June 2025 Paper 1 Q3
3. The first three terms of an arithmetic sequence are
\[6k,\ 10 \ \text{ and } \ 2k\]where \(k\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| Attempts to solve \(10 - 6k = 2k - 10 \Rightarrow k = \ldots\) | M1 | 3.1a |
| \((k =)\ \dfrac{5}{2}\) o.e. | A1 | 1.1b |
| (2) |
Notes
Condone using other letters for a and d (e.g. may use r for d)
M1: Attempts a valid method to solve the problem
- Uses the common difference to form a correct equation and attempts to solve to find a value for \(k\)
e.g. Attempts to solve \(10 - 6k = 2k - 10\) o.e such as \(2k - 6k = 2(10 - 6k)\) - Uses 10 as the mean of \(2k\) and \(6k\): \(\dfrac{2k + 6k}{2} = 10 \Rightarrow k = \ldots\)
- Sets up correct equations \(a = 6k,\ a + d = 10\) and \(a + 2d = 2k,\) or may be seen as \(6k + d = 10\) and \(10 + d = 2k\), and proceeds to find \(k\).
- Uses the summation formula \(S_3 = \dfrac{3}{2}(6k + 2k) = 6k + 10 + 2k\) and proceeds to find \(k\)
In each attempt the initial equation (or simultaneous equations) must be correct but do not be concerned by the mechanics of the rearrangement to find \(k\). May be implied by \(k = \dfrac{5}{2}\)
A1: \((k =)\ \dfrac{5}{2}\) o.e.
| Scheme | Marks | AO |
|---|---|---|
| Deduces the value of \(\text{``}d\text{''} = -5\) | B1 ft | 2.2a |
| \(S_{50} = \dfrac{50}{2}\left(2 \times \text{``}15\text{''} + 49 \times \text{``}{-}5\text{''}\right)\) | M1 | 1.1b |
| \(= -5375\) | A1 | 1.1b |
| (3) | ||
| (5 marks) |
Notes
Work seen in (a) can only be scored if seen or used in (b)
B1ft: Common difference \(= -5\) or ft on their value for \(k\) (even if \(k\) has been found from an incorrect method) e.g. \(10 - 6 \times \text{``}\dfrac{5}{2}\text{''}\) or e.g. \(2 \times \text{``}\dfrac{5}{2}\text{''} - 10\) if only a numerical value is seen.
May be implied or seen in a term or summation formula.
Note that some candidates may work in terms of \(k\) throughout so only allow B1ft to be scored when they substitute in their numerical value for \(k\), following \(10 - 6k\) o.e. correctly embedded in a correct formula. They may make arithmetical slips before they substitute in their numerical value for \(k\) which can be condoned.
M1: Attempts to use a correct formula. The expression is sufficient to score this mark but they must be using a correct value for \(a\) and \(\pm d\) (or ft on their value for \(k\) for \(a\) and \(d\)) which are correctly placed in the formula.
e.g. \(\left(S_{50} =\right) \dfrac{50}{2}\left(2 \times \text{``}6k\text{''} + 49 \times \text{``}\pm d\text{''}\right)\).
Alternatively, they may find the 50th term \(u_{50} = \text{``}15\text{''} + (50-1) \times \text{``}{-}5\text{''} = -230\) and use \(\left(S_{50} =\right) \dfrac{50}{2}\left(\text{``}15\text{''} + \text{``}{-}230\text{''}\right)\).
If working in terms of \(k\) they must substitute in their value for \(k\)
e.g. \(\left(S_{50} =\right) \dfrac{50}{2}\left(2 \times 6k + 49 \times (10 - 6k)\right) = -7050k + 12250 = -7050 \times \text{``}\dfrac{5}{2}\text{''} + 12250\)
Do not withhold this mark for omission of brackets around \((-5)\)
e.g. \(\dfrac{50}{2}\left(2 \times 15 + (49) - 5\right)\) scores M1
A1: \(-5375\) cao