Sequences — Paper 4 · IGCSE Mathematics 0580

9 The table shows some terms of several sequences. Term 1 2 3 4 8 Sequence P 7 5 3 1 p Sequence Q 1 8 27 64 q 1 2 3 4 Sequence R r 2 3 4 5 Sequence S 4 9 16 25 s Sequence T 1 3 9 27 t Sequence U 3 6 7 −2 u (a) Find the values of p, q, r, s, t and u. [6] (b) Find the nth term of sequence (i) P, [1] (ii) Q, [1] (iii) R, [1] (iv) S, [1] (v) T, [1] (vi) U. [1] (c) Which term in sequence P is equal to −777? [2] (d) Which term in sequence T is equal to 177 147? [2]

16 marks

Mark scheme: 9 (a) -7, 512, 8 , 81, 2187, -2106 B6 B1 each. Allow in any order ignore letters 9 (b) (i) (P) 9 – 2n B1 Accept correct expressions in any form e.g. 7 – 2(n – 1) (ii) (Q) n3 B1 If ‘n =’ withhold the first mark earned (iii) n B1 (R) n + 1 (iv) (S) (n + 1)2 B1 (v) n − 1 B1 (T) 3 (vi) (U) (n + 1)2 - 3 n − 1 B1ft their (iv)-their (v) dep on both algebraic expressions (c) their(b)(i) = – 777 M1 393 cao A1 www2 (d) 12 B2 SC1 for 11 or n - 1 = 11 or 312 3,11 seen [16]

This question in 0580/41 Oct/Nov 2007

11 For Examiner's Use Diagram 1 Diagram 2 Diagram 3 Diagram 4 The first four terms in a sequence are 1, 3, 6 and 10. They are shown by the number of dots in the four diagrams above. (a) Write down the next four terms in the sequence. Answer(a) , , , [2] (b) (i) The sum of the two consecutive terms 3 and 6 is 9. The sum of the two consecutive terms 6 and 10 is 16. Complete the following statements using different pairs of terms. The sum of the two consecutive terms and is . The sum of the two consecutive terms and is . [1] (ii) What special name is given to these sums? Answer(b)(ii) [1] n ( n +1) (c) (i) The formula for the nth term in the sequence 1, 3, 6, 10… is , k where k is an integer. Find the value of k. Answer(c)(i) k = [1] (ii) Test your formula when n = 4, showing your working. For Examiner's Answer (c)(ii) Use [1] (iii) Find the value of the 180th term in the sequence. Answer(c)(iii) [1] (d) (i) Show clearly that the sum of the nth and the (n + 1)th terms is (n + 1)2. Answer (d)(i) [3] (ii) Find the values of the two consecutive terms which have a sum of 3481. Answer(d)(ii) and [2]

12 marks

Mark scheme: 11 (a) 15, 21, 28, 36 B2 B1 for 3 correct (b) (i) 10 + 15 = 25, 15 + 21 = 36 etc B1 Any two complete and correct statements (ii) Square B1 (c) (i) 2 B1 4 × 5 (ii) = 10 o.e. E1 2 (iii) 16 290 c.a.o. B1 ( n + 1)( n + 2 ) n 2 + 3n + 2 (d) (i) or seen M1 Denominator could be their k 2 2 May be implied by next line n(n +)1 (n +1)(n + )2 n 2 + n n 2 + 3n + 2 + or + M1 This line must be seen and at least one 2 2 2 2 more step, without any error, to gain the E 2n2 + 4n + 2 mark ( n + )1 ( n + n + 2) 2 2 ( n + 1)(2n + 2) n2 + 2n +1 2 2 (n +)1 2( n + 1)(n + )1 = ( n + 2)1 E1 Dependent on M1M1. Fully established – 2 no errors (ii) 1711 and 1770 final answers c.a.o. B2 SC1 for 59 or 58 or 1711 or 1770 seen [12] Graph for Question 5 4444 3333 2222 1111 0000 -8-8-8-8 -7-7-7-7 -6-6-6-6 -5-5-5-5 -4-4-4-4 -3-3-3-3 -2-2-2-2 -1-1-1-1 0000 1111 2222 3333 4444 5555 6666 7777 8888 -1-1-1-1 -2-2-2-2 -3-3-3-3 4444

This question in 0580/41 May/June 2009

10 For Examiner's Total Use Row 1 1 = 1 Row 2 3 + 5 = 8 Row 3 7 + 9 + 11 = 27 Row 4 13 + 15 + 17 + 19 = 64 Row 5 Row 6 The rows above show sets of consecutive odd numbers and their totals. (a) Complete Row 5 and Row 6. [2] (b) What is the special name given to the numbers 1, 8, 27, 64…? Answer(b) [1] (c) Write down in terms of n, (i) how many consecutive odd numbers there are in Row n, Answer(c)(i) [1] (ii) the total of these numbers. Answer(c)(ii) [1] (d) The first number in Row n is given by n2 − n + 1. Show that this formula is true for Row 4. Answer(d) [1] (e) The total of Row 3 is 27. This can be calculated by (3 × 7) + 2 + 4. For Examiner's The total of Row 4 is 64. This can be calculated by (4 × 13) + 2 + 4 + 6. Use The total of Row 7 is 343. Show how this can be calculated in the same way. Answer(e) [1] (f) The total of the first n even numbers is n(n + 1). Write down a formula for the total of the first (n – 1) even numbers. Answer(f) [1] (g) Use the results of parts (d), (e) and (f) to show clearly that the total of the numbers in Row n gives your answer to part (c)(ii). Answer(g) [2]

10 marks

Mark scheme: 10 (a) 21 + 23 + 25 + 27 + 29 = 125 B1 31 + 33 + 35 + 37 + 39 + 41 = 216 B1 (b) Cubes B1 (c) (i) n oe B1 (ii) n3 oe B1 (d) 42 – 4 + 1 = 13 www E1 Allow 16 for 42, otherwise all must be seen (e) 7 × 43 + 2 + 4 + 6 + 8 + 10 + 12 B1 All must be seen (f) n(n – 1) final answer oe B1 (g) n(n2 – n + 1) + their (f) M1 n3 – n2 + n + n2 – n = n3 E1 All must be seen, no errors or omissions [10]

This question in 0580/41 Oct/Nov 2009

10 In all the following sequences, after the first two terms, the rule is to add the previous two terms to For find the next term. Examiner's Use (a) Write down the next two terms in this sequence. 1 1 2 3 5 8 13 [1] (b) Write down the first two terms of this sequence. 3 11 14 [2] (c) (i) Find the value of d and the value of e. 2 d e 10 Answer(c)(i) d = e = [3] (ii) Find the value of x, the value of y and the value of z. O33 x y z 18 Answer(c)(ii) x = y = z = [5]

