Higher January 2019 Paper 1 Q21
21 \((2x + 23)\), \((8x + 2)\) and \((20x - 52)\) are three consecutive terms of an arithmetic sequence.
Prove that the common difference of the sequence is 12
(4)
| Scheme | Marks |
|---|---|
| \((8x + 2) - (2x + 23)\) (\(= 6x - 21\)) or \((2x + 23) - (8x + 2)\) (\(= -6x + 21\)) or \((20x - 52) - (8x + 2)\) (\(= 12x - 54\)) or \((8x + 2) - (20x - 52)\) (\(= -12x + 54\)) | M1 |
| \((8x + 2) - (2x + 23) = (20x - 52) - (8x + 2)\) oe or \((2x + 23) - (8x + 2) = (8x + 2) - (20x - 52)\) oe | M1 |
| \(x = 5.5\) | A1 |
| Working required Eg \(2 \times 5.5 + 23\) (=34) and \(8 \times 5.5 + 2\) (=46) OR \(8 \times 5.5 + 2\) (=46) and \(20 \times 5.5 - 52\) (=58) Answer: shown | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a correct expression for the common difference in terms of \(x\) brackets must be present or removed correctly
M1: for a correct equation
A1: for 12 from correct working
| Scheme | Marks |
|---|---|
Alternative method – starts by assuming \(d = 12\) E.g. \((2x + 23) + 12 = (8x + 2)\) or \((8x + 2) + 12 = (20x - 52)\) or \((2x + 23) - 12 = (8x + 2)\) or \((8x + 2) - 12 = (20x - 52)\) or \((2x + 23) + (8x + 2) + (20x - 52) = \dfrac{3}{2}(2(2x + 23) + 2 \times 12)\) | M2 |
| \(x = 5.5\) or \(x = 1.5\) from \((2x + 23) - 12 = (8x + 2)\) or \(x = 3.5\) from \((8x + 2) - 12 = (20x - 52)\) | A1 |
| Working required \(2 \times 5.5 + 23\) (=34) and \(8 \times 5.5 + 2\) (=46) and \(20 \times 5.5 - 52\) (=58) OR \((2x + 23) + 12 = (8x + 2)\) and \((8x + 2) + 12 = (20x - 52)\) and gets \(x = 5.5\) both times Answer: shown | A1 |
Notes
M2: for a correct equation
If not M2 then award M1 for a correct expression for the common difference in terms of \(x\) brackets must be present or removed correctly e.g \((8x + 2) - (2x + 23)\) (\(= 6x - 21\)) or \((20x - 52) - (8x + 2)\) (\(= 12x - 54\))
A1: for explicitly showing both common differences are 12
OR
solves both \((2x + 23) + 12 = (8x + 2)\) and \((8x + 2) + 12 = (20x - 52)\) and gets \(x = 5.5\) both times