| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Periodic Sequences |
| Difficulty | Standard +0.8 This question requires understanding of modular arithmetic applied to sequences, proving periodicity by algebraic manipulation of a quadratic expression mod 10, and interpreting the result. While the algebra is manageable, the combination of modular arithmetic (not standard A-level), proof requirement, and conceptual understanding of periodic behavior makes this moderately challenging, especially for AS-level Further Maths students encountering these ideas. |
| Spec | 8.02e Finite (modular) arithmetic: integers modulo n |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | 𝑢 = = 3(n + 5)2 + 3(n + 5) + 7 = 3n2 + 33n + 97= |
| Answer | Marks |
|---|---|
| 𝑛 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | Expanding and considering at least one term mod 10 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | The sequence is periodic (with period 5) |
| Answer | Marks |
|---|---|
| possibilities. | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.4 | Allow repeating/cyclic |
Question 2:
2 | (a) | 𝑢 = = 3(n + 5)2 + 3(n + 5) + 7 = 3n2 + 33n + 97=
𝑛 + 5
= 3n2 + 3n + 7 (mod 10) 𝑢 (mod 10)
𝑛 | M1
A1
[2] | 3.1a
1.1 | Expanding and considering at least one term mod 10
or evaluating 𝑢 from n=1 to n=10
𝑛
Correct conclusion from correct algebraic work
2 | (b) | The sequence is periodic (with period 5)
Any shorter periodicity would have to be a factor of 5. Since
the sequence is not constant, there are no smaller
possibilities. | B1
B1
[2] | 1.1
2.4 | Allow repeating/cyclic
Allow statement only that sequence is non-constant or
sight of {3, 5, (3, 7, 7, …)}.2 For all positive integers $n$, the terms of the sequence $\left\{ u _ { n } \right\}$ are given by the formula $u _ { n } = 3 n ^ { 2 } + 3 n + 7 ( \bmod 10 )$.
\begin{enumerate}[label=(\alph*)]
\item Show that $u _ { n + 5 } = u _ { n }$ for all positive integers $n$.
\item Hence describe the behaviour of the sequence, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2023 Q2 [4]}}