AQA Further Paper 2 2019 June — Question 4 3 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeFinding n for given sum value
DifficultyStandard +0.3 This is a straightforward summation problem requiring splitting the sum into standard results (∑r and ∑1), then solving a quadratic equation. While it's Further Maths, the technique is mechanical and the algebra is simple, making it easier than average overall but not trivial due to the algebraic manipulation required.
Spec1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae
The positive integer \(k\) is such that $$\sum_{r=1}^{k} (3r - k) = 90$$ Find the value of \(k\). [3 marks]
Question 4:
AnswerMarks
4Removes ∑ sign and
creates an expression for
the sum in terms of k.
AnswerMarks Guidance
Allow max. 1 errorAO1.1a M1
3 2
�(3𝑟𝑟−𝑘𝑘)= 𝑘𝑘(𝑘𝑘+1)−𝑘𝑘
𝑟𝑟=1 2
1 2 3
𝑘𝑘 + 𝑘𝑘 = 90
2 2
𝑘𝑘 = 12
Forms a correct quadratic
AnswerMarks Guidance
equationAO1.1b A1
Obtains the correct
answer.
NMS 3/3
Treat trial and
AnswerMarks Guidance
improvement as NMSAO1.1b A1
Total3
QMarking Instructions AO 
Question 4:
4 | Removes ∑ sign and
creates an expression for
the sum in terms of k.
Allow max. 1 error | AO1.1a | M1 | 𝑘𝑘
3 2
�(3𝑟𝑟−𝑘𝑘)= 𝑘𝑘(𝑘𝑘+1)−𝑘𝑘
𝑟𝑟=1 2
1 2 3
𝑘𝑘 + 𝑘𝑘 = 90
2 2
𝑘𝑘 = 12
Forms a correct quadratic
equation | AO1.1b | A1
Obtains the correct
answer.
NMS 3/3
Treat trial and
improvement as NMS | AO1.1b | A1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
The positive integer $k$ is such that
$$\sum_{r=1}^{k} (3r - k) = 90$$

Find the value of $k$.
[3 marks]

\hfill \mbox{\textit{AQA Further Paper 2 2019 Q4 [3]}}
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Q1 1 Q2 1 Q3 1 Q4 3 Q5 4 Q6 6 Q7 6 Q8 9 Q9 13 Q10 7 Q11 8 Q12 5 Q13 10 Q14 12 Q15 14