| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find n satisfying a condition |
| Difficulty | Easy -1.2 This is a straightforward arithmetic sequence application requiring only basic formula recall (nth term and sum formulas) with simple arithmetic. The context is clear, steps are routine, and no problem-solving insight is needed—just direct application of standard AS-level formulas. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4n + 6\) (e.g. \(10 + 4(n-1)\)) | B1 | Correct expression for number of lengths swum on \(n\)th day; ACF, can be unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4n + 6 = \frac{3000}{25}\) | M1 | Forms linear equation \(25 \times\) their part(a) \(= 3000\); OE; condone incorrect inequalities |
| \(n = 28.5\); rounds/truncates to nearest positive integer | M1 | Condone incorrect inequalities |
| \(n = 29\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{29}{2}(2\times10 + (29-1)\times4)\times25 = 47850\) | M1 | Uses correct formula for sum to \(n\) terms of AP substituting \(a=10\) and \(d=4\) or \(l=122\) |
| 47850 metres or 1914 lengths or AWRT 29.7 days | A1 | OE; condone missing units |
| 47850 < 50 000; therefore the coach is not correct | R1F | Makes appropriate explicit comparison; FT \(n=28\) only with comparisons 44800 < 50 000, 1792 < 2000, 29.7 > 28 or 30 > 28 (max M1 A0 R1F) |
## Question 5:
**Part 5(a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4n + 6$ (e.g. $10 + 4(n-1)$) | B1 | Correct expression for number of lengths swum on $n$th day; ACF, can be unsimplified |
**Part 5(b)(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4n + 6 = \frac{3000}{25}$ | M1 | Forms linear equation $25 \times$ their part(a) $= 3000$; OE; condone incorrect inequalities |
| $n = 28.5$; rounds/truncates to nearest positive integer | M1 | Condone incorrect inequalities |
| $n = 29$ | A1 | CAO |
**Part 5(b)(ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{29}{2}(2\times10 + (29-1)\times4)\times25 = 47850$ | M1 | Uses correct formula for sum to $n$ terms of AP substituting $a=10$ **and** $d=4$ or $l=122$ |
| 47850 metres or 1914 lengths or AWRT 29.7 days | A1 | OE; condone missing units |
| 47850 < 50 000; therefore the coach is not correct | R1F | Makes appropriate explicit comparison; FT $n=28$ only with comparisons 44800 < 50 000, 1792 < 2000, 29.7 > 28 or 30 > 28 (max M1 A0 R1F) |
---5 Ziad is training to become a long-distance swimmer.
He trains every day by swimming lengths at his local pool.\\
The length of the pool is 25 metres.\\
Each day he increases the number of lengths that he swims by four.\\
On his first day of training, Ziad swims 10 lengths of the pool.\\
5
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for the number of lengths Ziad will swim on his $n$th day of training.
5
\item (i) Ziad's target is to be able to swim at least 3000 metres in one day.\\
Determine the minimum number of days he will need to train to reach his target.\\
5 (b) (ii) Ziad's coach claims that when he reaches his target he will have covered a total distance of over 50000 metres.
Determine if Ziad's coach is correct.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q5 [7]}}