| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Two related arithmetic progressions |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on arithmetic sequences requiring standard formulas (nth term and sum). Part (i) is given as 'show that' making it easier, parts (ii) and (iii) involve direct application of formulas with simple algebra. Slightly above average due to the equation-solving in part (iii), but still routine C2 material with no novel insight required. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a + d = 26\) and \(a + 4d = 41\), subtracting \(3d = 15\), so \(d = 5\) | M1, M1, A1 | |
| (ii) \(a = 21\), \(u_{12} = 21 + (11 \times 5) = 76\) | B1, M1, A1 | |
| (iii) \(\frac{2}{3}[42 + 5(n-1)] = \frac{2}{3}[-24 + 7(n-1)]\) leading to \(n(5n + 37) = n(7n - 31)\), so \(2n(n - 34) = 0\). Since \(n > 0\), \(n = 34\) | M1, A1, M1, A1 | (10) |
**(i)** $a + d = 26$ and $a + 4d = 41$, subtracting $3d = 15$, so $d = 5$ | M1, M1, A1 |
**(ii)** $a = 21$, $u_{12} = 21 + (11 \times 5) = 76$ | B1, M1, A1 |
**(iii)** $\frac{2}{3}[42 + 5(n-1)] = \frac{2}{3}[-24 + 7(n-1)]$ leading to $n(5n + 37) = n(7n - 31)$, so $2n(n - 34) = 0$. Since $n > 0$, $n = 34$ | M1, A1, M1, A1 | **(10)**
---(i) Show that the common difference is 5 .\\
(ii) Find the 12th term.
Another arithmetic sequence has first term -12 and common difference 7 .\\
Given that the sums of the first $n$ terms of these two sequences are equal,\\
(iii) find the value of $n$.\\
\hfill \mbox{\textit{OCR C2 Q7 [10]}}