11 marks

Mark scheme: 10 (a) 21 and 34 1 (b) –5 8 1 + 1 (c) (i) 4, 6 3 M1 for 2 + d = e oe or d + e = 10 oe seen and either M1 for a correct eqn in d or e seen e.g. 2e = 12 oe or 2d = 8 oe or B1 for either correct (ii) x = 28 5 B4 for any two correct y = –5 or M3 for any of 18 = 3x – 66 oe z = 23 or 3y + 33 = 18 oe or 33 – 3z = -36 oe or M1 for 2 of y = x – 33 oe or y + z = 18 oe or x + y = z oe and M1 for combining two of the previous equations correctly isw (does not have to be simplified) after 0 scored SC1 for –33 + their x = their y or their x + their y = their z or their y + their z = 18

This question in 0580/42 Oct/Nov 2010

10 The first and the nth terms of sequences A, B and C are shown in the table below. For Examiner's (a) Complete the table for each sequence. Use 1st term 2nd term 3rd term 4th term 5th term nth term Sequence A 1 n3 Sequence B 4 4n Sequence C 4 (n + 1)2 [5] (b) Find (i) the 8th term of sequence A, Answer(b)(i) [1] (ii) the 12th term of sequence C. Answer(b)(ii) [1] (c) (i) Which term in sequence A is equal to 15 625? Answer(c)(i) [1] (ii) Which term in sequence C is equal to 10 000? Answer(c)(ii) [1] (d) The first four terms of sequences D and E are shown in the table below. Use the results from part (a) to find the 5th and the nth terms of the sequences D and E. 1st term 2nd term 3rd term 4th term 5th term nth term Sequence D 5 16 39 80 Sequence E 0 1 4 9 [4]

13 marks

Mark scheme: 10 (a) (A 1) 8 27 64 125 2 B1 for 3 correct (B 4) 8 12 16 20 1 (C 4) 9 16 25 36 2 B1 for 3 correct (b) 512 1 169 1 (c) 25 1 99 1 (d) 145 n3 + 4n oe 1, 1 16 (n + 1)2 – 4n oe but isw 1, 1 Likely oe is (n – 1)2

This question in 0580/41 May/June 2011

9 (a) (i) Work out the first 3 terms of the sequence whose nth term is n(n + 2). For Examiner's Answer(a)(i) , , [2] Use (ii) Which term in this sequence is equal to 168? Answer(a)(ii) [3] (b) Find a formula for the nth term of the following sequences. (i) 5 8 11 14 17 …… Answer(b)(i) [2] (ii) 1 2 4 8 16 …… Answer(b)(ii) [2] (c) Diagram 1 Diagram 2 Diagram 3 A sequence of diagrams is formed by drawing equilateral triangles each of side one centimetre. Diagram 1 has 3 one centimetre lines. Diagram 2 has 9 one centimetre lines. The formula for the total number of one centimetre lines needed to draw all of the first n diagrams is an3 + bn2 + n. Find the values of a and b. Answer(c) a = b = [6]

15 marks

Mark scheme: 9 (a) (i) 3, 8, 15 in correct positions 2 B1 for 2 correct values in correct positions (ii) 12 3 M2 for 12 × (12 + 2) (= 168) or 12, (12 + 2) or M1 for n2 + 2n = 168 then M1 for (n + a)(n + b) where a and b are integers and ab = – 168 or a + b = 2 oe (b) (i) 2 + 3n oe 2 Allow unsimplified e.g. 5 + 3(n – 1) B1 for 3n oe seen (ii) 2n – 1 oe 2 B1 for 2k seen 1 1 (c) a = , b = 1 cao 6 B1 for 12 or 30 seen but if 30 clearly only from 2 2 Diagram 4 then B0. M1 for any 1 of a + b + 1 = 3 oe 8a + 4b + 2 = 12 oe 27a + 9b + 3 = 30 oe M1 for a 2nd of the above equations M1 (indep) for correctly eliminating a or b from pair of linear equations B1 for one correct value

This question in 0580/42 May/June 2011

12 (a) The nth term of a sequence is n(n+1). For Examiner's Use (i) Write the two missing terms in the spaces. 2, 6, , 20, [2] (ii) Write down an expression in terms of n for the (n + 1)th term. Answer(a)(ii) [1] (iii) The difference between the nth term and the (n + 1)th term is pn + q. Find the values of p and q. Answer(a)(iii) p = q = [2] (iv) Find the positions of the two consecutive terms which have a difference of 140. Answer(a)(iv) and [2] (b) A sequence u1, u2, u3, u4, …………. is given by the following rules. u1 = 2, u2 = 3 and un = 2u n − 2 + u n −1 for n [ 3. For example, the third term is u3 and u3 = 2u1 + u2 = 2 × 2 + 3 = 7. So, the sequence is 2, 3, 7, u4, u5, ….. (i) Show that u4 = 13. Answer(b)(i) [1] (ii) Find the value of u5 . Answer(b)(ii) u5 = [1] (iii) Two consecutive terms of the sequence are 3413 and 6827 . Find the term before and the term after these two given terms. Answer(b)(iii) , 3413, 6827, [2]

11 marks

Mark scheme: 12 (a) (i) 12, …, 30 2 B1 each (ii) (n + 1)(n + 2) oe 1 isw if expand incorrectly (iii) p = 2 1 q = 2 1 (iv) 69(th), 70(th) 2 M1 for their 2n + 2 = 140 soi (b) (i) 2 × 3 + 7 1 Accept 2 × 3 + 2 × 2 + 3 (ii) 27 1 (iii) 1707, …, 13 653 1,1

This question in 0580/43 Oct/Nov 2011

6 (a) For Examiner's 114° Use 2x° NOT TO SCALE x° (x – 10)° Find the value of x. Answer(a) x = [3] (b) (i) Write the four missing terms in the table for sequences A, B, C and D. Term 1 2 3 4 5 n Sequence A – 4 2 5 8 3n – 7 Sequence B 1 4 9 16 25 Sequence C 5 10 15 20 25 Sequence D 6 14 24 36 50 [4] (ii) Which term in sequence D is equal to 500? Answer(b)(ii) [2] x 2 − 16 (c) Simplify . 2 x 2 + 7 x − 4 Answer(c) [4]

13 marks

Mark scheme: 6 (a) (x =) 64 www 3 3 B2 for x + 2 x + x = 360 − 114 + 10 or better or M1 for x + 2 x + 114 + x − 10 = 360 (b) (i) –1 1 n 2 oe 1 5n oe 1 2 1 n + 5n oe (ii) 20 2 M1 for their n 2 + 5n = 500 or 20 and 25 seen x − 4 (c) Final answer cao www 4 4 B1 for (x – 4)(x + 4) 2 x − 1 B2 for (2x – 1)(x + 4) or SC1 for (2x + a)(x + b) where either a + 2b = 7 or ab = – 4

