| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Standard summation formula application |
| Difficulty | Moderate -0.8 This is a straightforward application of arithmetic sequence formulas and standard deviation. Part (a) requires showing a given result using the sum formula for an arithmetic sequence, which is routine. Part (b) applies standard summation formulae (Σr and Σr²) in a prescribed way with no problem-solving required—just careful algebraic manipulation. The question is easier than average because it's highly procedural with clear signposting and a 'show that' part that provides the answer. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\bar{x} = \frac{1}{n}\sum_{r=1}^{n}(7+3r)\) | M1 | 1.1a |
| \(\sum_{r=1}^{n}(7+3r) = 7n + 3\cdot\frac{n}{2}(n+1)\) | M1 | 1.1b |
| \(\bar{x} = 7+\frac{3}{2}(n+1) = \frac{14+3n+3}{2} = \frac{1}{2}(3n+17)\) | A1* | 2.1 |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Correct overall strategy to find the variance or standard deviation (attempt at mean, attempt at \(\sum(7+3r)^2\), attempt at variance formula) | M1 | 3.1a |
| mean \((\bar{x}) = 136\) | B1 | 1.1b |
| Way 1: \(\sum_{r=1}^{n}(7+3r)^2 = \sum_{r=1}^{n}(49+42r+9r^2) = 49n+42\cdot\frac{1}{2}n(n+1)+9\cdot\frac{1}{6}n(n+1)(2n+1)\) | M1 | 1.1b |
| Way 2: \(\sum_{r=1}^{n}(x_i-\bar{x})^2 = \sum_{r=1}^{n}(7+3r-\text{"136"})^2 = a\sum r^2 + b\sum r + c\sum 1\) | B1 | 1.1b |
| Way 1: \(\frac{\text{"2032690"}}{85} - 136^2 = \ldots\) or \(\frac{\text{"2032690"}}{84} - \frac{85}{84}\times 136^2 = \ldots\) | M1 | 1.1b |
| Way 2: \(\frac{\text{"460530"}}{85} = \ldots\) or \(\frac{\text{"460530"}}{84} = \ldots\) | ||
| So s.d. \(= \sqrt{5418} = 73.6\) (g) — Accept 74.0 (g) if sample s.d. used | A1 | 1.1b |
| (6) | (9 marks) |
## Question 6:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\bar{x} = \frac{1}{n}\sum_{r=1}^{n}(7+3r)$ | M1 | 1.1a |
| $\sum_{r=1}^{n}(7+3r) = 7n + 3\cdot\frac{n}{2}(n+1)$ | M1 | 1.1b |
| $\bar{x} = 7+\frac{3}{2}(n+1) = \frac{14+3n+3}{2} = \frac{1}{2}(3n+17)$ | A1* | 2.1 |
| **(3)** | | |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Correct overall strategy to find the variance or standard deviation (attempt at mean, attempt at $\sum(7+3r)^2$, attempt at variance formula) | M1 | 3.1a |
| mean $(\bar{x}) = 136$ | B1 | 1.1b |
| Way 1: $\sum_{r=1}^{n}(7+3r)^2 = \sum_{r=1}^{n}(49+42r+9r^2) = 49n+42\cdot\frac{1}{2}n(n+1)+9\cdot\frac{1}{6}n(n+1)(2n+1)$ | M1 | 1.1b |
| Way 2: $\sum_{r=1}^{n}(x_i-\bar{x})^2 = \sum_{r=1}^{n}(7+3r-\text{"136"})^2 = a\sum r^2 + b\sum r + c\sum 1$ | B1 | 1.1b |
| Way 1: $\frac{\text{"2032690"}}{85} - 136^2 = \ldots$ or $\frac{\text{"2032690"}}{84} - \frac{85}{84}\times 136^2 = \ldots$ | M1 | 1.1b |
| Way 2: $\frac{\text{"460530"}}{85} = \ldots$ or $\frac{\text{"460530"}}{84} = \ldots$ | | |
| So s.d. $= \sqrt{5418} = 73.6$ (g) — Accept 74.0 (g) if sample s.d. used | A1 | 1.1b |
| **(6)** | | **(9 marks)** |\begin{enumerate}
\item An art display consists of an arrangement of $n$ marbles.
\end{enumerate}
When arranged in ascending order of mass, the mass of the first marble is 10 grams. The mass of each subsequent marble is 3 grams more than the mass of the previous one, so that the $r$ th marble has mass $( 7 + 3 r )$ grams.\\
(a) Show that the mean mass, in grams, of the marbles in the display is given by
$$\frac { 1 } { 2 } ( 3 n + 17 )$$
Given that there are 85 marbles in the display,\\
(b) use the standard summation formulae to find the standard deviation of the mass of the marbles in the display, giving your answer, in grams, to one decimal place.
\hfill \mbox{\textit{Edexcel CP AS 2019 Q6 [9]}}