| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find specific terms |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring simple substitution for part (a) and working backwards through two iterations for part (b). The arithmetic is trivial (dividing by 3 and adding 1), and the problem-solving demand is minimal—students just apply the given formula mechanically. Easier than average for A-level, though not completely trivial since part (b) requires reversing the process. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(u_4 = \frac{5+1}{3} = 2\) | B1 | |
| (b) \(5 = \frac{u_3+1}{3}\), \(u_2 = 14\) | M1 A1 | |
| \(14 = \frac{u_1+1}{3}\), \(u_1 = 41\) | A1 | (4) |
**(a)** $u_4 = \frac{5+1}{3} = 2$ | B1 |
**(b)** $5 = \frac{u_3+1}{3}$, $u_2 = 14$ | M1 A1 |
$14 = \frac{u_1+1}{3}$, $u_1 = 41$ | A1 | (4)A sequence is defined by the recurrence relation
$$u_{n+1} = \frac{u_n + 1}{3}, \quad n = 1, 2, 3, ...$$
Given that $u_3 = 5$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $u_4$, [1]
\item find the value of $u_1$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q2 [4]}}