| Date | May 2016 | Marks available | 6 | Reference code | 16M.1.hl.TZ1.1 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
The fifth term of an arithmetic sequence is equal to 6 and the sum of the first 12 terms is 45.
Find the first term and the common difference.
Markscheme
use of either \({u_n} = {u_1} + (n - 1)d\) or \({S_n} = \frac{n}{2}\left( {2{u_1} + (n - 1)d} \right)\) M1
\({u_1} + 4d = 6\) (A1)
\(\frac{{12}}{2}(2{u_1} + 11d) = 45\) (A1)
\( \Rightarrow 4{u_1} + 22d = 15\)
attempt to solve simultaneous equations M1
\(4(6 - 4d) + 22d = 15\)
\(6d = - 9 \Rightarrow d = - 1.5\) A1
\({u_1} = 12\) A1
[6 marks]
Examiners report
Most candidates tackled this question through the use of the standard formula for arithmetic series. Others attempted a variety of trial and improvement approaches with varying degrees of success.
Syllabus sections
Topic 1 - Core: Algebra » 1.1 » Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.
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