Questions on arithmetic series

Question 1

An arithmetic sequence has first term 7 and common difference 5. Find the 15th term.

Accepted Answer

u₁₅ = 7 + 14 × 5 = 77. M1 for a + 14d, A1 for 77. The index in a + (n − 1)d is n − 1, not n: 7 + 15 × 5 = 82 is the value of the sixteenth term.

Question 2

Find the nth term of the sequence 4, 11, 18, …, and determine whether 200 is a term.

Accepted Answer

a = 4 and d = 7, so u_n = 4 + 7(n − 1) = 7n − 3. Setting 7n − 3 = 200 gives n = 29, a positive whole number, so 200 is the 29th term. B1 for u_n = 7n − 3, M1 for setting it equal to 200, A1 for the 29th term. A fractional n would have settled it the other way. Simplifying to 7n − 3 rather than leaving 4 + 7(n − 1) is expected.

Question 3

The 6th term of an arithmetic sequence is 23 and the common difference is 4. Find the first term.

Accepted Answer

Five steps run from the first term to the 6th, so a + 5 × 4 = 23 and a = 3. M1 for a + 5d = 23, A1 for a = 3. The index in a + (n − 1)d counts steps rather than terms, so six steps back from the 6th term is one step too far and reaches −1.

Question 4

Find the sum of all the positive multiples of 7 that are less than 200.

Accepted Answer

They run 7, 14, …, 196: arithmetic with a = 7 and d = 7. The last one is 7 × 28 = 196, because 7 × 29 = 203 has already passed 200, so there are 28 terms. M1 for counting the terms, A1 for 28. Then S₂₈ = 14 × (7 + 196) = 14 × 203 = 2842. A1 for 2842. As 200/7 is 28.57, it is the whole-number part that counts.

Question 5

The 4th term of an arithmetic sequence is 18 and the 9th term is 38. Find the first term and common difference, and hence the 25th term.

Accepted Answer

From the 4th term to the 9th is five steps, so 5d = 38 − 18 = 20 and d = 4. Then a + 3d = 18 gives a = 6, and u₂₅ = 6 + 24 × 4 = 102. M1 A1 for d = 4, A1 for a = 6, A1 for the 25th term. Counting the steps between two given terms is quicker than solving simultaneous equations.

Question 6

Find the sum of the first 30 terms of the series 8 + 11 + 14 + …

Accepted Answer

a = 8 and d = 3, so S₃₀ = 15 × (16 + 29 × 3) = 15 × 103 = 1545. Substitution in either standard form of the arithmetic-series formula gives 1545: the last-term form has l = 8 + 29 × 3 = 95 and 15 × (8 + 95) = 1545. M1 for the sum formula with a, d and n in place, A1 for the correct substitution, A1 for 1545.

Question 7

Prove that the sum of the first n natural numbers is n(n + 1)/2, and evaluate the sum of the first 200.

Accepted Answer

Write S = 1 + 2 + … + n forwards and backwards and add the two lines: each column pairs to n + 1, and there are n columns, so 2S = n(n + 1) and S = n(n + 1)/2. For n = 200: 200 × 201/2 = 20 100. M1 for reversing and adding, A1 for 2S = n(n + 1), A1 for the printed result, B1 for the value of the sum.

Question 8

A saver deposits £20 in the first month, then increases the deposit by £4 each month. Find the total saved after 24 months.

Accepted Answer

The deposits are arithmetic with a = 20 and d = 4, so S₂₄ = 12 × (40 + 23 × 4) = 12 × 132 = £1584. M1 for the sum formula with a = 20 and d = 4, A1 for the substitution, A1 for the total. This is a sum rather than a term, because every month's deposit stays in the pot. Answering £112, the 24th deposit alone, is the misreading the phrase total saved is there to prevent.

Question 9

The first three terms of an arithmetic sequence are 2k, 3k + 1 and k + 14, where k is a constant. Find the value of k, and hence find the sum of the first 20 terms.

Accepted Answer

Consecutive terms differ by the same amount, so (3k + 1) − 2k = (k + 14) − (3k + 1), that is k + 1 = 13 − 2k, giving 3k = 12 and k = 4. M1 for equating the two differences, A1 for k = 4. The terms are then 8, 13 and 18, so a = 8 and d = 5, and S₂₀ = 10 × (16 + 19 × 5) = 10 × 111 = 1110. M1 for the sum formula with a = 8 and d = 5, A1 for 1110. The equation is linear, so k = 4 is the only value that makes the three expressions arithmetic. Setting the middle term equal to the mean of the outer two, 2(3k + 1) = 2k + (k + 14), is the same equation written another way.

Question 10

A cyclist follows a training plan. In week 1 she rides 30 km, and in each week after that she rides 6 km further than in the week before. In the final week of the plan she rides 96 km. Find the number of weeks the plan lasts, and the total distance in kilometres she rides over the whole plan.

