Arithmetic series Problem set A Worked solutions

Find each sum.

1. (a) 4 + 9 + 14 + … (25 terms)

   • \(S_n = \frac{n}{2}(2u_1 + (n - 1)d)\)
   • \(u_1 = 4,\ d = 5,\ n = 25: S_{25} = \frac{25}{2}(2(4) + 24(5))\)
   • \(S_{25} = 1600\)

2. (b) 60 + 54 + 48 + … (30 terms)

   • \(u_1 = 60,\ d = -6,\ n = 30: S_{30} = \frac{30}{2}(2(60) + 29(-6))\)
   • \(S_{30} = -810\)

3. (c) The first 50 odd numbers

   • \(1, 3, 5, \ldots:\ u_1 = 1,\ d = 2,\ n = 50\)
   • \(S_{50} = \frac{50}{2}(2(1) + 49(2))\)
   • \(S_{50} = 2500\)

An arithmetic series has first term 2, and the sum of its first 10 terms is 155. Find the common difference.

• \(S_{10} = \frac{10}{2}(2(2) + 9d) = 155\)
• \(5(4 + 9d) = 155 \Rightarrow 4 + 9d = 31\)
• \(9d = 27 \Rightarrow d = 3\)

Find each sum.

1. (a) 12 terms, first term 5, last term 71

   • \(S_n = \frac{n}{2}(u_1 + u_n)\)
   • \(S_{12} = \frac{12}{2}(5 + 71) = 456\)

2. (b) 3 + 7 + 11 + … + 99

   • \(u_1 = 3,\ d = 4:\ 3 + (n - 1)(4) = 99 \Rightarrow n = 25\)
   • \(S_{25} = \frac{25}{2}(3 + 99) = 1275\)

Answer each part.

1. (a) How many terms of 2 + 6 + 10 + … are needed for the sum to first exceed 800?

   • \(u_1 = 2,\ d = 4:\ S_n = \frac{n}{2}(4 + 4(n - 1)) = 2n^2\)
   • \(S_{20} = 800\ (\text{not more than }800),\quad S_{21} = 882\)
   • n = 21 terms.

2. (b) The sum of the first \(n\) terms of 7 + 10 + 13 + … is 847. Find \(n\).

   • \(u_1 = 7,\ d = 3:\ S_n = \frac{n}{2}(14 + 3(n - 1)) = 847\)
   • \(\text{Calculator table: } S_{21} = 777,\ S_{22} = 847\)
   • \(n = 22\)

The 4th term of an arithmetic series is 19 and the 9th term is 44. Find the sum of the first 15 terms.

• \(u_1 + 3d = 19\) and \(u_1 + 8d = 44\)
• \(d = 5,\ u_1 = 4\)
• \(S_{15} = \frac{15}{2}\,[2(4) + 14(5)] = 585\)

Logs are stacked with 24 in the bottom row and one fewer in each row above. The top row has 9 logs. How many logs are there?

• \(u_1 = 24,\ d = -1:\ 24 - (n - 1) = 9 \Rightarrow n = 16\)
• \(S_{16} = \frac{16}{2}(24 + 9) = 8(33)\)
• 264 logs.

Noor saves £20 in the first month and £5 more each month than the month before. (a) How much has she saved in total after 24 months? (b) In which month does her total first pass £3000?

• \(u_1 = 20,\ d = 5\)
• \((a)\ S_{24} = \frac{24}{2}(2(20) + 23(5)) = 1860\)
• \((b)\ S_n = \frac{n}{2}(40 + 5(n - 1)) > 3000\)
• \(\text{Calculator table: } S_{31} = 2945,\ S_{32} = 3120\)
• (a) £1860 (b) Month 32.

The sum of the first \(n\) terms of a sequence is \(S_n = 3n^2 + 2n\). Find the first term and a rule for \(u_n\).

• \(u_1 = S_1 = 3(1)^2 + 2(1) = 5\)
• \(u_n = S_n - S_{n-1} = 3n^2 + 2n - [3(n - 1)^2 + 2(n - 1)]\)
• \(u_n = 3n^2 + 2n - 3n^2 + 6n - 3 - 2n + 2 = 6n - 1\)
• \(\text{Check: } 6(1) - 1 = 5 = u_1\)

Find the sum of the multiples of 6 from 6 to 300.

• \(u_1 = 6,\ d = 6:\ 6 + (n - 1)(6) = 300 \Rightarrow n = 50\)
• \(S_{50} = \frac{50}{2}(6 + 300) = 25(306)\)
• 7650.