| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find n given sum condition |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic sequence question requiring basic recurrence relation evaluation, pattern recognition, and summation formula application. Part (i) is simple iteration, parts (ii-iii) are direct recognition, and part (iv) uses standard arithmetic series formulas with simple algebra—all routine C2 techniques with no novel problem-solving required. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(u_5 = 8 + 4 \times 3\) | M1 |
| \(= 20\) A.G. | A1 | Obtain 20 |
| (ii) | \(u_n = 3n + 5\) ie \(p = 3, q = 5\) | B1 |
| B1 | Obtain correct \(3n + 5\), or \(p = 3, q = 5\) stated | |
| (iii) | arithmetic progression | B1 |
| (iv) | \(\frac{N}{2}(16 + (2N - 1)3) - \frac{N}{2}(6 + (N-1)3) = 1256\) | M1 |
| \(26N + 12N^2 - 13N - 3N^2 = 2512\) | M1 | Attempt \(S_{2N}\), using any correct formula, with 2N consistent (inc \(\sum(3n+5)\)) |
| \(9N^2 + 13N - 2512 = 0\) | M1* | Attempt subtraction (correct order) and equate to 1256 |
| \((9N + 157)(N - 16) = 0\) | M1dep* | Attempt to solve quadratic in N |
| \(N = 16\) | A1 | Obtain N = 16 only, from correct working |
| OR | M1 | Attempt given difference as single summation with N terms |
| M1 | Attempt \(a = u_{N+1}\) | |
| M1 | Attempt \(l = u_{2N}\) | |
| M1 | Equate to 1256 and attempt to solve quadratic | |
| A1 | Obtain N = 16 only, from correct working |
(i) | $u_5 = 8 + 4 \times 3$ | M1 | Attempt $a + (n-1)d$ or equiv inc list of terms |
| $= 20$ A.G. | A1 | Obtain 20 |
(ii) | $u_n = 3n + 5$ ie $p = 3, q = 5$ | B1 | Obtain correct expression, poss unsimplified, eg $8 + 3(n-1)$ |
| | B1 | Obtain correct $3n + 5$, or $p = 3, q = 5$ stated |
(iii) | arithmetic progression | B1 | Any mention of arithmetic |
(iv) | $\frac{N}{2}(16 + (2N - 1)3) - \frac{N}{2}(6 + (N-1)3) = 1256$ | M1 | Attempt $S_N$ using any correct formula with 2N consistent (inc $\sum(3n+5)$) |
| $26N + 12N^2 - 13N - 3N^2 = 2512$ | M1 | Attempt $S_{2N}$, using any correct formula, with 2N consistent (inc $\sum(3n+5)$) |
| $9N^2 + 13N - 2512 = 0$ | M1* | Attempt subtraction (correct order) and equate to 1256 |
| $(9N + 157)(N - 16) = 0$ | M1dep* | Attempt to solve quadratic in N |
| $N = 16$ | A1 | Obtain N = 16 only, from correct working |
**OR** | M1 | Attempt given difference as single summation with N terms |
| | M1 | Attempt $a = u_{N+1}$ |
| | M1 | Attempt $l = u_{2N}$ |
| | M1 | Equate to 1256 and attempt to solve quadratic |
| | A1 | Obtain N = 16 only, from correct working |8 A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by
$$u _ { 1 } = 8 \quad \text { and } \quad u _ { n + 1 } = u _ { n } + 3 .$$
(i) Show that $u _ { 5 } = 20$.\\
(ii) The $n$th term of the sequence can be written in the form $u _ { n } = p n + q$. State the values of $p$ and $q$.\\
(iii) State what type of sequence it is.\\
(iv) Find the value of $N$ such that $\sum _ { n = 1 } ^ { 2 N } u _ { n } - \sum _ { n = 1 } ^ { N } u _ { n } = 1256$.
\hfill \mbox{\textit{OCR C2 2010 Q8 [10]}}