| Date | November 2007 | Marks available | 3 | Reference code | 07N.1.sl.TZ0.3 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The fifth term of an arithmetic sequence is 20 and the twelfth term is 41.
(i) Find the common difference.
(ii) Find the first term of the sequence.
Calculate the eighty-fourth term.
Calculate the sum of the first 200 terms.
Markscheme
(i) \(u_5 = u_1 + 4d = 20\)
\(u_{12} = u_1 + 11d = 41\) (M1)
(M1) for both equations correct (or (M1) for \(20 + 7d = 41\))
\(7d = 21\)
\(d = 3\) (A1) (C2)
(ii) \(u_1 + 12 = 20\)
\(u_1 = 8\) (A1)(ft) (C1)
[3 marks]
\(u_{84} = 8 + (84 - 1)3\)
\(= 257\) (A1)(ft) (C1)
[1 mark]
\(S_{200} = 100(16 + 199 \times 3)\) (M1)
\( = 61300\) (A1)(ft) (C2)
[2 marks]
Examiners report
This question was generally answered well. Most of the candidates had a good understanding of how to use the formulae for an arithmetic sequence.
This question was generally answered well. Most of the candidates had a good understanding of how to use the formulae for an arithmetic sequence.
This question was generally answered well. Most of the candidates had a good understanding of how to use the formulae for an arithmetic sequence.
