June 2025 Paper 2 Q7
7 An arithmetic progression has first term \(a\) and common difference \(d\), where \(a\) and \(d\) are non-zero. The first, third and fourth terms of the arithmetic progression are consecutive terms of a geometric progression with common ratio \(r\).
| Scheme | Marks | AO |
|---|---|---|
| (i) 3rd term of AP is \(a + 2d\) \(ar = a + 2d \Rightarrow r = \dfrac{a + 2d}{a}\) AG | B1 | 2.1 |
| [1] | ||
| (ii) \(\dfrac{a + 3d}{a + 2d} = \dfrac{a + 2d}{a}\) oe | M1 | 3.1a |
| \(\left[a^2 + 3ad = a^2 + 4ad + 4d^2\right]\) \(\Rightarrow d = -\dfrac{a}{4}\) | A1 | 1.1 |
| [2] |
Notes
(a)(i)
B1: Must see \(a + 2d\) (either on its own or as one side of an equation) before this answer.
(a)(ii)
M1: May be implied by A1.
May see \(a(a + 3d) = (a + 2d)^2\) oe or \(a\left(\frac{a + 2d}{a}\right)^2 = a + 3d\)
A1: Must be \(d\) in terms of \(a\) so do not accept \(a = -4d\) for this mark.
Alternative method
| Scheme | Marks | AO |
|---|---|---|
| \(ar - a = 2\left(ar^2 - ar\right)\) | M1 | |
| \(r - 1 = 2r(r - 1)\) or \(1 = 2r\) \(\Rightarrow r = \frac{1}{2} \Rightarrow d = a \times \left(\frac{1}{2}\right)^2 - a \times \frac{1}{2}\) \(\Rightarrow d = -\dfrac{a}{4}\) | A1 |
M1: May be implied by A1.
(NB \(ar - a = 2d\) with (a)(i) substituted is not sufficient)
A1: Must be \(d\) in terms of \(a\) so do not accept \(a = -4d\) for this mark.
| Scheme | Marks | AO |
|---|---|---|
| \(r = \dfrac{a - \frac{a}{2}}{a}\) or \(\dfrac{a - \frac{3a}{4}}{a - \frac{a}{2}}\) or \(\dfrac{-4d + 2d}{-4d}\) oe | M1 | 1.1 |
| \(= \dfrac{1}{2}\) | A1 | 1.1 |
| [2] |
Notes
M1: FT their expression from (a)(ii) with \(r = \frac{a + 2d}{a}\) or \(r = \frac{a + 3d}{a + 2d}\). May be implied by A1.
A1: Must see \(\frac{1}{2}\) in their answer for (b) to score these marks.