This question in 0580/42 May/June 2012

11 For Examiner's Use Diagram 1 Diagram 2 Diagram 3 The diagrams show a sequence of dots and circles. Each diagram has one dot at the centre and 8 dots on each circle. The radius of the first circle is 1 unit. The radius of each new circle is 1 unit greater than the radius of the previous circle. (a) Complete the table for diagrams 4 and 5. Diagram 1 2 3 4 5 Number of dots 9 17 25 Area of the largest circle π 4π 9π Total length of the circumferences of the circles 2π 6π 12π [4] (b) (i) Write down, in terms of n, the number of dots in diagram n. Answer(b)(i) [2] (ii) Find n, when the number of dots in diagram n is 1097. Answer(b)(ii) n = [2] (c) Write down, in terms of n and π, the area of the largest circle in (i) diagram n, Answer(c)(i) [1] (ii) diagram 3n. Answer(c)(ii) [1] (d) Find, in terms of n and π, the total length of the circumferences of the circles in diagram n. Answer(d) [2]

12 marks

Mark scheme: 11 (a) 33, 41 1 16 ,π 25 π 1 2 B1 each 20 ,π30 π (b) (i) 8 n + 1 oe final answer 2 e.g. 9 + 8(n –1), condone n = 8n + 1 SC1 for 8 n + k (ii) 137 www2 2 M1 for their (b)(i) = 1097 (c) (i) n 2 π oe final answer 1 (ii) 9 n 2 π oe final answer 1 Allow (3n ) 2 π (d) n ( n + )1 π oe final answer 2 SC1 for a quadratic expression e.g. n ( n + )1 , n2 + 5, n2 + n π

This question in 0580/43 May/June 2012

9 The fi rst four diagrams in a sequence are shown below. For Examiner′s Use Diagram 1 Diagram 2 Diagram 3 Diagram 4 The diagrams are made from dots ( ) and squares ( ) joined by lines. (a) Complete the table. Diagram 1 2 3 4 5 n Number of dots 6 9 12 1 Number of squares 0 1 3 n(n – 1) 2 Number of triangles 4 9 16 3 Number of lines 9 18 30 45 63 (n + 1)(n + 2) 2 [9] (b) Which diagram has 360 lines? Answer(b) … [2] (c) The total number of lines in the fi rst n diagrams is For Examiner′s Use 1 2 n3 + pn2 + qn. 1 (i) When n = 1, show that p + q = 8 . 2 Answer(c)(i) [1] (ii) By choosing another value of n and using the equation in part (c)(i), fi nd the values of p and q. Answer(c)(ii) p = … q = … [5] _____________________________________________________________________________________ Question 10 is printed on the next page.

17 marks

Mark scheme: 9 (a) 15 18 3n + 3 or 3(n + 1) 9 B2 for 15, 6, 25 6 10 or B1 for two correct values 25 36 (n + 1)2 B3 for 18, 10, 36 or B1 for each correct value B2 for 3n + 3 oe or M1 for 3n + k, for any k B2 for (n + 1)2 oe or M1 for a quadratic expression (b) 14 2 M1 for (n + 1)(n + 2) = 240 or better or 15 × 16 = 240 (c) (i) ½ + p + q = 9 1 (ii) [p = ] 3 5 B2 for 4p + 2q = 23 11 or B1 for ½ × 23 + p × 22 + q × 2 oe [q = ] 2 M1 for correct multiplication and subtraction of their equations 11 A1 for [p = ] 3 or [q = ] 2 If 0 scored then SC1 for either correct IGCSE – October/November 2013 0580 43 x

This question in 0580/43 Oct/Nov 2013

9 Diagram 1 Diagram 2 Diagram 3 Diagram 4 The fi rst four diagrams in a sequence are shown above. The diagrams are drawn using white squares and grey squares . (a) Complete the columns in the table for Diagram 4 and Diagram n. Diagram 1 2 3 4 n Number of white squares 12 20 28 Number of grey squares 0 1 4 Total number of squares 12 21 32 (n + 1)(n + 5) [6] (b) Work out the number of the diagram which has a total of 480 squares. Answer(b) … [2] (c) The total number of squares in the fi rst n diagrams is 1 3 n3 + pn2 + qn. 2 (i) Use n = 1 in this expression to show that p + q = 11 3 . Answer(c)(i) [1] 1 (ii) Use n = 2 in the expression to show that 4p + 2q = 30 3 . Answer(c)(ii) [2] (iii) Find the values of p and q. Answer(c)(iii) p = … q = … [3] __________________________________________________________________________________________

14 marks

Mark scheme: 9 (a) 36, 9, 45 2 B1 for two correct values 8n + 4 oe 2 M1 for 8n + k, for any k (n – 1)2 oe 2 M1 for a quadratic expression of form n2 [+ an + b ] oe (b) 19 2 M1 for (n + 1)(n + 5) = 480 or better or 20 × 24 seen 1 1 1 3 2 (c) (i) + p + q = 12 and no errors 1 Accept p + q = 12 – after [1 ] + p [1 ] + q []1 shown 3 3 3 seen 1 (ii) × 8 + 4p + 2q = 12 + 21 2 M1 for 12 + 21 seen or 33 seen 3 7 (iii) [p =] oe 3 M1 for correct multiplication and subtraction or 2 substitution using the correct given equations 49 7 49 [q =] oe B1 for [p =] or [q =] 6 2 6 After 0 scored, SC1 for 2 values satisfying one of the original correct given equations

This question in 0580/41 Oct/Nov 2014

9 Layer 1 Layer 2 Layer 3 The diagrams show layers of white and grey cubes. Khadega places these layers on top of each other to make a tower. (a) Complete the table for towers with 5 and 6 layers. Number of layers 1 2 3 4 5 6 Total number of white cubes 0 1 6 15 Total number of grey cubes 1 5 9 13 Total number of cubes 1 6 15 28 [4] (b) (i) Find, in terms of n, the total number of grey cubes in a tower with n layers. Answer(b)(i) … [2] (ii) Find the total number of grey cubes in a tower with 60 layers. Answer(b)(ii) … [1] (iii) Khadega has plenty of white cubes but only 200 grey cubes. How many layers are there in the highest tower that she can build? Answer(b)(iii) … [2] (c) The expression for the total number of white cubes in a tower with n layers is pn2 + qn + 3. Find the value of p and the value of q. Show all your working. Answer(c) p = … q = … [5] (d) Find an expression, in terms of n, for the total number of cubes in a tower with n layers. Give your answer in its simplest form. Answer(d) … [2] __________________________________________________________________________________________

