| Date | November 2013 | Marks available | 7 | Reference code | 13N.2.hl.TZ0.2 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The fourth term in an arithmetic sequence is 34 and the tenth term is 76.
(a) Find the first term and the common difference.
(b) The sum of the first n terms exceeds 5000. Find the least possible value of n.
Markscheme
(a) METHOD 1
\(34 = a + 3d\) and \(76 = a + 9d\) (M1)
\(d = 7\) A1
\(a = 13\) A1
METHOD 2
\(76 = 34 + 6d\) (M1)
\(d = 7\) A1
\(34 = a + 21\)
\(a = 13\) A1
[3 marks]
(b) \(\frac{n}{2}\left( {26 + 7(n - 1)} \right) > 5000\) (M1)(A1)
\(n > 36.463 \ldots \) (A1)
Note: Award M1A1A1 for using either an equation, a graphical approach or a numerical approach.
\(n = 37\) A1 N3
[4 marks]
Total [7 marks]
Examiners report
Both parts were very well done. In part (a), a few candidates made a careless algebraic error when attempting to find the value of a or d.
In part (b), a few candidates attempted to find the value of n for which \({u_n} > 5000\). Some candidates used the incorrect formula \({S_n} = \frac{n}{2}\left[ {{u_1} + (n - 1)d} \right]\). A number of candidates unnecessarily attempted to simplify \({S_n}\). Most successful candidates in part (b) adopted a graphical approach and communicated their solution effectively. A few candidates did not state their value of n as an integer.
