth term.
- 2.The first four terms of a sequence are 4, 9, 14, 19. Work out an expression, in terms of n, for the nth term.
- 3.Aisha saves money each week. In week 1 she has saved £15 in total, in week 2 she has saved £23 in total, in week 3 she has saved £31 in total and in week 4 she has saved £39 in total, with the total increasing by the same amount each week. Work out an expression, in terms of n, for the total amount she has saved, in pounds, after week n.
- 4.A car park charges by the hour. Parking for 1 hour costs £5, 2 hours costs £9, 3 hours costs £13 and 4 hours costs £17, with the cost increasing by the same amount for each extra hour. Work out an expression, in terms of n, for the cost, in pounds, of parking for n hours.
- 5.The first four terms of a sequence are 9, 14, 19, 24. Work out an expression, in terms of n, for the nth term.
- 6.The first four terms of a sequence are 18, 15, 12, 9. Work out an expression, in terms of n, for the nth term.
- 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.The first four terms of a sequence are 4, 6, 8, 10. Work out an expression, in terms of n, for the nth term.
- 10.The first five terms of a sequence are 2, 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
- 11.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.
- 12.A pattern is made from tiles. Pattern 1 has 7 tiles, pattern 2 has 11 tiles, pattern 3 has 15 tiles, and pattern 4 has 19 tiles, with each new pattern having 4 more tiles than the one before it. Work out an expression, in terms of n, for the number of tiles in pattern n.
- 13.A coach company charges a booking fee plus a price for each passenger. A booking for 1 passenger costs £9, for 2 passengers costs £15, for 3 passengers costs £21, and for 4 passengers costs £27. Work out an expression, in terms of n, for the cost in pounds of a booking for n passengers.
- 14.A pattern is made from tiles. Pattern 1 uses 4 tiles, pattern 2 uses 7 tiles, pattern 3 uses 10 tiles and pattern 4 uses 13 tiles, with each pattern using 3 more tiles than the one before. Work out an expression, in terms of n, for the number of tiles used in pattern n.
- 15.The first four terms of a sequence are 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
Answer key
- (d) 4n + 2 — Method: find the common difference, then find the constant by adjusting the first term. Working: 10 − 6 = 4, 14 − 10 = 4, so the common difference is 4 and the coefficient of n is 4. The constant is the first term minus the common difference: 6 − 4 = 2. Answer: the nth term is 4n + 2. 4n + 6 comes from using the first term, 6, as the constant without subtracting the common difference. 4n − 2 comes from working out the constant the wrong way round, as the common difference minus the first term (4 − 6 = −2) instead of the first term minus the common difference. 6n + 4 comes from swapping the common difference and the first term.
- (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).
- (c) 8n + 7 — Method: find the weekly increase, then find the constant by adjusting the week 1 total. Working: the total goes up by £8 each week (23 − 15 = 8), so the coefficient of n is 8. The constant is the week 1 total minus the common difference: 15 − 8 = 7. Answer: the nth term is 8n + 7. 8n + 15 comes from using the week 1 total, 15, as the constant without subtracting the common difference. 8n − 1 comes from a slip in working out the constant, subtracting the common difference twice (15 − 8 − 8 = −1) instead of once. 7n + 8 comes from swapping the weekly increase and the constant.
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (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.
- (d) 21 − 3n — Method: find the common difference, then find the constant that fits the first term. Working: 15 − 18 = −3, 12 − 15 = −3, 9 − 12 = −3, so the terms decrease by 3 each time and the nth term has the form c − 3n. Substituting n = 1: c − 3(1) = 18, so c = 21. Answer: the nth term is 21 − 3n. The value 18 − 3n comes from using the first term as c directly, without adding back the difference that was removed. The value 3n − 21 comes from a sign error that flips the whole expression. The value −3n comes from using only the common difference and leaving out the constant.
- (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.
- (c) 2n + 2 — Method: find the common difference between consecutive terms, then find the constant that fits the first term. Working: 6 − 4 = 2, 8 − 6 = 2, 10 − 8 = 2, so the terms increase by 2 each time and the nth term has the form 2n + c. Substituting n = 1: 2(1) + c = 4, so c = 2. Answer: the nth term is 2n + 2. The value 2n comes from using only the common difference and leaving out the constant c entirely. The value 2n + 4 comes from using the first term itself as c, without subtracting the common difference to find the true constant. The value 2n − 2 comes from a sign error when working out c, giving −2 instead of +2.
- (a) 3n − 1 — Method: find the common difference between consecutive terms, then find the constant by adjusting the first term. Working: 5 − 2 = 3, 8 − 5 = 3, 11 − 8 = 3, so the common difference is 3 and the coefficient of n is 3. The constant is the first term minus the common difference: 2 − 3 = −1. Answer: the nth term is 3n − 1. 3n + 2 comes from using the first term, 2, as the constant without subtracting the common difference. 3n − 2 comes from a slip when working out the constant, treating 2 − 3 as −2 instead of −1. 2n + 3 comes from swapping the common difference and the first term.
- (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.
- (c) 4n + 3 — Method: use the given number of extra tiles per pattern as the coefficient of n, then find the constant that fits pattern 1. Working: each pattern adds 4 tiles, so the expression has the form 4n + c. Substituting n = 1: 4(1) + c = 7, so c = 3. Answer: the number of tiles in pattern n is 4n + 3. The value 4n comes from leaving out the constant c altogether. The value 4n + 7 comes from using pattern 1's total, 7, as the constant directly instead of subtracting the difference first. The value 7n + 4 comes from swapping the roles of the number of tiles in pattern 1 and the number added each time.
- (d) 6n + 3 — Method: find the price per passenger, then find the booking fee that fits a booking for 1 passenger. Working: the cost rises by £6 for each extra passenger (15 − 9 = 6, 21 − 15 = 6, 27 − 21 = 6), so the cost has the form 6n + c. Substituting n = 1: 6(1) + c = 9, so c = 3. Answer: the cost in pounds is 6n + 3. 6n ignores the booking fee altogether. 6n + 9 uses the cost of one passenger, £9, as the fee without first subtracting the £6 per-passenger price. 9n + 6 swaps the two numbers, using the first cost as the price per passenger and the difference as the fee.
- (b) 3n + 1 — Method: find how many more tiles each pattern uses, then find the constant by adjusting pattern 1's total. Working: each pattern uses 3 more tiles than the last, so the coefficient of n is 3. The constant is pattern 1's total minus the common difference: 4 − 3 = 1. Answer: the nth term is 3n + 1. 3n + 4 comes from using pattern 1's total, 4, as the constant without subtracting the common difference. 3n − 2 comes from a slip in working out the constant, subtracting the common difference twice (4 − 3 − 3 = −2) instead of once. n + 3 comes from swapping the common difference and the constant.
- (d) 3n + 2 — The common difference is 3 (8−5=3), so the expression starts 3n. To match the first term when n=1, 3×1+c=5, so c=2: the nth term is 3n+2. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 3n+5 (giving 8, 11, 14, 17 — one term too high throughout). A candidate who omits the constant term altogether would write just 3n (giving 3, 6, 9, 12, not matching the sequence at all). A candidate who adds the common difference to n instead of multiplying would write n+3 (giving 4, 5, 6, 7, far too small).
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