16 marks

Mark scheme: 9 (a) 28 45 1, 1 17 21 1 45 66 1 (b) (i) 4n – 3 oe 2 M1 for 4n + k (ii) 237 1 (iii) 50 2FT FT their (b)(i) = 200 solved and then answer truncated dep on linear expression of form an + k M1 for their 4n – 3 = 200 or their 4n – 3 Y 200 (c) p = 2 and q = –5 with some 5 M2 for any 2 of p + q + 3 = 0 oe, correct supporting working 22 p + 2q + 3 = 1 oe, 32 p + 3q + 3 = 6 oe, leading to the solutions 42 p + 4q + 3 = 15 oe , 52 p + 5q + 3 = their 28 oe, etc. or M1 for any one of these M1 indep for correctly eliminating p or q from pair of linear equations A1 for one correct value If 0 scored SC1 for 2 values that satisfy one of their original equations After M0, 2 correct answers SC1 (d) 2n2 – n or n(2n – 1) 2 B1 for answer 2n2 + k[n] or M1 for their quadratic from (c) + their linear from (b)(i) 1 1 1

This question in 0580/42 Oct/Nov 2014

11 Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 1 shows two lines of length 1 unit at right angles forming an . Two s are added to Diagram 1 to make Diagram 2. This forms one small square. Three s are added to Diagram 2 to make Diagram 3. This forms three small squares. The sequence of Diagrams continues. (a) Draw Diagram 5. [1] (b) Complete the table. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Number of lines of length 1 unit 2 6 12 20 Number of small squares 0 1 3 6 [2] (c) Find an expression, in terms of n, for the number of lines of length 1 unit in Diagram n. Answer(c) … [2] (d) Find an expression, in terms of n, for the number of small squares in Diagram n. Answer(d) … [2]

7 marks

Mark scheme: 11 (a) 1 (b) 30 1 10 1 2 + bn + c a, b, c numeric a ≠ 0 (c) n (n + )1 oe 2 B1 for an 1 (d) n (n − )1 oe 2 1 2 B1 for using oe in expression of form 2 1 (an 2 + bn + c ) a ≠ 0 or kn(n − )1 k ≠ 0 2

This question in 0580/42 Feb/March 2015

11 The first four terms of sequences A, B, C and D are shown in the table. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term 1 2 3 4 A 3 4 5 6 B 3 4 5 6 C –1 0 1 2 D –3 0 5 12 (a) Complete the table. [8] 36 (b) Which term in sequence A is equal to 37 ? Answer(b) … [2] (c) Which term in sequence D is equal to 725? Answer(c) … [2]

12 marks

Mark scheme: 5 n 11 (a) oe 8 B1 each 7 n + 2 7 n + 2 oe 3 n − 2 oe 21 n 2 − 4 oe 72 n 36 (b) 72 2 M1 for or their = 74 n + 2 37 (c) 27 2 M1 for their (n2 – 4) = 725 or 25 × 29 [= 725]

This question in 0580/43 May/June 2015

10 The table shows the first five terms of sequences A, B and C. Sequence 1st term 2nd term 3rd term 4th term 5th term 6th term A 3 4 5 6 7 B 0 1 4 9 16 C –3 –3 –1 3 9 (a) Complete the table for the 6th term of each sequence. [2] (b) Write down the nth term of sequence A. Answer(b) … [1] (c) (i) Find the nth term of sequence B. Answer(c)(i) … [2] (ii) Find the value of n when the nth term of sequence B is 8281. Answer(c)(ii) n = … [2] (d) (i) Find the nth term of sequence C in its simplest form. Answer(d)(i) … [2] (ii) Find the 8th term of sequence C. Answer(d)(ii) … [1] n - 1 1 (e) The nth term of another sequence D is - . e 2 o Complete the table for the first four terms of sequence D. Sequence 1st term 2nd term 3rd term 4th term D [3]

13 marks

Mark scheme: 10 (a) 8 2 B1 for 2 correct 25 17 (b) n + 2 oe 1 (c) (i) (n − 1)2 oe 2 M1 for (n + k)2 for integer k (ii) 92 2 M1 for 8281 or 91 seen (d) (i) n2 − 3n − 1 final answer 2 M1 for their (n − 1)2 − their (n + 2) soi (ii) 39 1 (e) 1 1 and −2 oe 1 1 oe 1 4 1 oe 1 −8

This question in 0580/42 Oct/Nov 2015

10 Complete the table for each sequence. Sequence 1st term 2nd term 3rd term 4th term 5th term 6th term nth term A 15 8 1 –6 5 6 7 8 B 18 19 20 21 C 2 5 10 17 D 2 6 18 54 [11]

11 marks

Mark scheme: 10 A –13, –20 1 –7n + 22 oe 2 SC1 for –7n + k or kn + 22 oe 9 10 B , 1 22 23 n + 4 oe 2 B1 for n + 4 oe or n + 17 oe seen, but not in n + 17 wrong position C 26, 37 1 n2 + 1 oe 1 D 162, 486 1 2 × 3 n −1 oe 2 SC1 for k × 3 n + p [k, p integers] 3 n Accept 2 × 3

This question in 0580/43 Oct/Nov 2015

10 Diagram 1 Diagram 2 Diagram 3 Each diagram is made from tiles in the shape of equilateral triangles and rhombuses. The length of a side of each tile is 1 unit. (a) Complete the table below for this sequence of diagrams. Diagram 1 2 3 4 5 Number of equilateral triangle shaped tiles 2 3 4 5 6 Number of rhombus shaped tiles 1 3 6 Total number of tiles 3 6 10 Number of 1 unit lengths 8 15 24 [6] (b) (i) The number of 1 unit lengths in Diagram n is n2 + 4n + p. Find the value of p. p = … [2] (ii) Calculate the number of 1 unit lengths in Diagram 10. … [1] (c) The total number of tiles in Diagram n is an2 + bn + 1. Find the value of a and the value of b. a = … b = … [5] (d) Part of the Louvre museum in Paris is in the shape of a square-based pyramid made from glass tiles. Each of the triangular faces of the pyramid is represented by Diagram 17 in the sequence. (i) Calculate the total number of glass tiles on one triangular face of this pyramid. … [2] (ii) 11 tiles are removed from one of the triangular faces to create an entrance into the pyramid. Calculate the total number of glass tiles used to construct this pyramid. … [1]

