4.5 Arithmetic sequences and series — Practise

Work through these on your own. Write $a$ and $d$ on the page before you reach for a formula. Read the question again to decide whether it wants one term or a total.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

An arithmetic sequence begins $7, 11, 15, \dots$ . Work out the 15th term.
An arithmetic sequence begins $100, 92, 84, \dots$ . Work out the 20th term.
Work out the sum of the first 20 terms of $5, 9, 13, \dots$ .
For the series $3 + 7 + 11 + \dots + 83$ , work out how many terms there are and then the total.
Work out the sum of the first 30 terms of $2, 6, 10, \dots$ .
The 8th term of an arithmetic sequence is $45$ and the 12th term is $65$ . Find the first term and the common difference.
An arithmetic sequence begins $40, 36, 32, \dots$ . Find the first term that is negative.
An arithmetic series has first term $3$ and common difference $7$ . Work out the sum of the first 12 terms.
Work out the sum of the multiples of $4$ from $12$ to $96$ inclusive.
The quiz night raises $150$ pounds in month 1 and $50$ more each month. Work out the total raised in the first 18 months.
A savings club pays in £200 in month 1 and £15 more each month after that. Work out the total paid in over 24 months.
The sum of the first $n$ terms of an arithmetic series is $S_n = n^2 + 3n$ . Work out the 5th term and the common difference.

Answers and concise solutions

1. The term

Answer: Here $a = 7$ and $d = 4$ . Then $u_{15} = 7 + (15 - 1)(4) = 7 + 56 = 63$ . The number of steps is $14$ , one less than the position.

2. The term

Answer: The sequence falls by $8$ each step, so $d = -8$ . Then $u_{20} = 100 + (20 - 1)(-8) = 100 - 152 = -52$ . A negative term is expected because the sequence passes zero after the 13th term.

3. The sum

Answer: Here $a = 5$ and $d = 4$ . Then $S_{20} = \frac{20}{2}(2(5) + 19(4)) = 10(10 + 76) = 10(86) = 860$ . The bracket uses the $19$ steps from the first term to the twentieth.

4. The number of terms and the total

Answer: Set the term equal to the last term: $3 + (n - 1)(4) = 83$ , so $4(n - 1) = 80$ and $n = 21$ . There are 21 terms. Then $S_{21} = \frac{21}{2}(3 + 83) = \frac{21}{2}(86) = 903$ . The negative root of any quadratic in $n$ is rejected, because a number of terms must be a positive whole number.

5. The sum

Answer: Here $a = 2$ and $d = 4$ . Then $S_{30} = \frac{30}{2}(2(2) + 29(4)) = 15(4 + 116) = 15(120) = 1800$ . The sixteenth term is $62$ , so the average term is about $60$ and the total is about $30 \times 60$ , which matches.

6. The first term and the common difference

Answer: Write two equations: $a + 7d = 45$ and $a + 11d = 65$ . Subtracting gives $4d = 20$ , so $d = 5$ . Substituting back gives $a + 35 = 45$ , so $a = 10$ . Check the 12th term: $10 + 11(5) = 65$ .

7. The first negative term

Answer: Here $a = 40$ and $d = -4$ , so $u_n = 40 - 4(n - 1) = 44 - 4n$ . The eleventh term is $44 - 44 = 0$ , which is not negative. The twelfth term is $44 - 48 = -4$ , so the first negative term is the 12th, $u_{12} = -4$ .

8. The sum

Answer: Substitute into $S_n = \frac{n}{2}(2a + (n - 1)d)$ : $S_{12} = \frac{12}{2}(2(3) + 11(7)) = 6(6 + 77) = 6(83) = 498$ . The last term is $3 + 11(7) = 80$ , so the average term is about $41$ and the total is about $12 \times 41$ , which agrees.

9. The sum

Answer: The multiples form an arithmetic series with $a = 12$ , $d = 4$ and last term $96$ . From $12 + (n - 1)(4) = 96$ we get $n - 1 = 21$ , so $n = 22$ . Then $S_{22} = \frac{22}{2}(12 + 96) = 11 \times 108 = 1188$ . There are 22 terms, because both ends are included.

10. The total

Answer: Here $a = 150$ and $d = 50$ . Then $S_{18} = \frac{18}{2}(2(150) + 17(50)) = 9(300 + 850) = 9(1150) = 10{,}350$ . The quiz night alone raises £10,350 in 18 months, which is more than the £9,000 the roof costs. The month 18 income is $150 + 17(50) = 1000$ pounds.

11. The total

Answer: The payments form an arithmetic series with $a = 200$ and $d = 15$ . Then $S_{24} = \frac{24}{2}(2(200) + 23(15)) = 12(400 + 345) = 12(745) = 8940$ . The club pays in £8940 over two years. The final payment is $200 + 23(15) = 545$ pounds.

12. The term and the difference

Answer: The $n$ th term is the difference between consecutive sums: $u_n = S_n - S_{n-1}$ . For $n = 5$ , $u_5 = S_5 - S_4 = 40 - 28 = 12$ . The common difference is $u_6 - u_5 = (S_6 - S_5) - (S_5 - S_4) = 54 - 2(40) + 28 = 2$ . Working with differences of sums turns a sum formula into term information.