| Date | November 2009 | Marks available | 6 | Reference code | 09N.2.sl.TZ0.1 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
In an arithmetic sequence, \({S_{40}} = 1900\) and \({u_{40}} = 106\) . Find the value of \({u_1}\) and of d .
Markscheme
METHOD 1
substituting into formula for \({S_{40}}\) (M1)
correct substitution A1
e.g. \(1900 = \frac{{40({u_1} + 106)}}{2}\)
\({u_1} = - 11\) A1 N2
substituting into formula for \({u_{40}}\) or \({S_{40}}\) (M1)
correct substitution A1
e.g. \(106 = - 11 + 39d\) , \(1900 = 20( - 22 + 39d)\)
\(d = 3\) A1 N2
METHOD 2
substituting into formula for \({S_{40}}\) (M1)
correct substitution A1
e.g. \(20(2{u_1} + 39d) = 1900\)
substituting into formula for \({u_{40}}\) (M1)
correct substitution A1
e.g. \(106 = {u_1} + 39d\)
\({u_1} = - 11\) , \(d = 3\) A1A1 N2N2
[6 marks]
Examiners report
Most candidates answered this question correctly. Those who chose to solve with a system of equations often did so algebraically, using a fair bit of time doing so and sometimes making a careless error in the process. Few candidates took advantage of the system solving features of the GDC.