Accepted Answer

The weekly distances are arithmetic with a = 30 and d = 6, so 30 + (n − 1)6 = 96 gives (n − 1)6 = 66 and n = 12: the plan lasts 12 weeks. M1 for 30 + (n − 1)6 = 96, A1 for 12 weeks. With the first and last distances both known, the last-term form of the sum is quickest: S₁₂ = 6 × (30 + 96) = 756 km. M1 for the sum formula, A1 for 756. Dividing 66 by 6 gives 11, the number of steps, and week 1 is added to that because it is a term rather than a step.

Question 11

Evaluate Σ(4r + 1) for r running from 8 to 30.

Accepted Answer

The terms are arithmetic. At r = 8 the term is 33 and at r = 30 it is 121. M1 for the first and last terms, 33 and 121. The counter takes every whole value between, so there are 30 − 8 + 1 = 23 terms, not 30 − 8: both ends are included. B1 for 23 terms. Then the sum is (23/2)(33 + 121) = 23 × 77 = 1771. M1 for the sum formula with those three numbers, A1 for 1771. Subtracting two sums that both start at r = 1 gives the same total, 1890 − 119 = 1771: the terms from r = 1 to r = 7 are exactly the ones the shifted counter leaves out.

Question 12

Find the smallest number of terms of the series 5 + 9 + 13 + … whose sum exceeds 1000.

Accepted Answer

S_n = n(2 × 5 + 4(n − 1))/2 = n(2n + 3) must exceed 1000. Trying the boundary: S₂₁ = 945, too small, and S₂₂ = 1034, past the mark. So 22 terms are needed. M1 for S_n in terms of n, M1 for comparing it with 1000, A1 for locating the boundary, A1 for 22 terms. Solving the quadratic 2n² + 3n − 1000 > 0 gives n > 21.6, the same verdict; either route must end with whole terms, rounded up.

Question 13

An arithmetic series has first term a and common difference d, and S_n denotes the sum of its first n terms. Show that S_2n − 2S_n = n²d, and verify the result for the series 3 + 7 + 11 + … with n = 5.

Accepted Answer

Putting 2n terms into the sum formula, S_2n = (2n/2)(2a + (2n − 1)d) = n(2a + (2n − 1)d) = 2an + 2n²d − nd. M1 for the sum formula with 2n in place of n, A1 for 2an + 2n²d − nd. Doubling the n-term sum, 2S_n = n(2a + (n − 1)d) = 2an + n²d − nd. Subtracting, the 2an and the nd cancel in pairs and S_2n − 2S_n = 2n²d − n²d = n²d. M1 for subtracting the two expressions, A1 for n²d. For 3 + 7 + 11 + … the first term is 3 and the common difference is 4, so S₅ = (5/2)(6 + 4 × 4) = 55 and S₁₀ = 5(6 + 9 × 4) = 210, and 210 − 2 × 55 = 100, which is 5² × 4 as claimed. M1 for both sums, A1 for 100 matching n²d.

Question 14

A charity is offered two twelve-month donation schemes. Under scheme P the donor gives £50 in month 1 and £6 more in each month than in the month before. Under scheme Q the donor gives £85 in every month. Find the total given under each scheme over the twelve months, state which scheme gives more and by how much, and find the first month in which the monthly gift under scheme P is larger than the monthly gift under scheme Q.

Accepted Answer

Scheme P is arithmetic with a = 50 and d = 6, so its twelve-month total is S₁₂ = 6 × (100 + 11 × 6) = 6 × 166 = £996. M1 for the sum formula with a = 50 and d = 6, A1 for £996. Scheme Q gives 12 × 85 = £1020, so scheme Q gives more, by £24. B1 for £1020 with the comparison. The monthly gift under scheme P is 50 + 6(n − 1), which beats 85 when 6(n − 1) > 35, that is n > 6.83, so the first such month is month 7, at £86 against £85; in month 6 it is £80. M1 for 50 + 6(n − 1) > 85, A1 for month 7. The two comparisons point different ways: scheme P gives the larger monthly amount from month 7 onwards and is still £24 behind on the twelve-month total, because it spent the first six months below £85.

Question 15

An arithmetic series has first term 43 and common difference −5. Show that the sum of the first n terms is n(91 − 5n)/2, find the number of terms that gives the greatest sum, and find that greatest sum. Explain why taking more terms than that reduces the total.

Accepted Answer

S_n = (n/2)(2 × 43 + (n − 1)(−5)) = (n/2)(86 − 5n + 5) = n(91 − 5n)/2. M1 for the sum formula with a = 43 and d = −5, A1 for n(91 − 5n)/2. As a quadratic in n the expression has a negative n² coefficient, so it has a maximum, halfway between its roots 0 and 18.2, at n = 9.1. M1 for locating the turning point. Since n counts terms it has to be a whole number, and the two either side give S₉ = 9 × 46/2 = 207 and S₁₀ = 10 × 41/2 = 205, with S₈ = 8 × 51/2 = 204 behind as well. So 9 terms give the greatest sum, 207. A1 for 9 terms, A1 for 207. The nth term is 43 − 5(n − 1) = 48 − 5n, which is 3 at n = 9 and −2 at n = 10, so every term from the 10th onwards is negative and each one taken lowers the running total. B1 for the terms turning negative after the 9th.