June 2022 Paper 1 Q9

AQACurrent spec9 marksSequences & Series

9 The first three terms of an arithmetic sequence are given by

\[2x + 5 \qquad 5x + 1 \qquad 6x + 7\]

(a) Show that \(x = 5\) is the only value which gives an arithmetic sequence. [3 marks]

(b)

(i) Write down the value of the first term of the sequence. [1 mark]

(ii) Find the value of the common difference of the sequence. [1 mark]

(c) The sum of the first \(N\) terms of the arithmetic sequence is \(S_N\) where\[S_N \lt 100\,000\]\[S_{N+1} \gt 100\,000\]

Find the value of \(N\). [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Forms an appropriate equation in \(x\) only by either using the differences of at least one pair of terms or Using the mean of the first and third term = the second term Condone missing brackets or Forms two simultaneous equations in \(x\) and \(d\) or Substitutes \(x\) = 5 and demonstrates that the three terms obtained, 15, 26 and 37 have a common difference of 11 or Shows that the sum formula for an arithmetic series works when \(x\) = 5 The approaches that substitute \(x\) = 5 score a maximum of M1 A0 R0	M1	3.1a
Obtains a correct equation or Obtains two correct simultaneous equations in \(x\) and \(d\) Need not be simplified	A1	1.1b
Solves to conclude that \(x = 5\) is the only solution Must include the word ‘only’ OE	R1	2.1
	(3)

\[5x + 1 - (2x + 5) = 6x + 7 - (5x + 1)\]\[3x - 4 = x + 6\]\[x = 5\]

Therefore \(x = 5\) is the only solution

Mark scheme (c)

Scheme	Marks	AO
Forms an expression for the sum to \(N\) or \(N\) +1 terms using their \(a\) and \(d\) values Need not be simplified Condone missing brackets or use of \(n\) or Uses a trial and improvement method obtaining sums for two different values of \(n\)	M1	3.1a
Forms an equation or inequality using their expression and \(100000 \pm k\) where \(0 \leqslant k \leqslant 11\) or Uses trial and improvement to obtain one sum below 100000 and one sum above 100000 for consecutive integers	M1	1.1a
Obtains either 133.9.. or 132.9.. or \(N \gt 132\) or \(N \lt 134\) or Obtains the sum of 98553 when \(n\) = 133 and obtains the sum of 100031 when \(n\) = 134	A1	1.1b
Obtains 133 having solved a correct quadratic This mark can be recovered if \(N\) = 133 and \(N\) = 134 are correctly checked	A1	3.2a
	(4)
	(9 marks)

\[S_N = \frac{N}{2}\big(2 \times 15 + 11(N - 1)\big)\]\[\frac{N}{2}\big(2 \times 15 + 11(N - 1)\big) = 100000\]\[N = 133.9\ldots\]\[N = 133\]