17 marks

Mark scheme: 10 (a) 10 15 15 21 35 48 6 B1 for each correct entry (b) (i) 3 2 M1 for any correct substitution in n2 + 4n + p = number of tiles eg 22 + 4(2) + p = 15 (ii) 143 1FT FT 140 + their (b)(i) 1 3 (c) a = oe b = oe nfww 5 B1 for a correct simplified equation 2 2 e.g. a + b + 1 = 3, 4a + 2b + 1 = 6, 9a + 3b + 1 = 10 etc B1 for a 2nd correct simplified equation M1 for correctly eliminating one variable for their equations in a and b 1 A1 for a = nfww 2 3 A1 for b = nfww 2 (d) (i) 171 2FT FT their a ×172 + their b ×17 + 1 M1 for their a ×172 + their b ×17 + 1 (ii) 673 1FT FT their (d)(i) × 4 – 11

This question in 0580/43 May/June 2016

10 (a) Complete the table for the four sequences A, B, C and D. Sequence Next term nth term A 2 5 8 11 B 20 14 8 2 C 1 4 9 16 D 0 2 6 12 [10] n ^3 n+ 1h (b) The sum of the first n terms of a sequence is 2 . (i) When the sum of the first n terms is 155, show that 3n2 + n – 310 = 0. [2] (ii) Solve 3n2 + n – 310 = 0. n = … or n = … [3] (iii) Complete the statement. The sum of the first … terms of this sequence is 155. [1]

16 marks

Mark scheme: 10 (a) A: 14 3n – 1 oe 3 B1 for 14 B2 for 3n – 1 oe or M1 for 3n + k, for any k oe B: –4 26 – 6n oe 3 B1 for –4 B2 for 26 – 6n oe or M1 for k – 6n, for any k oe C: 25 n2 oe 2 B1 for 25 B1 for n2 oe D: 20 n2 – n oe 2 B1 for 20 B1 for n2 – n oe 2 + n n ( 3n + 1) 3n (b) (i) = 155 M1 Accept = 155 2 2 3n2 + n = 310 Intermediate step must include elimination of fraction eg n(3n + 1) = 310 3n2 + n – 310 = 0 A1 With no errors or omissions 31 (ii) 10 , – oe 3 M2 for (3n + 31)(n – 10) [= 0] 3 or M1 for 3n(n – 10) + 31(n – 10) or n(3n + 31) – 10(3n + 31) or (3n + a)(n + b) where ab = –310 or a + 3b = 1 (iii) 10 1FT FT their b(ii) if only one positive integer solution

This question in 0580/43 Oct/Nov 2016

11 On Monday, Ankuri sent this text message to two friends. Today is Day Number 1. Tomorrow, please add 1 to the Day Number and send this text message to two friends. All the friends who receive a text message follow the instructions. (a) Complete the table. Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday Day Number 1 2 3 Number of text messages sent 2 4 today [4] (b) Write down an expression for the number of text messages sent on Day Number n. … [1] (c) Ankuri thinks that, by the end of Day Number 3, the total number of text messages that have been sent is 2 4 - 2 . (i) Show that she is correct. [2] (ii) Complete the statement. The total number of text messages sent by the end of Day Number 5 is … which is equal to 2k – 2 where k = … . [2] (iii) Write down an expression for the total number of text messages sent by the end of Day Number n. … [1] (iv) Find the Day Number when the total number of text messages sent by the end of the day is 1022. … [1]

11 marks

Mark scheme: 11 (a) 4 5 6 7 1 8 16 32 64 128 3 B2 for 3 or 4 correct or B1 for first 2 correct If 0 scored, SC1 for 4 values correctly doubled FT one error (b) 2n oe 1 (c) (i) 2 + 4 + 8 = 14 1 16 − 2 = 14 1 or for 14 + 2 = 16 = 24 (ii) 62 2 B1 for each and 6 (iii) 2n + 1 − 2 oe 1 (iv) 9 1

This question in 0580/42 Feb/March 2017

9 (a) The nth term of a sequence is 8n - 3 . (i) Write down the first two terms of this sequence. … , … [1] (ii) Show that the number 203 is not in this sequence. [2] (b) Find the nth term of these sequences. (i) 13, 19, 25, 31, … … [2] (ii) 4, 8, 14, 22, … … [2] (c) … , 20, 50, … The second term of this sequence is 20 and the third term is 50. The rule for finding the next term in this sequence is subtract y then multiply by 5. Find the value of y and work out the first term of this sequence. y = … First term = … [4]

11 marks

Mark scheme: 9(a)(i) 5 and 13 1 9(a)(ii) 8n – 3 = 203 M1 Evaluation of 25th or 26th term with supporting evidence or explanation 3 A1 Second evaluation of 25th or 26th terms 25.75 or 25 with supporting evidence or explanation 4 If zero scored, SC1 for 25.75 or 197 and 205 with partial evidence or explanation 9(b)(i) 6n + 7 oe final answer 2 B1 for 6n + c or kn + 7 k ≠ 0 9(b)(ii) n2 + n + 2 oe final answer 2 B1 for a quadratic expression or second difference = 2 9(c) [y = ] 10 2 M1 for 5(20 – y) = 50 [First term = ] 14 2 M1 for 5(x – their y) = 20 or for 20 ÷ 5 + their y

This question in 0580/41 May/June 2017

11 The table shows the first four terms in sequences A, B, C and D. Complete the table. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 16 25 36 49 B 5 8 11 14 C 11 17 25 35 3 4 5 6 D 2 3 4 5 [12]

12 marks

Mark scheme: 11 2 1, 2 M1 for a quadratic expression seen or 64 ( n + 3 ) oe final answer second differences 2 17 3n + 2 oe final answer 1, 2 B1 for 3n + k (any k) or kn + 2 (k ≠ 0) 2 1, 2FT FT their difference expressions A – B 47 ( n + 3 ) − (3n + 2) oe isw M1 for expression an2 + bn + c seen or second differences 2 7 n + 2 1, 2 n + k + 1 oe final answer B1 for seen 6 n + 1 n + k

This question in 0580/43 May/June 2017

6 Diagram 1 Diagram 2 Diagram 3 Diagram 4 These are the first four diagrams in a sequence. Each diagram is made from small squares and crosses. (a) Complete the table. Diagram 1 2 3 4 5 n Number of crosses 6 10 14 Number of small squares 2 5 10 [6] (b) Find the number of crosses in Diagram 60. … [1] (c) Which diagram has 226 squares? Diagram … [1] (d) The side of each small square has length 1 cm. The number of lines of length 1 cm in Diagram n is 2n2 + 2n + q. Find the value of q. q = … [2]

10 marks

Mark scheme: 6(a) 18 22 4n + 2 oe 6 B2 for 18, 22, 17, 26 17 26 n2 + 1 oe or B1 for two or three correct values AND B2 for 4n + 2 oe or B1 for 4n + k oe or pn + 2 (p ≠ 0) AND B2 for n2 + 1 oe or B1 for n2 + k oe 6(b) 242 1 FT their 4n + 2 provided a linear expression 6(c) 15 1 6(d) 3 2 M1 for 2 × 12 + 2 × 1 + q = 7 oe

