| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Recurrence Relations with Given Sum Condition |
| Difficulty | Standard +0.3 This is a straightforward recurrence relation question requiring systematic substitution to find u₂ and u₃, then algebraic manipulation using the given sum condition. The steps are mechanical (substitute, expand, collect terms) with no conceptual difficulty or novel insight required. Slightly easier than average due to its routine nature, though the algebra in part (b) requires careful bookkeeping. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_2 = b - 3a\) | B1 | oe |
| \(u_3 = b - a(b-3a)\) \(\left(= b - ab + 3a^2\right)\) | B1 | oe isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3 + b - 3a + b - a(b-3a) = 153\) oe | M1 | Attempt to add \(u_1\), \(u_2\), \(u_3\) and set equal to 153; condone copying slips; invisible brackets may be implied |
| Substitute \(b = a+9\): \(3+(a+9)-3a+(a+9)-a((a+9)-3a)=153 \Rightarrow 2a^2-10a-132=0\) | dM1 | Dependent on M1; substitute \(b=a+9\) and collect terms to 3TQ in \(a\) |
| \(\Rightarrow a^2 - 5a - 66 = 0\) | A1* | Given answer; no arithmetic errors; must see at least one intermediate stage |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = -6\) | B1 | Seen or used; ignore positive root |
| \(b = 9 + (-6) = 3 \Rightarrow u_2 = 3 - 3(-6) = 21\) or \(u_2 = 9 - 2a = 9 - 2\times(-6) = 21\) | M1, A1 | M1: substitute negative root into \(b=a+9\) then into \(u_2\) expression. A1: 21 only cao; if two answers reached must select 21; correct answer alone scores full marks |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_2 = b - 3a$ | B1 | oe |
| $u_3 = b - a(b-3a)$ $\left(= b - ab + 3a^2\right)$ | B1 | oe isw |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3 + b - 3a + b - a(b-3a) = 153$ oe | M1 | Attempt to add $u_1$, $u_2$, $u_3$ and set equal to 153; condone copying slips; invisible brackets may be implied |
| Substitute $b = a+9$: $3+(a+9)-3a+(a+9)-a((a+9)-3a)=153 \Rightarrow 2a^2-10a-132=0$ | dM1 | Dependent on M1; substitute $b=a+9$ and collect terms to 3TQ in $a$ |
| $\Rightarrow a^2 - 5a - 66 = 0$ | A1* | Given answer; no arithmetic errors; must see at least one intermediate stage |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = -6$ | B1 | Seen or used; ignore positive root |
| $b = 9 + (-6) = 3 \Rightarrow u_2 = 3 - 3(-6) = 21$ or $u_2 = 9 - 2a = 9 - 2\times(-6) = 21$ | M1, A1 | M1: substitute negative root into $b=a+9$ then into $u_2$ expression. A1: 21 only cao; if two answers reached must select 21; correct answer alone scores full marks |\begin{enumerate}
\item A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by
\end{enumerate}
$$\begin{aligned}
u _ { n + 1 } & = b - a u _ { n } \\
u _ { 1 } & = 3
\end{aligned}$$
where $a$ and $b$ are constants.\\
(a) Find, in terms of $a$ and $b$,\\
(i) $u _ { 2 }$\\
(ii) $u _ { 3 }$
Given
\begin{itemize}
\item $\sum _ { n = 1 } ^ { 3 } u _ { n } = 153$
\item $b = a + 9$\\
(b) show that
\end{itemize}
$$a ^ { 2 } - 5 a - 66 = 0$$
(c) Hence find the larger possible value of $u _ { 2 }$
\hfill \mbox{\textit{Edexcel P2 2023 Q11 [8]}}