| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find term or common difference |
| Difficulty | Moderate -0.8 This is a straightforward C1 arithmetic series question requiring direct application of standard formulas. Part (a) involves simple algebra to find d from a₁ and a₃, then using the sum formula. Part (b) requires identifying the arithmetic sequence (56, 64, ..., 144) and applying the same formula. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(a = 3, a + 2d = 27\) | B1 | |
| \(2d = 24, d = 12\) | M1 A1 | |
| (ii) \(= \frac{11}{2}[6 + (10 \times 12)]\) | M1 | |
| \(= \frac{11}{2} \times 126 = 693\) | A1 | |
| (b) \(a = 56, l = 144\) | B1 | |
| \(56 + 8(n-1) = 144, n = 12\) | M1 A1 | |
| \(S_{12} = \frac{12}{2}(56 + 144) = 6 \times 200 = 1200\) | M1 A1 | (10) |
(a) (i) $a = 3, a + 2d = 27$ | B1 |
$2d = 24, d = 12$ | M1 A1 |
(ii) $= \frac{11}{2}[6 + (10 \times 12)]$ | M1 |
$= \frac{11}{2} \times 126 = 693$ | A1 |
(b) $a = 56, l = 144$ | B1 |
$56 + 8(n-1) = 144, n = 12$ | M1 A1 |
$S_{12} = \frac{12}{2}(56 + 144) = 6 \times 200 = 1200$ | M1 A1 | (10)8. (a) The first and third terms of an arithmetic series are 3 and 27 respectively.
\begin{enumerate}[label=(\roman*)]
\item Find the common difference of the series.
\item Find the sum of the first 11 terms of the series.\\
(b) Find the sum of the integers between 50 and 150 which are divisible by 8 .
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q8 [10]}}