| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Easy -1.2 This is a straightforward arithmetic series question with standard bookwork proof in part (a) and routine applications in parts (b) and (c). The proof is a standard derivation expected at this level, while parts (b) and (c) involve direct substitution into formulas with no problem-solving insight required. Significantly easier than average A-level questions. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \((S=)a + (a+d) + \ldots + [a+(n-1)d]\) | B1 | Requires at least 3 terms, must include first and last terms, an adjacent term and dots |
| \((S=)[a+(n-1)d] + \ldots + a\) | M1 | For reversing series (dots needed) |
| \(2S = [2a+(n-1)d] + \ldots + [2a+(n-1)d]\) | dM1 | For adding; must have \(2S\) and be a genuine attempt; dependent on 1st M1 |
| \(S = \frac{n}{2}[2a+(n-1)d]\) cso | A1 | NB: allow first 3 marks for use of \(l\) for last term but not for final mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(600 = 200 + (N-1)20 \Rightarrow N = \ldots\) | M1 | Use of 600 with a correct formula in an attempt to find \(N\) |
| \(N = 21\) | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = \frac{21}{2}(2 \times 200 + 20 \times 20)\) or \(\frac{21}{2}(200+600)\) | M1 A1 | M1: Use of correct sum formula with their integer \(n = N\) or \(N-1\) from part (b) where \(3 < N < 52\), \(a=200\), \(d=20\) |
| \(S = \frac{20}{2}(2 \times 200 + 19 \times 20)\) or \(\frac{20}{2}(200+580)\) \((= 8400\) or \(7800)\) | M1 A1 | M1: same condition as above |
| \(600 \times (52 - \text{"N"})\ (= 18600)\) | M1 | M1: \(600 \times k\) where \(k\) is an integer and \(3 < k < 52\) |
| Correct un-simplified follow through expression with their \(k\) consistent with \(n\) so that \(n + k = 52\) | A1ft | |
| Total is \(27000\) | A1 cao |
## Question 5:
### Part (a):
| $(S=)a + (a+d) + \ldots + [a+(n-1)d]$ | B1 | Requires at least 3 terms, must include first and last terms, an adjacent term and dots |
|---|---|---|
| $(S=)[a+(n-1)d] + \ldots + a$ | M1 | For reversing series (dots needed) |
| $2S = [2a+(n-1)d] + \ldots + [2a+(n-1)d]$ | dM1 | For adding; must have $2S$ and be a genuine attempt; dependent on 1st M1 |
| $S = \frac{n}{2}[2a+(n-1)d]$ **cso** | A1 | NB: allow first 3 marks for use of $l$ for last term but not for final mark |
### Part (b):
| $600 = 200 + (N-1)20 \Rightarrow N = \ldots$ | M1 | Use of 600 with a **correct** formula in an attempt to find $N$ |
|---|---|---|
| $N = 21$ | A1 cso | |
### Part (c):
| $S = \frac{21}{2}(2 \times 200 + 20 \times 20)$ or $\frac{21}{2}(200+600)$ | M1 A1 | M1: Use of correct sum formula with their integer $n = N$ or $N-1$ from part (b) where $3 < N < 52$, $a=200$, $d=20$ |
|---|---|---|
| $S = \frac{20}{2}(2 \times 200 + 19 \times 20)$ or $\frac{20}{2}(200+580)$ $(= 8400$ or $7800)$ | M1 A1 | M1: same condition as above |
| $600 \times (52 - \text{"N"})\ (= 18600)$ | M1 | M1: $600 \times k$ where $k$ is an integer and $3 < k < 52$ |
| Correct un-simplified follow through expression with their $k$ consistent with $n$ so that $n + k = 52$ | A1ft | |
| Total is $27000$ | A1 cao | |
---An arithmetic series has first term $a$ and common difference $d$.
\begin{enumerate}[label=(\alph*)]
\item Prove that the sum of the first $n$ terms of the series is
$$\frac { 1 } { 2 } n [ 2 a + ( n - 1 ) d ]$$
A company, which is making 200 mobile phones each week, plans to increase its production.
The number of mobile phones produced is to be increased by 20 each week from 200 in week 1 to 220 in week 2, to 240 in week 3 and so on, until it is producing 600 in week $N$.
\item Find the value of $N$
The company then plans to continue to make 600 mobile phones each week.
\item Find the total number of mobile phones that will be made in the first 52 weeks starting from and including week 1.\\
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-16_2673_1948_107_118}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2018 Q5 [11]}}