| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Arithmetic progression with parameters |
| Difficulty | Moderate -0.3 This is a straightforward arithmetic sequence question requiring basic algebraic manipulation to find x (equating common differences), then standard formula application for sum of arithmetic series. The multi-part structure and final inequality adds slight complexity, but all techniques are routine A-level content with no novel problem-solving required. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5x+1-(2x+5) = 6x+7-(5x+1)\), giving \(3x-4=x+6\), so \(x=5\) | M1 | AO 3.1a. Forms equation in \(x\) only using differences of at least one pair of terms; or uses mean of first and third term = second term; or forms two simultaneous equations in \(x\) and \(d\). Approaches that substitute \(x=5\) score M1 A0 R0 |
| Correct equation obtained | A1 | AO 1.1b. Need not be simplified |
| Concludes \(x=5\) is the only solution | R1 | AO 2.1. Must include the word 'only' OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 15\) | B1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(d = 11\) | B1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_N = \frac{N}{2}(2\times15 + 11(N-1))\) | M1 | AO 3.1a. Forms expression for sum to \(N\) or \(N+1\) terms using their \(a\) and \(d\); need not be simplified |
| \(\frac{N}{2}(2\times15+11(N-1)) = 100000\) | M1 | AO 1.1a. Forms equation/inequality using expression and \(100000 \pm k\) where \(0 \leq k \leq 11\) |
| \(N = 133.9\ldots\), so \(N = 133\) | A1 | AO 1.1b. Obtains either \(133.9\ldots\) or \(132.9\ldots\); or \(N>132\) or \(N<134\) |
| \(N = 133\) confirmed | A1 | AO 3.2a. Obtains 133 having solved correct quadratic; recoverable if \(N=133\) and \(N=134\) correctly checked |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5x+1-(2x+5) = 6x+7-(5x+1)$, giving $3x-4=x+6$, so $x=5$ | M1 | AO 3.1a. Forms equation in $x$ only using differences of at least one pair of terms; or uses mean of first and third term = second term; or forms two simultaneous equations in $x$ and $d$. Approaches that substitute $x=5$ score M1 A0 R0 |
| Correct equation obtained | A1 | AO 1.1b. Need not be simplified |
| Concludes $x=5$ is the **only** solution | R1 | AO 2.1. Must include the word 'only' OE |
---
## Question 9(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 15$ | B1 | AO 1.1b |
---
## Question 9(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $d = 11$ | B1 | AO 1.1b |
---
## Question 9(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_N = \frac{N}{2}(2\times15 + 11(N-1))$ | M1 | AO 3.1a. Forms expression for sum to $N$ or $N+1$ terms using their $a$ and $d$; need not be simplified |
| $\frac{N}{2}(2\times15+11(N-1)) = 100000$ | M1 | AO 1.1a. Forms equation/inequality using expression and $100000 \pm k$ where $0 \leq k \leq 11$ |
| $N = 133.9\ldots$, so $N = 133$ | A1 | AO 1.1b. Obtains either $133.9\ldots$ or $132.9\ldots$; or $N>132$ or $N<134$ |
| $N = 133$ confirmed | A1 | AO 3.2a. Obtains 133 having solved correct quadratic; recoverable if $N=133$ and $N=134$ correctly checked |
---9 The first three terms of an arithmetic sequence are given by
$$2 x + 5 \quad 5 x + 1 \quad 6 x + 7$$
9
\begin{enumerate}[label=(\alph*)]
\item Show that $x = 5$ is the only value which gives an arithmetic sequence.\\
9
\item (i) Write down the value of the first term of the sequence.\\
9 (b) (ii) Find the value of the common difference of the sequence.\\
9
\item The sum of the first $N$ terms of the arithmetic sequence is $S _ { N }$ where
$$\begin{array} { r }
S _ { N } < 100000 \\
S _ { N + 1 } > 100000
\end{array}$$
Find the value of $N$.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2022 Q9 [9]}}