A2 June 2019 Paper 2 Q4

AQACurrent spec3 marksSeries

4 The positive integer \(k\) is such that

\[\sum_{r=1}^{k} (3r - k) = 90\]

Find the value of \(k\). [3 marks]

Mark scheme

Scheme	Marks	AO
Removes \(\sum\) sign and creates an expression for the sum in terms of \(k\). Allow max. 1 error	M1	1.1a
Forms a correct quadratic equation	A1	1.1b
Obtains the correct answer. NMS 3/3 Treat trial and improvement as NMS	A1	1.1b
	(3 marks)

\[\sum_{r=1}^{k} (3r - k) = \frac{3}{2}k(k + 1) - k^2\]\[\frac{1}{2}k^2 + \frac{3}{2}k = 90\]\[k = 12\]