AQA Paper 1 2021 June — Question 6 7 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2021
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeTwo related arithmetic progressions
DifficultyStandard +0.3 This is a straightforward two-part arithmetic sequence problem requiring standard formula application. Part (a) involves solving simultaneous equations using a₉=3 and S₂₁=42 with familiar formulas. Part (b) requires equating two sum formulas and solving a quadratic equation. All techniques are routine for AS-level with no novel insight needed, making it slightly easier than average.
Spec1.04h Arithmetic sequences: nth term and sum formulae

6
  1. The ninth term of an arithmetic series is 3 The sum of the first \(n\) terms of the series is \(S _ { n }\) and \(S _ { 21 } = 42\) Find the first term and common difference of the series.
    [0pt] [4 marks]
    6
  2. A second arithmetic series has first term - 18 and common difference \(\frac { 3 } { 4 }\) The sum of the first \(n\) terms of this series is \(T _ { n }\) Find the value of \(n\) such that \(T _ { n } = S _ { n }\) [0pt] [3 marks]

Question 6(a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(a + 8d = 3\)B1 (1.1b) OE
\(\frac{21}{2}(2a + 20d) = 42\)B1 (1.1b) OE
Solve simultaneously: \(a + 10d = 2\), \(a = 7\), \(d = -0.5\)M1 (3.1a) Elimination of one variable or better; condone slips \(a+9d=3\) or \(\frac{21}{2}(2a+20d)=21\)
Correct \(a\) and \(d\)A1 (1.1b)
Question 6(b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{n}{2}(14 - 0.5(n-1)) = \frac{n}{2}(-36 + 0.75(n-1))\) giving \(n = 0\) or \(41\)B1F (1.1b) At least one correct unsimplified expression for \(S_n\) or \(T_n\); FT non-zero values of \(a\) and \(d\); PI by simplified correct equation
Equates \(S_n\) and \(T_n\) expressionsM1 (3.1a) FT non-zero \(a\), \(d\); finds non-zero value of \(n\); PI by \(n = 41\)
\(n = 41\)R1 (2.2a) Deduces correct value
## Question 6(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + 8d = 3$ | B1 (1.1b) | OE |
| $\frac{21}{2}(2a + 20d) = 42$ | B1 (1.1b) | OE |
| Solve simultaneously: $a + 10d = 2$, $a = 7$, $d = -0.5$ | M1 (3.1a) | Elimination of one variable or better; condone slips $a+9d=3$ or $\frac{21}{2}(2a+20d)=21$ |
| Correct $a$ and $d$ | A1 (1.1b) | |

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## Question 6(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{n}{2}(14 - 0.5(n-1)) = \frac{n}{2}(-36 + 0.75(n-1))$ giving $n = 0$ or $41$ | B1F (1.1b) | At least one correct unsimplified expression for $S_n$ or $T_n$; FT non-zero values of $a$ and $d$; PI by simplified correct equation |
| Equates $S_n$ and $T_n$ expressions | M1 (3.1a) | FT non-zero $a$, $d$; finds non-zero value of $n$; PI by $n = 41$ |
| $n = 41$ | R1 (2.2a) | Deduces correct value |

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6
\begin{enumerate}[label=(\alph*)]
\item The ninth term of an arithmetic series is 3

The sum of the first $n$ terms of the series is $S _ { n }$ and $S _ { 21 } = 42$\\
Find the first term and common difference of the series.\\[0pt]
[4 marks]\\

6
\item A second arithmetic series has first term - 18 and common difference $\frac { 3 } { 4 }$\\
The sum of the first $n$ terms of this series is $T _ { n }$\\
Find the value of $n$ such that $T _ { n } = S _ { n }$\\[0pt]
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2021 Q6 [7]}}