October 2021 Paper 3 Q3
3 An arithmetic progression has first term 2 and common difference \(d\), where \(d \neq 0\). The first, third and thirteenth terms of this progression are also the first, second and third terms, respectively, of a geometric progression.
By determining \(d\), show that the arithmetic progression is an increasing sequence. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(2 + 2d = 2r\) | B1 | 1.1 |
| \(2 + 12d = 2r^2\) | B1 | 1.1 |
| \(1 + 6d = (1 + d)^2\) or \(2 + 12d = 2(1 + d)^2\) | M1* | 1.1 |
| \(d^2 - 4d = 0 \Rightarrow d = \ldots\) | M1dep* | 1.1 |
| \(d = 4\) and as the common difference is positive the progression is an increasing sequence | A1 | 2.4 |
| [5] |
Notes
B1: Or for \(a + 2d = ar\)
B1: Or for \(a + 12d = ar^2\)
M1*: Setting up an equation in \(d\) or \(r\) only – dependent on one B mark
\(2 + 12(r - 1) = 2r^2\)
M1dep*: Solving their two-term quadratic equation in \(d\) (or three-term quadratic in \(r\))
\(r^2 - 6r + 5 = 0\)
\((r - 5)(r - 1) = 0\)
\(\Rightarrow r = \ldots\)
A1: Correct value for \(d\) and link to increasing sequence – must either say that \(d\) is positive (oe) or state at least the correct first four terms and comment that they are increasing
Condone no mention of \(d \neq 0\)
Alternative method
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{u_3}{u_2} =\right)\dfrac{2 + 12d}{2 + 2d}\) | B1 |
| \(\left(\dfrac{u_2}{u_1} =\right)\dfrac{2 + 2d}{2}\) | B1 |
| \(\dfrac{2 + 12d}{2 + 2d} = \dfrac{2 + 2d}{2}\) | M1* |
| \(d^2 - 4d = 0 \Rightarrow d = \ldots\) | M1dep* |
| \(d = 4\) and as the common difference is positive the progression is an increasing sequence | A1 |
Notes
B1: or for \(\dfrac{u_3}{u_1}\)
M1*: Setting up an equation in \(d\) only – dependent on one B mark
M1dep*: Solving their two-term quadratic equation in \(d\)
A1: As above