| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sigma notation: arithmetic series evaluation |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic series question requiring basic sigma notation interpretation and simultaneous equations. Part (a) involves simple substitution (r=5 gives first term 21, common difference is 4, count gives 16 terms). Part (b) requires setting up two equations from given conditions and solving—standard A-level technique with no conceptual challenges or novel problem-solving required. |
| Spec | 1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| 10 |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| First term \(= 21\) | B1 (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Common difference \(= 4\) | B1 (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Number of terms \(= 16\) | B1 (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n=91\), \(a=10b+c\), \(d=b\), \(L=100b+c\) | B1 (AO 1.1b) | Finds/uses at least one of: first term, common difference, last term, number of terms correctly; or expresses as difference of two series for \(n=1\) to 100 and \(n=1\) to 9 |
| \(\dfrac{91}{2}(2(10b+c)+90b) = 7735\) leading to \(91(55b+c)=7735\) hence \(55b+c=85\) | M1 (AO 3.1a) | Forms equation in \(b\) and \(c\) using their first term, number of terms, and either their common difference or last term |
| \(5050b + 100c - 45b - 9c = 7735\) or \(5005b + 91c = 7735\) | A1 (AO 1.1b) | Correct equation (ACF) |
| Completes rigorous argument to show required result, including at least one single step of correct working between initial correct formula and given answer | R1 (AO 2.1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4(11b + c) = 49b + c\), leading to \(5b - 3c = 0\) | B1 | Uses or writes down \(a + 39d\) or \(a + d\) with their expressions for \(a\) and \(d\); must be in terms of \(b\) and \(c\) |
| \(b = 1.5\), \(c = 2.5\) | M1 | Uses their \(a + 39d\) and \(a + d\) consistently to form equation \(u_{40} = 4u_2\) in terms of \(b\) and \(c\); condone use of \(50b + c\) for fortieth term; condone \(11b + c = 4(49b + c)\) |
| M1 | Solves \(55b + c = 85\) with their other equation involving \(b\) and \(c\); PI by obtaining correct values of \(b\) and \(c\); or obtains \(b = -12.75\) and \(c = 786.25\) from using \(11b + c = 4(49b + c)\) | |
| Correct values of \(b\) and \(c\) | A1 |
## Question 10(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| First term $= 21$ | B1 (AO 1.1b) | |
---
## Question 10(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Common difference $= 4$ | B1 (AO 1.1b) | |
---
## Question 10(a)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Number of terms $= 16$ | B1 (AO 1.1b) | |
---
## Question 10(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=91$, $a=10b+c$, $d=b$, $L=100b+c$ | B1 (AO 1.1b) | Finds/uses at least one of: first term, common difference, last term, number of terms correctly; or expresses as difference of two series for $n=1$ to 100 and $n=1$ to 9 |
| $\dfrac{91}{2}(2(10b+c)+90b) = 7735$ leading to $91(55b+c)=7735$ hence $55b+c=85$ | M1 (AO 3.1a) | Forms equation in $b$ and $c$ using their first term, number of terms, and either their common difference or last term |
| $5050b + 100c - 45b - 9c = 7735$ or $5005b + 91c = 7735$ | A1 (AO 1.1b) | Correct equation (ACF) |
| Completes rigorous argument to show required result, including at least one single step of correct working between initial correct formula and given answer | R1 (AO 2.1) | |
## Question 10(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4(11b + c) = 49b + c$, leading to $5b - 3c = 0$ | B1 | Uses or writes down $a + 39d$ or $a + d$ with their expressions for $a$ and $d$; must be in terms of $b$ and $c$ |
| $b = 1.5$, $c = 2.5$ | M1 | Uses their $a + 39d$ and $a + d$ consistently to form equation $u_{40} = 4u_2$ in terms of $b$ and $c$; condone use of $50b + c$ for fortieth term; condone $11b + c = 4(49b + c)$ |
| | M1 | Solves $55b + c = 85$ with their other equation involving $b$ and $c$; PI by obtaining correct values of $b$ and $c$; or obtains $b = -12.75$ and $c = 786.25$ from using $11b + c = 4(49b + c)$ |
| Correct values of $b$ and $c$ | A1 | |
---10
\begin{enumerate}[label=(\alph*)]
\item An arithmetic series is given by
$$\sum _ { r = 5 } ^ { 20 } ( 4 r + 1 )$$
10 (a) (i) Write down the first term of the series.\\
10 (a) (ii) Write down the common difference of the series.\\
10 (a) (iii) Find the number of terms of the series.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
10
\item & \begin{tabular}{l}
A different arithmetic series is given by \(\sum _ { r = 10 } ^ { 100 } ( b r + c )\) \\
where $b$ and $c$ are constants. \\
The sum of this series is 7735 \\
\end{tabular} & \\
\hline
\end{tabular}
\end{center}
10 (b) (ii) The 40th term of the series is 4 times the 2nd term.
Find the values of $b$ and $c$.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2020 Q10 [12]}}