| Date | May 2009 | Marks available | 3 | Reference code | 09M.1.sl.TZ2.4 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Consider the following sequence:
57, 55, 53 ..., 5, 3
Find the number of terms of the sequence.
Find the sum of the sequence.
Markscheme
3 = 57 + (n − 1) × (−2)
OR
57 = 3 + (n − 1) × (2) (A1)(M1)
Note: Award (A1) for 3 or 57 seen as un, (M1) for correctly substituted formula or list of values seen.
n = 28 (A1) (C3)
[3 marks]
\({{\text{S}}_{28}} = \frac{{28}}{2}(57 + 3)\)
OR
\({{\rm{S}}_{28}} = \frac{{28}}{2}(2(57) + (28 - 1) \times - 2)\)
OR
\({{\rm{S}}_{28}} = \frac{{28}}{2}(2(3) + (28 - 1) \times 2)\) (M1)(A1)(ft)
Note: (A1)(ft) for 28 seen.
Award (M1) for correctly substituted formula or list of values seen.
\({{\rm{S}}_{28}} = 840\) (A1)(ft) (C3)
[3 marks]
Examiners report
Most candidates recognised the arithmetic sequence and used the correct formula, although some used a list to find the answers. A common error was to use the common difference as 2 rather than −2. Many candidates were awarded follow through marks in part (b), correctly using their incorrect value of n from part (a).
Most candidates recognised the arithmetic sequence and used the correct formula, although some used a list to find the answers. A common error was to use the common difference as 2 rather than −2. Many candidates were awarded follow through marks in part (b), correctly using their incorrect value of n from part (a).
