| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sum of first n terms |
| Difficulty | Moderate -0.5 This is a straightforward arithmetic sequence question requiring basic formula application. Part (i) involves simple substitution into u_n = 7 + 4(n-1). Part (ii) requires recognizing the arithmetic series formula and comparing two sums, but the algebra is routine with no conceptual challenges. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(7 + 16 \times 4 = 71\) AG | M1 | Attempt to find 17th term in the given AP. Attempt to use \(u_n = a+(n-1)d\) with \(a=7\) and \(d=4\). Allow a more informal method, including writing out the sequence with \(a=7\) and \(d=4\). Could also attempt \(u_{17}\) from attempt at \(u_n = 4n+3\) – must be seen explicitly |
| A1 | Show clear detail to obtain \(u_{17} = 71\). If listing terms, 71 must either be last number in list or clearly identified e.g. underlined | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{35} = \frac{35}{2}(2 \times 7 + 34 \times 4) = 2625\) | M1 | Attempt sum of first 35 terms of given AP. Must use correct formula, with \(a=7\) and \(d=4\). If using \(\frac{1}{2}n(a+l)\) then must be valid attempt at \(l\). Could use \(4\sum n + \sum 3\), but M0 for \(4\sum n+3\) |
| A1 | Obtain 2625. Must be evaluated. Allow M1A1 for 2625 from no working | |
| either \(S_{50} = \frac{50}{2}(2 \times 7 + 49 \times 4) = 5250\), \(5250-2625=2625\) AG or \(S_{36-50} = \frac{15}{2}(2 \times 147 + 14 \times 4) = 2625\) AG | M1 | Attempt a correct method to show given relationship. Must show explicit calculation so M0 for just stating e.g. \(S_{50}=5250\). Could sum first 50 terms of AP and find the difference between this and the sum of the first 35 terms, or equiv. Could attempt to sum terms from \(u_{36}\) to \(u_{50}\) but M0 if summing from \(u_{35}\) (= 143) |
| A1 | Show given equality convincingly. No need for explicit conclusion once both sums shown to be 2625 | |
| [4] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $7 + 16 \times 4 = 71$ **AG** | M1 | Attempt to find 17th term in the given AP. Attempt to use $u_n = a+(n-1)d$ with $a=7$ and $d=4$. Allow a more informal method, including writing out the sequence with $a=7$ and $d=4$. Could also attempt $u_{17}$ from attempt at $u_n = 4n+3$ – must be seen explicitly |
| | A1 | Show clear detail to obtain $u_{17} = 71$. If listing terms, 71 must either be last number in list or clearly identified e.g. underlined |
| | [2] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{35} = \frac{35}{2}(2 \times 7 + 34 \times 4) = 2625$ | M1 | Attempt sum of first 35 terms of given AP. Must use correct formula, with $a=7$ and $d=4$. If using $\frac{1}{2}n(a+l)$ then must be valid attempt at $l$. Could use $4\sum n + \sum 3$, but M0 for $4\sum n+3$ |
| | A1 | Obtain 2625. Must be evaluated. Allow M1A1 for 2625 from no working |
| either $S_{50} = \frac{50}{2}(2 \times 7 + 49 \times 4) = 5250$, $5250-2625=2625$ **AG** or $S_{36-50} = \frac{15}{2}(2 \times 147 + 14 \times 4) = 2625$ **AG** | M1 | Attempt a correct method to show given relationship. Must show explicit calculation so M0 for just stating e.g. $S_{50}=5250$. Could sum first 50 terms of AP **and** find the difference between this and the sum of the first 35 terms, or equiv. Could attempt to sum terms from $u_{36}$ to $u_{50}$ but M0 if summing from $u_{35}$ (= 143) |
| | A1 | Show given equality convincingly. No need for explicit conclusion once both sums shown to be 2625 |
| | [4] | |
---2 A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by
$$u _ { 1 } = 7 \text { and } u _ { n + 1 } = u _ { n } + 4 \text { for } n \geqslant 1 .$$
(i) Show that $u _ { 17 } = 71$.\\
(ii) Show that $\sum _ { n = 1 } ^ { 35 } u _ { n } = \sum _ { n = 36 } ^ { 50 } u _ { n }$.
\hfill \mbox{\textit{OCR C2 2013 Q2 [6]}}