| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find parameter from given term |
| Difficulty | Moderate -0.8 This is a straightforward C1 recurrence relation question requiring simple substitution to find x₂ and x₃, followed by solving a quadratic equation. All steps are routine with no problem-solving insight needed—easier than average A-level questions. |
| 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 |
| \((x_2 =)\ a + 5\) | B1 | Accept \(a1+5\) or \(1\times a+5\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x_3) = a''(a+5)''+5\) | M1 | Must see \(a\)(their \(x_2\)) \(+ 5\) |
| \(= a^2 + 5a + 5\) | A1cso | Must have seen \(a(a[1]+5)+5\); must have both brackets and no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(41 = a^2 + 5a + 5 \Rightarrow a^2 + 5a - 36(=0)\) or \(36 = a^2 + 5a\) | M1 | Forming suitable equation using \(x_3\) and 41, collecting like terms, reducing to 3TQ. Allow one sign error. If completing square should get \((a\pm\tfrac{5}{2})^2 = 36 + \tfrac{25}{4}\) |
| \((a+9)(a-4) = 0\) | M1 | Attempting to solve their 3TQ |
| \(a = 4\) or \(-9\) | A1 | Both 4 and \(-9\). No working or T&I leading to both answers scores 3/3 but no marks for only one answer |
## Question 4:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x_2 =)\ a + 5$ | B1 | Accept $a1+5$ or $1\times a+5$ etc |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x_3) = a''(a+5)''+5$ | M1 | Must see $a$(their $x_2$) $+ 5$ |
| $= a^2 + 5a + 5$ | A1cso | Must have seen $a(a[1]+5)+5$; must have both brackets and no incorrect working |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $41 = a^2 + 5a + 5 \Rightarrow a^2 + 5a - 36(=0)$ or $36 = a^2 + 5a$ | M1 | Forming suitable equation using $x_3$ and 41, collecting like terms, reducing to 3TQ. Allow one sign error. If completing square should get $(a\pm\tfrac{5}{2})^2 = 36 + \tfrac{25}{4}$ |
| $(a+9)(a-4) = 0$ | M1 | Attempting to solve their 3TQ |
| $a = 4$ or $-9$ | A1 | Both 4 and $-9$. No working or T&I leading to both answers scores 3/3 but no marks for only one answer |
---4. A sequence $x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots$ is defined by
$$\begin{aligned}
x _ { 1 } & = 1 \\
x _ { n + 1 } & = a x _ { n } + 5 , \quad n \geqslant 1
\end{aligned}$$
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for $x _ { 2 }$ in terms of $a$.
\item Show that $x _ { 3 } = a ^ { 2 } + 5 a + 5$
Given that $x _ { 3 } = 41$
\item find the possible values of $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2012 Q4 [6]}}