This question in 0580/41 Oct/Nov 2017

12 Marco is making patterns with grey and white circular mats. Pattern 1 Pattern 2 Pattern 3 Pattern 4 The patterns form a sequence. Marco makes a table to show some information about the patterns. Pattern number 1 2 3 4 5 Number of grey mats 6 9 12 15 Total number of mats 6 10 15 21 (a) Complete the table for Pattern 5. [2] (b) Find an expression, in terms of n, for the number of grey mats in Pattern n. … [2] (c) Marco makes a pattern with 24 grey mats. Find the total number of mats in this pattern. … [2] (d) Marco needs a total of 6 mats to make the first pattern. He needs a total of 16 mats to make the first two patterns. 1 3 2 He needs a total of n + an + bn mats to make the first n patterns. 6 Find the value of a and the value of b. a = … b = … [6]

12 marks

Mark scheme: 12(a) 18 2 B1 for each 28 12(b) 3n + 3 oe 2 B1 for 3n + k oe or cn + 3 oe c ≠ 0 12(c) 45 2 M1 for identifying 7th pattern or M1 for their ( 3n + 3 ) = 24 12(d) 3 13 6 M1 for any correct substitution [ a = ] oe [b = ] oe 1 2 3 e.g. (2)³ + 2²a + 2b 6 A1 for one of e.g. 1 + a + b = 6 oe 6 8 + 4a + 2b = 16 oe 6 27 + 9a + 3b = 31 oe 6 64 + 16 a + 4b = 52 oe 6 A1 for another of the above M1 for correctly eliminating one variable from their equations 3 A1 for a = 2 13 A1 for b = oe 3

This question in 0580/41 May/June 2018

10 (a) Find the next term and the nth term of this sequence. 3 4 5 6 7 , , , , , … 5 7 9 11 13 Next term = … nth term = … [3] (b) Find the nth term of each sequence. (i) –1, –3, –5, –7, –9, … … [2] (ii) 2, 9, 28, 65, 126, … … [2]

7 marks

Mark scheme: 10(a) 8 B1 15 n + 2 B2 B1 for n + 2 as numerator or 2n + 3 as oe denominator 2 n + 3 10(b)(i) 1 − 2 n oe 2 B1 for −2 n + k oe or pn + 1 (p ≠ 0) oe 10(b)(ii) n 3 + 1 oe 2 M1 for cubic expression

This question in 0580/43 Oct/Nov 2018

11 (a) The table shows the first five terms of sequence A and sequence B. Term 1 2 3 4 5 6 Sequence A 7 13 23 37 55 Sequence B 1 3 9 27 81 (i) Complete the table for the 6th term of each sequence. [2] (ii) Find the nth term of (a) sequence A, … [2] (b) sequence B. … [2] (b) The nth term of another sequence is 4n 2 + n + 3 . Find (i) the 2nd term, … [1] (ii) the value of n when the nth term is 498. n = … [3]

10 marks

Mark scheme: 11(a)(i) 77 243 2 B1 for each 11(a)(ii)(a) 2 n 2 + 5 oe 2 M1 for a quadratic expression as the answer or B1 for common 2nd difference of 4 11(a)(ii)(b) 3n −1 oe 2 B1 for 3k oe where k is a linear function of n 11(b)(i) 21 1 11(b)(ii) 11 3 B2 for ( 4 n + 45 )( n − 11) seen or B1 for 4n2 + n + 3 = 498 oe

This question in 0580/42 Feb/March 2019

11 Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 The sequence of diagrams above is made up of small lines and dots. (a) Complete the table. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Number of 4 10 18 28 small lines Number of 4 8 13 19 dots [4] (b) For Diagram n find an expression, in terms of n, for the number of small lines. … [2] (c) Diagram r has 10 300 small lines. Find the value of r. r = … [2] (d) The number of dots in Diagram n is an 2 + bn + 1. Find the value of a and the value of b. a = … b = … [2]

10 marks

Mark scheme: 11(a) 40 54 4 B1 for each 26 34 11(b) 2 2 B1 for a quadratic expression n + 3n or n (n + 3) oe or for 2nd common difference 2 (at least 2 shown) or for 2 correct equations seen or for subtracting n2 11(c) 100 2 M1 for their (b) = 10300 seen 11(d) 1 2 B1 for each [a = ] oe or M1 for one correct equation 2 or for 2nd difference = 1 soi (at least 2 shown) and 5 [b =] oe 2

This question in 0580/42 May/June 2019

10 (a) Complete the table for the 5th term and the nth term of each sequence. 1st 2nd 3rd 4th 5th nth term term term term term term 9 5 1 -3 4 9 16 25 1 8 27 64 8 16 32 64 [11] (b) 0, 1, 1, 2, 3, 5, 8, 13, 21, … This sequence is a Fibonacci sequence. After the first two terms, the rule to find the next term is “add the two previous terms”. For example, 5 + 8 = 13 . Use this rule to complete each of the following Fibonacci sequences. 2 4 … … … 1 … … … 11 … -1 … … 1 [3] 1 3 4 7 11 (c) , , , , , … 3 4 7 11 18 p (i) One term of this sequence is . q Find, in terms of p and q, the next term in this sequence. … [1] (ii) Find the 6th term of this sequence. … [1]

16 marks

Mark scheme: 10(a) – 7 11 B1 13 – 4n oe B2 or B1 for 13 – kn (k ≠ 0) or for k – 4n 36 B1 (n + 1)2 oe B2 or B1 for any quadratic 125 B1 n 3 oe B1 B1 128 2n + 2 oe B2 or B1 for 2k oe 10(b) … , … , 6, 10, 16 3 B1 for each correct row … , 3, 4, 7, … 2, … , 1, 0, … 10(c)(i) q 1 p + q 10(c)(ii) 18 1 29

This question in 0580/41 Oct/Nov 2019

7 (a) Naga has n marbles. Panav has three times as many marbles as Naga. Naga loses 5 marbles and Panav buys 10 marbles. Together they now have more than 105 marbles. Write down and solve an inequality in n. … [3] (b) y is inversely proportional to x2. When x = 4 , y = 7.5 . Find y when x = 5 . y = … [3] (c) Find the nth term of each sequence. (i) 4 2 0 - 2 - 4 … … [2] (ii) 1 7 17 31 49 … … [2]

