| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sum of specific range of terms |
| Difficulty | Moderate -0.3 This is a straightforward arithmetic series question requiring standard formula application. Parts (a)-(b) are routine bookwork (finding d, a, and S₂₀). Part (c) requires the insight that S₅₀ - S₂₀ gives the desired sum, which is a small step beyond pure recall but still a common textbook exercise type. Slightly easier than average due to minimal problem-solving demand. |
| Spec | 1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(a + 19d = 181\); \(a + 4d = 46\); \(\Rightarrow 15d = 181 - 46\); \(\Rightarrow d = 9\) | M1, A1, A1 | \(a + (n-1)d\) used; PI; AG (be convinced) |
| Total for 3(a)(i): 3 marks |
| Answer | Marks |
|---|---|
| \(a = 10\) | B1 |
| Total for 3(a)(ii): 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{20} = \frac{20}{2}[2a + (20-1)d]\); \(.... = 1910\) | M1, A1 | OE; |
| Total for 3(b): 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{n=1}^{50}u_n - \sum_{n=1}^{20}u_n\); \(.... = 11525 - "1910" = 9615\) | M1, A1√ | OE; ft on 11525 − c's \(S_{20}\) |
| Total for 3(c): 2 marks | ||
| Total for Question 3: 8 marks |
**3(a)(i)**
| $a + 19d = 181$; $a + 4d = 46$; $\Rightarrow 15d = 181 - 46$; $\Rightarrow d = 9$ | M1, A1, A1 | $a + (n-1)d$ used; PI; AG (be convinced) |
| Total for 3(a)(i): 3 marks |
**3(a)(ii)**
| $a = 10$ | B1 | |
| Total for 3(a)(ii): 1 mark |
**3(b)**
| $S_{20} = \frac{20}{2}[2a + (20-1)d]$; $.... = 1910$ | M1, A1 | OE; |
| Total for 3(b): 2 marks |
**3(c)**
| $\sum_{n=1}^{50}u_n - \sum_{n=1}^{20}u_n$; $.... = 11525 - "1910" = 9615$ | M1, A1√ | OE; ft on 11525 − c's $S_{20}$ |
| Total for 3(c): 2 marks |
| **Total for Question 3: 8 marks** |
---3 An arithmetic series has fifth term 46 and twentieth term 181.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the common difference is 9 .
\item Find the first term.
\end{enumerate}\item Find the sum of the first 20 terms of the series.
\item The $n$th term of the series is $u _ { n }$.
Given that the sum of the first 50 terms of the series is 11525 , find the value of
$$\sum _ { n = 21 } ^ { 50 } u _ { n }$$
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q3 [6]}}