| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Easy -1.2 This is a standard textbook question on arithmetic series requiring routine application of the sum formula. Part (i) is a bookwork proof that should be memorized. Part (ii) involves straightforward substitution into the formula and simple algebraic manipulation to reach a given quadratic, followed by solving using the quadratic formula or factorization. No problem-solving insight is required—just methodical application of a well-practiced technique. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| States \(S = a + (a+d) + \ldots + (a+(n-1)d)\) | B1 | 1.1a — minimum 3 correct terms including first and last |
| Reverses series: \(S = (a+(n-1)d) + (a+(n-2)d) + \ldots + a\) and adds | M1 | 3.1a |
| Reaches \(2S = n \times (2a + (n-1)d)\) and proves \(S = \frac{n}{2}[2a+(n-1)d]\) | A1* | 2.1 — no errors, all steps shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(64 = \frac{n}{2}(20 - 0.8(n-1))\) o.e. | M1 | 3.1b |
| \(0.8n^2 - 20.8n + 128 = 0 \Rightarrow n^2 - 26n + 160 = 0\) | A1* | 2.1 — correct derivation shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(n = 10, 16\) | B1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(n = 10\) weeks with minimal correct reason, e.g. he has already reached £64 by week 10 so would not save for another 6 weeks; or negative amounts saved from week 14 so 16 does not make sense | B1 | 2.3 |
# Question 13:
## Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| States $S = a + (a+d) + \ldots + (a+(n-1)d)$ | B1 | 1.1a — minimum 3 correct terms including first and last |
| Reverses series: $S = (a+(n-1)d) + (a+(n-2)d) + \ldots + a$ and adds | M1 | 3.1a |
| Reaches $2S = n \times (2a + (n-1)d)$ and proves $S = \frac{n}{2}[2a+(n-1)d]$ | A1* | 2.1 — no errors, all steps shown |
## Part (ii)(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $64 = \frac{n}{2}(20 - 0.8(n-1))$ o.e. | M1 | 3.1b |
| $0.8n^2 - 20.8n + 128 = 0 \Rightarrow n^2 - 26n + 160 = 0$ | A1* | 2.1 — correct derivation shown |
## Part (ii)(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $n = 10, 16$ | B1 | 1.1b |
## Part (ii)(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $n = 10$ weeks with minimal correct reason, e.g. he has already reached £64 by week 10 so would not save for another 6 weeks; or negative amounts saved from week 14 so 16 does not make sense | B1 | 2.3 |\begin{enumerate}
\item (i) In an arithmetic series, the first term is $a$ and the common difference is $d$.
\end{enumerate}
Show that
$$S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]$$
(ii) James saves money over a number of weeks to buy a printer that costs $\pounds 64$
He saves $\pounds 10$ in week $1 , \pounds 9.20$ in week $2 , \pounds 8.40$ in week 3 and so on, so that the weekly amounts he saves form an arithmetic sequence.
Given that James takes $n$ weeks to save exactly $\pounds 64$\\
(a) show that
$$n ^ { 2 } - 26 n + 160 = 0$$
(b) Solve the equation
$$n ^ { 2 } - 26 n + 160 = 0$$
(c) Hence state the number of weeks James takes to save enough money to buy the printer, giving a brief reason for your answer.
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q13 [7]}}