10 marks

Mark scheme: 7(a) n − 5 + 3n + 10 > 105 or better B1 n > 25 final answer B2 M1 for 4n > 100 7(b) 4.8 3 k M1 for y = or better x 2 their k M1 for [ y = ] 5 2 OR M2 for y × 5 2 = 7.5 × 4 2 7(c)(i) 6 − 2n oe final answer 2 B1 for answer 6 − kn (k ≠ 0) oe or answer j−2n oe or for correct expression shown in working and then spoilt 7(c)(ii) 2 n 2 − 1 oe final answer 2 B1 for 2nd diff = 4 or a quadratic expression or for correct expression shown in working and then spoilt

This question in 0580/42 Feb/March 2020

7 Diagram 1 Diagram 2 Diagram 3 Diagram 4 These are the first four diagrams of a sequence. The diagrams are made from white dots and black dots. (a) Complete the table for Diagram 5 and Diagram 6. Diagram 1 2 3 4 5 6 Number of white dots 1 4 9 16 Number of black dots 0 1 3 6 Total number of dots 1 5 12 22 [2] (b) Write an expression, in terms of n, for the number of white dots in Diagram n. … [1] 1 2 (c) The expression for the total number of dots in Diagram n is ( 3n - n ) . 2 (i) Find the total number of dots in Diagram 8. … [1] (ii) Find an expression for the number of black dots in Diagram n. Give your answer in its simplest form. … [2] (d) T is the total number of dots used to make all of the first n diagrams. T = an 3 + bn 2 Find the value of a and the value of b. You must show all your working. a = … b = … [5]

11 marks

Mark scheme: 7(a) 25 36 2 B1 for 3, 4 or 5 correct 10 15 35 51 7(b) n2 1 7(c)(i) 92 1 7(c)(ii) 1 2 1 (n2 – n) oe M1 for (3n2 – n) – n2 oe 2 2 1 or for final quadratic answer with n2 oe 2 1 2 or − n oe but not both 2 7(d) 1 1 5 B2 for 2 correct equations eg a = , b = a + b = 1, 8a + 4b = 6 2 2 or B1 for 1 correct equation B2 for one correct value or M1 (dep on at least B1) for correctly eliminating one variable from two linear equations in a and b OR 1 B2 for a = 2 or B1 for 6a = 3 or for 3rd difference = 3 1 B2 for b = 2 or M1 for substituting their a into a correct equation of first differences

This question in 0580/41 Oct/Nov 2020

11 Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 13 9 5 1 B 0 7 26 63 7 8 9 10 C 8 16 32 64 (a) Complete the table for the three sequences. [10] (b) One term in Sequence C is p. q Write down the next term in Sequence C in terms of p and q. … [2]

12 marks

Mark scheme: 11(a) A : –3 17 – 4n oe 3 B1 for –3 B2 for 17 – 4n oe or B1 for k – 4n oe or 17 – pn oe, p ≠ 0 B : 124 n3 – 1 oe 3 B1 for 124 B2 for n3 – 1 oe or B1 for any cubic 11 n + 6 4 11 C : oe B1 for 128 2 n + 2 128 n + 6 B3 for oe 2 n + 2 or B2 for 2 n + 2 oe seen or B1 for 2k oe or n + 6 seen 11(b) p + 1 2 B1 for p + 1 or 2q oe oe 2 q

This question in 0580/42 Oct/Nov 2020

11 The table shows the first four terms in sequences A, B, and C. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 4 9 14 19 B 3 10 29 66 C 1 4 16 64 Complete the table. [9]

9 marks

This question in 0580/43 Oct/Nov 2020

10 The table shows four sequences A, B, C and D. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 1 8 27 64 B 5 11 17 23 C 0.25 0.5 1 2 4 D 4.75 10.5 16 21 Complete the table. [9]

9 marks

Mark scheme: 10 125 n3 oe final ans B2 B1 for 125 B1 for n3 29 6n – 1 oe final ans B3 B1 for 29 B2 for 6n – 1 oe or B1 for 6n + k or an – 1 (a ≠ 0) 2n – 3 oe final ans B2 B1 for 2n [+ k] oe 25 6n – 1 – 2n – 3 oe final ans B2 FT their 29 – 4 and their 6n – 1 – their 2n – 3 OR B1FT for each 1 3 1 2 17 OR 25.25 − n + n + n − 1 oe B1 for each 24 8 3 final ans

This question in 0580/42 May/June 2021

11 (a) These are the first four terms of a sequence. 11 7 3 -1 (i) Write down the next term. … [1] (ii) Write down the term to term rule for this sequence. … [1] (iii) Find the nth term of this sequence. … [2] 2n (b) The nth term of a different sequence is . n + 1 (i) Find the difference between the 5th term and the 6th term of this sequence. Give your answer as a fraction. … [2] 3 (ii) Is a term in this sequence? 4 Show how you decide. [3]

9 marks

Mark scheme: 11(a)(i) –5 1 11(a)(ii) Subtract 4 oe 1 11(a)(iii) 15 – 4n oe final answer 2 B1 for k – 4n or 15 – jn j ≠ 0 11(b)(i) 1 2 12 10 or equivalent fraction B1 for and 21 7 6 11(b)(ii) 3 M2 3 2 n n = oe M1 for = oe 5 4 n + 1 or or 2n ⩾ n + 1 but 3 < 4. M1 for 2n > n + 1 but 3 < 4 No, n is not an integer oe A1 or 3 No, is less than 1, oe 4

This question in 0580/43 May/June 2021

2 (a) Simplify fully. (i) p 3 # p 11 … [1] 18 m 6 (ii) 2 3m … [2] 1 27x 9 y 27 - 3 (iii) e 64 o … [3] (b) A sequence has nth term 3n 2. Write down the first 3 terms of this sequence. … , … , … [2] (c) Find the nth term for each of these sequences. (i) 13, 16, 19, 22, 25, … … [2] (ii) 3, 17, 55, 129, 251, … … [2] (d) Solve. 3x - 22 = 23 4 x = … [3] (e) Use the quadratic formula to solve 3x 2 + 8x - 20 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … , x = … [4]

19 marks

Mark scheme: 2(a)(i) p14 final answer 1 2(a)(ii) 6m4 final answer 2 B1 for 6mk or km4 in final answer or correct answer seen and spoilt 2(a)(iii) 4 4 x −3 y −9 3 B2 for correct answer seen and spoilt or final answer or 2 correct elements in final answer 3 9 3x y 3 4 3 or B1 for one of or oe or x3 or y9 seen 3 4 2(b) 3, 12, 27 2 B1 for 12 or 27 2(c)(i) 3n + 10 oe final answer 2 B1 for 3n + k oe or jn + 10 oe (j ≠ 0) or for correct expression shown in working and then spoilt 2(c)(ii) 2n3 + 1 oe final answer 2 B1 for 3rd diff = 12 (both needed) or for cubic answer or for correct expression shown in working and then spoilt 2(d) 38 3 M2 for 3x = 4 × 23 + 22 or M1 for 3x – 22 = 4 × 23 3 x 22 or for = 23 + oe 4 4 2(e) 2 B2 2 −8 8 − 4(3)( −20) B1 for 8 − 4(3)( −20) oe 2  3 −+8 q −−8 q 2 or oe or oe or both −8 8 ( −20) 2  3 2  3 or  − 2  3 4  32 3 or better – 4.24, 1.57 final answers B2 B1 for each If B0, SC1 for answers – 4.2 or –4.23 or –4.240 to – 4.239 and 1.6 or 1.572 to 1.573 or – 4.24 and 1.57 seen in working or for –1.57 and 4.24 as final answer

