| Date | November 2012 | Marks available | 4 | Reference code | 12N.2.hl.TZ0.1 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Find the sum of all the multiples of 3 between 100 and 500.
Markscheme
METHOD 1
102 + 105 + … + 498 (M1)
so number of terms = 133 (A1)
EITHER
\( = \frac{{133}}{2}(2 \times 102 + 132 \times 3)\) (M1)
= 39900 A1
OR
\( = (102 + 498) \times \frac{{133}}{2}\) (M1)
= 39900 A1
OR
\(\sum\limits_{n = 34}^{166} {3n} \) (M1)
= 39900 A1
METHOD 2
\(500 \div 3 = 166.666...{\text{ and }}100 \div 3 = 33.333...\)
\(102 + 105 + ... + 498 = \sum\limits_{n = 1}^{166} {3n} - \sum\limits_{n = 1}^{33} {3n} \) (M1)
\(\sum\limits_{n = 1}^{166} {3n} = 41583\) (A1)
\(\sum\limits_{n = 1}^{33} {3n} = 1683\) (A1)
the sum is 39900 A1
[4 marks]
Examiners report
Most candidates got full marks in this question. Some mistakes were detected when trying to find the number of terms of the arithmetic sequence, namely the use of the incorrect value n = 132 ; a few interpreted the question as the sum of multiples between the 100th and 500th terms. Occasional application of geometric series was attempted.
