| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sequence defined by formula |
| Difficulty | Moderate -0.8 This is a straightforward C1 question on arithmetic series requiring only direct substitution (part a), recognition of the common difference from the formula (part b), and a standard summation proof using either the arithmetic series formula or splitting into Σr terms (part c). All techniques are routine with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required in part (c). |
| Spec | 1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(-3, -1, 1\) | B1 B1 | (2 marks) One correct |
| (b) \(2\) | B1ft | (1 mark) (R only if terms in (a) are in arithmetic progression) |
| (c) Sum \(= \frac{1}{2}n\{2(-3) + (n-1)(2)\}\) or \(\frac{1}{2}n\{(-3) + (2n-5)\}\) | M1 A1ft | |
| \(= \frac{1}{2}n\{2n - 8\} = n(n-4)\) | A1 | (3 marks) |
**(a)** $-3, -1, 1$ | B1 B1 | (2 marks) One correct
**(b)** $2$ | B1ft | (1 mark) (R only if terms in (a) are in arithmetic progression)
**(c)** Sum $= \frac{1}{2}n\{2(-3) + (n-1)(2)\}$ or $\frac{1}{2}n\{(-3) + (2n-5)\}$ | M1 A1ft |
$= \frac{1}{2}n\{2n - 8\} = n(n-4)$ | A1 | (3 marks)
**Total: 6 marks**
---5. The $r$ th term of an arithmetic series is ( $2 r - 5$ ).
\begin{enumerate}[label=(\alph*)]
\item Write down the first three terms of this series.
\item State the value of the common difference.
\item Show that $\sum _ { r = 1 } ^ { n } ( 2 r - 5 ) = n ( n - 4 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2005 Q5 [6]}}