| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| 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 sequence question requiring only basic substitution to find terms, recognition of the period-2 pattern (alternating between 3 and -2), and simple arithmetic to sum 101 terms. The recurrence relation is given explicitly, no algebraic manipulation or proof is needed, and the summation is just counting pairs plus one extra term. Below average difficulty for A-level. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04f Sequence types: increasing, decreasing, periodic |
| Answer | Marks |
|---|---|
| 12(a) | −6 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| is obvious. | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 12(b) | States 2 | 2.2a |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 12(c) | Shows that pairs of consecutive |
| Answer | Marks | Guidance |
|---|---|---|
| sum of 2 s | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| = 1 | 2.2a | R1 |
| Subtotal | 2 | |
| Question 12 Total | 5 | |
| Q | Marking instructions | AO |
Question 12:
--- 12(a) ---
12(a) | −6
Substitutes u =3 into
1 u
n
PI u = − 2
2 | 1.1a | M1 | u =−2
2
u =3
3
u =−2
4
Obtains u = − 2 , u = 3 , u = − 2
2 3 4
Condone missing labels if order
is obvious. | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 12(b) ---
12(b) | States 2 | 2.2a | B1 | 2
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 12(c) ---
12(c) | Shows that pairs of consecutive
terms sum to 1 in a series
Or
Considers a sum of 3s and a
sum of 2 s | 3.1a | M1 | 1n 0 1
u = 3 −= 2 + 3 − 2 + ... + 3 − 2 + 3
n
= 1 1
= 5 0
= 53
1n 0 1
Deduces u = 53
n
= 1 | 2.2a | R1
Subtotal | 2
Question 12 Total | 5
Q | Marking instructions | AO | Marks | Typical solutionThe terms, $u_n$, of a periodic sequence are defined by
$$u_1 = 3 \quad \text{and} \quad u_{n+1} = \frac{-6}{u_n}$$
\begin{enumerate}[label=(\alph*)]
\item Find $u_2$, $u_3$ and $u_4$
[2 marks]
\item State the period of the sequence.
[1 mark]
\item Find the value of $\sum_{n=1}^{101} u_n$
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2024 Q12 [5]}}