| Date | May 2010 | Marks available | 3 | Reference code | 10M.1.sl.TZ2.7 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
A concert choir is arranged, per row, according to an arithmetic sequence. There are 20 singers in the fourth row and 32 singers in the eighth row.
Find the common difference of this arithmetic sequence.
There are 10 rows in the choir and 11 singers in the first row.
Find the total number of singers in the choir.
Markscheme
20 = u1 + 3d (A1)
32 = u1 + 7d (A1)
Note: Award (A1) for each equation, (A1) for correct answer.
OR
\(d = \frac{{32 - 20}}{4}\) (A1)(A1)
Note: Award (A1) for numerator, (A1) for denominator.
d = 3 (A1) (C3)
[3 marks]
\(\frac{{10}}{2}(2 \times 11 + 9 \times 3)\) or \(\frac{{10}}{2}(11 + 38)\) (M1)(A1)(ft)
Note: Award (M1) for correct substituted formula, (A1) for correct substitution, follow through from their answer to part (a).
OR
11 + 14 + ... + 38 (M1)(A1)(ft)
Note: Award (M1) for attempt at the sum of a list, (A1)(ft) for all correct numbers, follow through from their answer to part (a).
= 245 (A1)(ft) (C3)
[3 marks]
Examiners report
This question was very well answered with most candidates finding the common difference and the total number of singers. Most candidates used the given formulae, rather than making lists. A common mistake was to find the number of singers in the back row, rather than find the total number of singers in the choir.
This question was very well answered with most candidates finding the common difference and the total number of singers. Most candidates used the given formulae, rather than making lists. A common mistake was to find the number of singers in the back row, rather than find the total number of singers in the choir.
