| Date | November 2010 | Marks available | 1 | Reference code | 10N.2.sl.TZ0.3 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Write down | Question number | 3 | Adapted from | N/A |
Question
The nth term of an arithmetic sequence is given by \({u_n} = 5 + 2n\) .
Write down the common difference.
(i) Given that the nth term of this sequence is 115, find the value of n .
(ii) For this value of n , find the sum of the sequence.
Markscheme
\(d = 2\) A1 N1
[1 mark]
(i) \(5 + 2n = 115\) (A1)
\(n = 55\) A1 N2
(ii) \({u_1} = 7\) (may be seen in above) (A1)
correct substitution into formula for sum of arithmetic series (A1)
e.g. \({S_{55}} = \frac{{55}}{2}(7 + 115)\) , \({S_{55}} = \frac{{55}}{2}(2(7) + 54(2))\) , \(\sum\limits_{k = 1}^{55} {(5 + 2k)} \)
\({S_{55}} = 3355\) (accept \(3360\)) A1 N3
[5 marks]
Examiners report
The majority of candidates could either recognize the common difference in the formula for the nth term or could find it by writing out the first few terms of the sequence.
Part (b) demonstrated that candidates were not familiar with expression, "nth term". Many stated that the first term was 5 and then decided to use their own version of the nth term formula leading to a great many errors in (b) (ii).
