Higher November 2020 Paper 1 Q24
24 Here are the first five terms of an arithmetic sequence.
8 15 22 29 36
Work out the sum of all the terms from the 50th term to the 100th term inclusive.
(4)
| Scheme | Marks |
|---|---|
| \(a = 8 \quad d = 7\) | M1 |
\((S_{100} =)\; \dfrac{100}{2}(2 \times 8 + (100 - 1) \times 7)\) (= 35 450) or \((S_{49} =)\; \dfrac{49}{2}(2 \times 8 + (49 - 1) \times 7)\) (= 8624) or \((S_{50} =)\; \dfrac{50}{2}(2 \times 8 + (50 - 1) \times 7)\) (= 8975) | M1 |
| ‘35450’ – ‘8624’ or ‘35450’ – ‘8975’ + (8 + (50 – 1)×7) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 26 826 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: can be implied
| Scheme | Marks |
|---|---|
| \((u_n =)\; 7n + 1\) or \(a = 8\) and \(d = 7\) | M1 |
| \((u_{50} =)\; 7 \times 50 + 1\) (= 351) or \((u_{100} =)\; 7 \times 100 + 1\) (= 701) or \((u_{50} =)\; 8 + (50 - 1) \times 7\) (= 351) | M1 |
\(\dfrac{51}{2}(\text{‘}351\text{’} + \text{‘}701\text{’})\) or \(\dfrac{51}{2}(2 \times 351 + (51 - 1) \times 7)\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 26 826 | A1 |
Notes
M1: can be implied