| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find term or common difference |
| Difficulty | Easy -1.2 This is a straightforward two-part question requiring only direct application of standard arithmetic sequence formulas (a_n = a + (n-1)d and S_n = n/2(2a + (n-1)d)). Part (a) involves simple algebraic manipulation to find d, and part (b) is direct substitution. No problem-solving insight or multi-step reasoning required—purely procedural recall. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(16 + (21-1) \times d = 24 \Rightarrow d = ...\) | M1 | Correct strategy to find common difference using \(a = 16\), \(n = 21\) and \(24\). Method may be implied by working. If AP term formula quoted it must be correct; use of \(u_n = a + nd\) scores M0 |
| \(d = 0.4\) | A1 | Accept equivalents e.g. \(\frac{8}{20}, \frac{4}{10}, \frac{2}{5}\) etc. Answer only scores both marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(S_n = \frac{1}{2}n\{2a+(n-1)d\} \Rightarrow S_{500} = \frac{1}{2} \times 500\{2 \times 16 + 499 \times \text{"0.4"}\}\) | M1 | Attempts correct sum formula with \(a=16\), \(n=500\) and numerical \(d\) from part (a). Formula must be correct if quoted. |
| \(= 57900\) | A1 | Correct value. Answer only scores both marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(l = 16 + (500-1) \times \text{"0.4"} = 215.6 \Rightarrow S_{500} = \frac{1}{2} \times 500\{16 + \text{"215.6"}\}\) | M1 | Correct method for 500th term then uses \(S_n = \frac{1}{2}n\{a+l\}\) with their \(l\) |
| \(= 57900\) | A1 | Correct value |
## Question 1:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $16 + (21-1) \times d = 24 \Rightarrow d = ...$ | M1 | Correct strategy to find common difference using $a = 16$, $n = 21$ and $24$. Method may be implied by working. If AP term formula quoted it must be correct; use of $u_n = a + nd$ scores M0 |
| $d = 0.4$ | A1 | Accept equivalents e.g. $\frac{8}{20}, \frac{4}{10}, \frac{2}{5}$ etc. Answer only scores both marks. |
**(2 marks)**
---
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $S_n = \frac{1}{2}n\{2a+(n-1)d\} \Rightarrow S_{500} = \frac{1}{2} \times 500\{2 \times 16 + 499 \times \text{"0.4"}\}$ | M1 | Attempts correct sum formula with $a=16$, $n=500$ and numerical $d$ from part (a). Formula must be correct if quoted. |
| $= 57900$ | A1 | Correct value. Answer only scores both marks. |
**Alternative using $S_n = \frac{1}{2}n\{a+l\}$:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $l = 16 + (500-1) \times \text{"0.4"} = 215.6 \Rightarrow S_{500} = \frac{1}{2} \times 500\{16 + \text{"215.6"}\}$ | M1 | Correct method for 500th term then uses $S_n = \frac{1}{2}n\{a+l\}$ with their $l$ |
| $= 57900$ | A1 | Correct value |
**(2 marks) — Total: 4 marks**
> **Note:** Candidates implying $u_n = a + nd$ by showing $d = \frac{24-16}{21} = \frac{8}{21}$ scores **(a) M0A0**; subsequent $S_{500} = \frac{1}{2}\times 500\left\{2\times16 + 499 \times \frac{8}{21}\right\}$ scores **(b) M1A0**\begin{enumerate}
\item In an arithmetic series
\end{enumerate}
\begin{itemize}
\item the first term is 16
\item the 21 st term is 24\\
(a) Find the common difference of the series.\\
(b) Hence find the sum of the first 500 terms of the series.
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q1 [4]}}