Foundation November 2019 Paper 2 Q29
29 The 5th term of a linear sequence is 17
The 6th term of the sequence is 21
Work out the 100th term of the sequence. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(21 - 17\) or \(17 - 21\) or \(17 + 4\) or \(21 - 4\) or (difference is) 4 or (7th term =) \(21 + 4\) or 25 or (4th term =) \(17 - 4\) or 13 | M1 | may be seen as 17 21 with 4 between allow (difference is) \(-4\) |
| \(17 + (100 - 5) \times 4\) or \(17 + 95 \times 4\) or 17 + 380 or \(21 + (100 - 6) \times 4\) or \(21 + 94 \times 4\) or 21 + 376 or \(17 - 4 \times 4 + 99 \times 4\) or \(1 + 99 \times 4\) or 1 + 396 or \(17 - 5 \times 4 + 100 \times 4\) or \(-3 + 100 \times 4\) or \(-3 + 400\) | M1dep | must be using 4 oe calculation that would evaluate to 397 5th term \(+ 95 \times 4\) 6th term \(+ 94 \times 4\) 1st term \(+ 99 \times 4\) 0th term \(+ 100 \times 4\) |
| 397 | A1 | |
| Alternative method 2 | ||
| \(4n\) | M1 | oe eg \(n \times 4\) |
| \(4n - 3\) | A1 | oe |
| 397 | A1 | |
Additional guidance
| Term to term rule described eg Add on 4 each time | M1 |
| \(a + 5d = 21\), \(a + 4d = 17\) only | M0 |
| Difference shown as 4 then eg \(n + 4\) | M1 |
| Only eg \(n + 4\) or \(3n + 4\) | M0 |
| \(4n - 3\) seen even if not subsequently used | M1A1 |
| \(4n\) seen eg \(4n + 13\) even if not subsequently used | M1 |
| Correct list going up in 4s stopping at 397 | M1M1A1 |
| List going up in 4s with an error or not reaching 397 | M1M0A0 |
| No subtraction seen and incorrect difference eg 17 21 with +3 between | M0 |
| Alt 2 allow \(n4\) | M1 |
| \(4n - 3 = 100\) | M1A1A0 |
| Allow M1 even if not subsequently used |