Foundation June 2024 Paper 1 Q17
17 A linear sequence has
- 1st term \(= 10\)
- 1st term \(+\) 2nd term \(= 39\)
Work out the 5th term. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(39 - 10\) or 29 | M1 | oe |
| their \(29 - 10\) or 19 or \(19n\) | M1dep | oe \(39 - 10 - 10\) implies M1M1 (3rd term \(=\)) 48 implies M1M1 may be implied by the difference, after their 2nd term, consistently being the correct 19 \(19n\) may be seen as part of \(19n + b\) |
| their \(29 + 3 \times\) their 19 or \(10 + 4 \times\) their 19 or substitutes \(n = 5\) into expression of the form their \(19n + b\) | M1dep | oe (4th term \(=\)) 67 implies M1M1M1 \(b\) must be an integer |
| 86 | A1 | SC1 107 or 137 using Fibonacci SC1 126 using difference of 29 |
Additional guidance
| 3rd mark must be a correct method for working out the 5th term | |
| Going past the 5th term eg 10, 29, 48, 67, 86, 105, without answer 86 | M1M1M1A0 |
| \(10 + 19 = 39\) 10, 39, 58, 77, 96 (not the correct 19 being added) | M0 |