Higher June 2024 Paper 1R Q23
23 Here are the first three terms of an arithmetic sequence.
\[(4x - 14) \quad , \quad (x + 2) \quad , \quad (7x - 9)\]Find, as an integer, the sum of the first 40 terms of the sequence.
Show clear algebraic working.
(4)
| Scheme | Marks |
|---|---|
| \((7x - 9) - (x + 2) = (x + 2) - (4x - 14)\) oe eg \(6x - 11 = 16 - 3x\) OR \(x + 2 = 4x - 14 + d\) and \(7x - 9 = 4x - 14 + 2d\) oe eg \(-3x + 16 = d\) and \(3x + 5 = 2d\) | M1 |
\(x = 3\) and \(a = -2\) and \(d = 7\) OR \(x = 3\) and eg \((S_{40} =)\, \dfrac{40}{2}\left[2(4x - 14) + 39(-3x + 16)\right]\) or \(x = 3\) and eg \((S_{40} =)\, \dfrac{40}{2}\left[2(4x - 14) + 39(6x - 11)\right]\) | M1 |
\((S_{40} =)\, \dfrac{40}{2}\left(2 \times \text{``}{-2}\text{''} + 39 \times \text{``}{7}\text{''}\right)\) or eg \((S_{40} =)\, \dfrac{40}{2}\left[2\left(4 \times \text{``}{3}\text{''} - 14\right) + 39\left(-3 \times \text{``}{3}\text{''} + 16\right)\right]\) or \((S_{40} =)\, \dfrac{40}{2}\left[2\left(4 \times \text{``}{3}\text{''} - 14\right) + 39\left(6 \times \text{``}{3}\text{''} - 11\right)\right]\) | M1 |
| Working required Answer: 5380 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for setting up an equation in \(x\)
OR two simultaneous equations in \(x\) and \(d\)
M1: correct values or values from correct substitution OR \(x = 3\) and \(S_{40}\) expressed in terms of \(x\)
allow (40 – 1) for 39
M1: allow use of their \(a\) and their \(d\) or their \(x\) as long as clearly stated
allow (40 – 1) for 39
A1: (dep on M1)