Cambridge IGCSE Mathematics - International 0607 — 2013 Oct/Nov Paper 5 · Variant 1

This document consists of 6 printed pages and 2 blank pages. IB13 11_0607_05/2RP © UCLES 2013 [Turn over *8541720444* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/05 Paper 5 (Core) October/November 2013 1 hour Candidates answer on the Question Paper Additional Materials: Graphics Calculator READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. Do not use staples, paper clips, highlighters, glue or correction fluid. You may use a pencil for any diagrams or graphs. DO NOT WRITE IN ANY BARCODES. Answer all the questions. You must show all relevant working to gain full marks for correct methods, including sketches. In this paper you will also be assessed on your ability to provide full reasons and communicate your mathematics clearly and precisely. At the end of the examination, fasten all your work securely together. The total number of marks for this paper is 24. www.XtremePapers.com

2 © UCLES 2013 0607/05/O/N/13 For Examiner's Use Answer all the questions. INVESTIGATION SUMS OF SEQUENCES Here is the method to construct a sequence for this investigation. Method Example Write down any two numbers for the first two terms. 3 and 7 Add these two terms to make the third term. 3 + 7 = 10 Add the second and third terms to make the fourth term. 7 + 10 = 17 Add the third and fourth terms to make the fifth term. 10 + 17 = 27 Continue in this way to construct the sequence. This example makes the sequence: 3, 7, 10, 17, 27, 44, 1 Show that the sum of the first six terms in this sequence, divided by the fifth term, is 4. 2 (a) The first two terms of a new sequence are 272 and 412. (i) Use the method to write down the next four terms in this sequence. 272, 412, , , , (ii) Work out the sum of the first six terms in this sequence and divide it by the fifth term.

3 © UCLES 2013 0607/05/O/N/13 [Turn over For Examiner's Use (b) The first two terms of another sequence are 4.52 and 16.9 . (i) Use the method to write down the next four terms in this sequence. Do not round any of your numbers. 4.52, 16.9, , , , (ii) Work out the sum of the first six terms in this sequence and divide it by the fifth term. (c) (i) Choose two negative numbers to be the first two terms of a sequence. Use the method to work out the next four terms in this sequence. Write down the first six terms in your sequence. , , , , , (ii) Work out the sum of the first six terms in your sequence and divide it by the fifth term. (d) Describe the connection between the fifth term and the sum of the first six terms in each of these sequences.

4 © UCLES 2013 0607/05/O/N/13 For Examiner's Use 3 The first two terms of a new sequence are p and q. The table shows the working for the first five terms. Working Term p p q q p + q p + q q + p + q p + 2q p + q + p + 2q 2p + 3q (a) Complete the table. (b) Find an expression for the sum of these first six terms. Simplify your answer. (c) Find an equation to connect the fifth term and the sum of the first six terms.

5 © UCLES 2013 0607/05/O/N/13 [Turn over For Examiner's Use 4 (a) 3, 7, 10, 17, 27, 44, (i) Write down the next four terms in this sequence. 3, 7, 10, 17, 27, 44, , , , (ii) Work out the sum of the first ten terms in this sequence and divide it by the seventh term. (b) (i) Find the next four terms in the sequence in question 3. 7th term 8th term 9th term 10th term (ii) Find an expression for the sum of the first ten terms in this sequence. Simplify your answer. (iii) Find an equation to connect the seventh term and the sum of the first ten terms.

6 © UCLES 2013 0607/05/O/N/13 For Examiner's Use 5 (a) Find the next four terms in the sequence in question 4(b). 11th term 12th term 13th term 14th term (b) Find an expression for the sum of the first fourteen terms. Simplify your answer. (c) This sum is a multiple of one of the terms in question 4(b)(i). Find this multiple. (d) Describe this connection using algebra.

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CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2013 series 0607 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/05 Paper 5 (Core), maximum raw mark 24 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.XtremePapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0607 05 © Cambridge International Examinations 2013 1 108 ÷ 27 [= 4] 1 2 (a) (i) (ii) (b) (i) (ii) (c) (i) (ii) (d) 684, 1096, 1780, 2876 4 www 21.42, 38.32, 59.74, 98.06 4 www Candidates own negative sequence correct 4 www 5th term = sum of first 6 terms divided by 4 OR sum of first 6 terms divided by 5th term = 4 OR 5th term multiplied by 4 = sum of first 6 terms ORthe 5th term is always 4 times smaller than the sum of the first 6 terms oe 1 1FT 1 1FT 1 1 1 FT their total ÷ their 5th term FT their total ÷ their 5th term 3 (a) (b) (c) p + 2q + 2p + 3q 3p + 5q 8p + 12q oe isw OR 5p + 7q plus their 3p + 5q 2p + 3q = 4 12 8 q p + OR 4(2p + 3q) = 8p + 12q OR 4 3 2 12 8 = + + q p q p 1,1 1FT 1 Accept different order FT their 6th term in 3(a) C opportunity

Page 3 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0607 05 © Cambridge International Examinations 2013 4 (a) (i) (ii) (b) (i) (ii) (iii) 71, 115, 186, 301 11 www 5p + 8q 8p + 13q 13p + 21q 21p + 34q 55p + 88q oe isw 5p + 8q = 11 88 55 q p + OR 11(5p + 8q) = 55p + 88q OR 11 8 5 88 55 = + + q p q p 1 1FT 2FT 1 1 FT their sum ÷ their 7th term FT their previous 6th term in p and q in 3(a) B1 for any two correct including after incorrect FT C opportunity 5 (a) (b) (c) (d) 34p + 55q, 55p + 89q, 89p + 144q, 144p + 233q 377p + 609q oe isw 29 soi 13p + 21q = (377p + 609q) ÷ 29 OR (377p + 609q) ÷ (13p + 21q) = 29 OR 29(13p + 21q) = 377p + 609q oe 2FT 1 1 1 FT their previous 9th and 10th terms in p and q in 4(b)(i) B1 for any two correct including after incorrect FT C opportunity C opportunity Communication seen in one of 3(b) 4(b)(ii) 5(b) 5(c) 1 Total 24