| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Simple recurrence evaluation |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring only iterative calculation and pattern recognition. Part (a)(i) needs just two iterations, (a)(ii) requires spotting a cycle (which emerges quickly), and (b) asks for another value that enters the same cycle—all routine procedures with no conceptual depth or proof required. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_2 = -1\), \(u_3 = 2\) | M1 (AO 1.1a) | Substitutes 2 into formula correctly to obtain \(u_2 = -1\); PI by correct \(u_3 = 2\) |
| Obtains correct \(u_3 = 2\) with no further working resulting in a contradictory value for \(u_3\) | A1 (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_{50} = -1\) | B1 (AO 2.2a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_1 = -2\) | B1 (AO 2.2a) | Accept any correct value e.g. \(\sqrt{2}\) or \(-\sqrt{2}\); condone if \(\pm 2\) seen |
## Question 7(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_2 = -1$, $u_3 = 2$ | M1 (AO 1.1a) | Substitutes 2 into formula correctly to obtain $u_2 = -1$; PI by correct $u_3 = 2$ |
| Obtains correct $u_3 = 2$ with no further working resulting in a contradictory value for $u_3$ | A1 (AO 1.1b) | |
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## Question 7(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_{50} = -1$ | B1 (AO 2.2a) | |
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## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_1 = -2$ | B1 (AO 2.2a) | Accept any correct value e.g. $\sqrt{2}$ or $-\sqrt{2}$; condone if $\pm 2$ seen |
---7 Consecutive terms of a sequence are related by
$$u _ { n + 1 } = 3 - \left( u _ { n } \right) ^ { 2 }$$
7
\begin{enumerate}[label=(\alph*)]
\item In the case that $u _ { 1 } = 2$\\
7 (a) (i) Find $u _ { 3 }$
7 (a) (ii) Find $u _ { 50 }$
7
\item State a different value for $u _ { 1 }$ which gives the same value for $u _ { 50 }$ as found in part (a)(ii).
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2020 Q7 [4]}}