| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Periodic or repeating sequence |
| Difficulty | Moderate -0.8 This question tests basic arithmetic with a repeating pattern. Part (i) requires simple modular arithmetic (55 mod 4) to identify position in the cycle, and part (ii) involves recognizing 13 complete cycles plus 3 terms, then calculating 13×(1+3+5+3) + (1+3+5). These are straightforward pattern-recognition tasks with minimal problem-solving demand, well below average A-level difficulty. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| 11.4 oe | 2 | M1 for \(\frac{12}{3} + \frac{12}{4} + \frac{12}{5} + \frac{12}{6}\) oe; M0 unless four terms summed |
## Question 12:
| 11.4 oe | 2 | **M1** for $\frac{12}{3} + \frac{12}{4} + \frac{12}{5} + \frac{12}{6}$ oe; **M0** unless four terms summed |12 Calculate $\sum _ { r = 3 } ^ { 6 } \frac { 12 } { r }$.
12 A sequence begins
$$\begin{array} { l l l l l l l l l l l }
1 & 3 & 5 & 3 & 1 & 3 & 5 & 3 & 1 & 3 & \ldots
\end{array}$$
and continues in this pattern.\\
(i) Find the 55th term of this sequence, showing your method.\\
(ii) Find the sum of the first 55 terms of the sequence.
\hfill \mbox{\textit{OCR MEI C2 Q12 [5]}}