Higher June 2018 Paper 1R Q23
23 Work out the sum of the multiples of 3 between 1 and 1000
(4)
| Scheme | Marks |
|---|---|
| (First term = 3 and last term = 999) or \(a = 3\) and \(d = 3\) | M1 |
| 999 ÷ 3 (= 333) | M1 |
Sum = \(\dfrac{333}{2}(3 + 999)\) or Sum = \(\dfrac{333}{2}(2 \times 3 + (333 - 1)3)\) | M1 |
| 166 833 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for finding the number of terms
Allow 1000÷3 = 333.3 = 333
M1: for using a correct method to find the sum