Date | November 2012 | Marks available | 2 | Reference code | 12N.1.sl.TZ0.2 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Calculate | Question number | 2 | Adapted from | N/A |
Question
The first term of an arithmetic sequence is 3 and the seventh term is 33.
Calculate the common difference;
Calculate the 95th term of the sequence;
Calculate the sum of the first 250 terms.
Markscheme
33 = 3 + d(6) (M1)
Note: Award (M1) for correctly substituted formula or a correct numerical expression to find the common difference.
(d =)5 (A1) (C2)
[2 marks]
u95 = 3 + 94(5) (M1)
Note: Award (M1) for their correctly substituted formula.
= 473 (A1)(ft) (C2)
Note: Follow through from their part (a).
[2 marks]
S250 = 125[2(3) + 249(5)] (M1)
Note: Award (M1) for correctly substituted formula.
S250 = 156375 (A1)(ft) (C2)
Note: Follow through from their part (a).
[2 marks]
Examiners report
Much good work was seen in parts (a) and (b) showing that many centres had well prepared their students for questions on arithmetic sequences. In part (c) however there was poor use of \({S_n} = \frac{n}{2}\){first term + last term} with the incorrect calculation \(\frac{{250}}{2}\{ 3 + 250\} \) seen on a significant number of scripts.
Much good work was seen in parts (a) and (b) showing that many centres had well prepared their students for questions on arithmetic sequences. In part (c) however there was poor use of \({S_n} = \frac{n}{2}\){first term + last term} with the incorrect calculation \(\frac{{250}}{2}\{ 3 + 250\} \) seen on a significant number of scripts.
Much good work was seen in parts (a) and (b) showing that many centres had well prepared their students for questions on arithmetic sequences. In part (c) however there was poor use of \({S_n} = \frac{n}{2}\){first term + last term} with the incorrect calculation \(\frac{{250}}{2}\{ 3 + 250\} \) seen on a significant number of scripts.