This question in 0580/43 Oct/Nov 2022

n6 (a) A sequence has nth term . 2n + 3 (i) Find the first three terms of this sequence. Give your answers as fractions. … , … , … [2] 12 (ii) The kth term of this sequence is . 25 Find the value of k. k = … [2] (b) Find the nth term of each sequence. (i) 6, 13, 32, 69, 130, … … [2] (ii) 100, 50, 25, 12.5, 6.25, … … [2]

8 marks

Mark scheme: 6(a)(i) 1 2 3 2 B1 for 2 correct terms isw , , final answer or for 0.2 and (0.286 or 0.2857…) and 5 7 9 0.333… 6(a)(ii) 36 2 12(2 k  3) M1 for k = or better 25 6(b)(i) n3 + 5 oe final answer 2 B1 for any cubic or common third differences of 6 (at least 2) or for correct answer seen and spoilt 6(b)(ii) 100 × 21–n oe final answer 2 n   k  [+k]  1  B1 for 2–n oe or   oe in answer  2  or for correct answer seen and spoilt

This question in 0580/42 May/June 2023

6 (a) Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A -7 -3 1 5 B 7 13 23 37 C 2 3 4 5 27 81 243 729 Complete the table for the three sequences. [10] (b) In a sequence, the sum of the first 49 terms is 7644. The sum of the first 50 terms is 7975. Find the 50th term of this sequence. … [1]

11 marks

Mark scheme: 6(a) A 9 B1 4n – 11 oe final answer B2 B1 for 4n – k or jn – 11 oe j  0 B 55 B1 2n2 + 5 oe final answer B2 B1 for any quadratic or second differences = 4 6 B1 C oe 2187 n + 1 B3 B2 for 3n + 2 oe oe final answer 3n + 2 OR B1 for 3n + k seen oe B1 for n + 1 as the numerator of a fraction 6(b) 331 cao 1

This question in 0580/41 Oct/Nov 2023

3 5 (a) Simplify 25x 6 2 . ` j … [2] (b) These are the first five terms of a sequence. 1 1 6 36 216 6 Find the nth term of the sequence. … [2] (c) Expand and simplify. ( x + 4)( x - 3)( 3x - 1) … [3] 7 2(d) (i) Show that ( 3x + 5) + = x simplifies to 2x + x - 3 = 0 . x - 2 [4] (ii) Solve by factorisation 2x 2 + x - 3 = 0 . x = … or x = … [3] (e) A solid cylinder has base radius x and height 3x. The total surface area of the cylinder is the same as the total surface area of a solid hemisphere of radius 5y. 2 75y 2 Show that x = . 8 [The surface area, A, of a sphere with radius r is A = 4rr 2 .] [4]

18 marks

Mark scheme: 5(a) 125x 9 final answer 2 B1 for answer 125 kx or m x 9 or for correct answer seen then spoilt 5(b) 6 n  2 oe final answer 2 B1 for answer of form 6k oe  k  1  or answer of the form   oe  6  or for correct answer seen 5(c) 3x3 + 2x2 –37x + 12 final answer 3 B2 for correct expansion of three brackets unsimplified or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of two brackets with at least 3 terms out of 4 correct 5(d)(i) eliminates the fraction correctly M1 eg (3x + 5) (x – 2) + 7 = x (x – 2) 3x2 + 5x –6x – 10 + 7 = x2 – 2x oe B2 B1 for 3x2 + 5x –6x –10 [ + 7] oe seen with at least 3 terms correct leading to 2x2 + x – 3 = 0 A1 dep on M1 B2 with no errors or omissions 5(d)(ii)  2 x  3  x  1 M2 or M1 for (2x + a)(x + b) where ab = −3 or 2b + a = [+]1 or for partial factors 2x(x – 1) + 3(x – 1) or x(2x + 3) –[1](2x + 3) −1.5 oe and +1 B1 5(e) 2 M1 [TSA cylinder =] 2x  2x  3 x 2 4(5 y ) 2 M1 [TSA hemisphere=] (5 y )  2 Leading to M1 dep M1M1 2x 2  6x 2  50y 2  25y 2 oe 2 75 y 2 A1 dep on M1M1M1 x  8

This question in 0580/42 May/June 2024

2 (a) The nth term of a sequence is 120 - n 3 . (i) Find the 4th term of this sequence. … [1] (ii) Find the value of n when the nth term is -1211. n = … [2] (b) The nth term of a different sequence is 3 # ( 0 .2 ) n - 1 . Find the 5th term of this sequence. … [1] (c) The table shows the first four terms of sequences A, B and C. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 7 4 1 -2 1 2 3 4 B 4 5 6 7 C 0 2 6 12 Complete the table for each sequence. [8]

12 marks

Mark scheme: 2(a)(i) 56 1 2(a)(ii) 11 2 M1 for 120 – n3 = – 1211 or 120 – 113 = – 1211 2(b) 3 1 0.0048 or oe 625 2(c) 8 B1 for –5 A –5 10 – 3n B2 for 10 – 3n oe or B1 for k – 3n or for 10 – kn 5 B1 for 8 5 n n B B1 for oe 8 n  3 n  3 B1 for 20 2 B2 for n 2  n oe C 20 n  n or B1 for any quadratic or for at least two second differences of 2

This question in 0580/43 May/June 2024

11 The table shows three sequences. 1st 2nd 3rd 4th 5th 6th nth term term term term term term term Sequence 28 22 16 10 4 A Sequence 1 2 3 4 1 B 6 7 8 9 2 Sequence 1 1 1 2 4 C 4 2 Complete the table. [8]

8 marks

Mark scheme: 11 8 Mark the table first 6th term nth term B1 for –2 –2 34 – 6n B2 for 34 – 6n or B1 for answer –6n + c or 34 – kn (k ≠ 0) 6 n or for a correct answer seen and spoilt 11 n + 5 –3 6 8 2n B1 for 11 n B1 for n + 5 B1 for 8 B2 for 2n –3 or B1 for answer 2n – k (k can be 0) or for a correct answer seen and spoilt

This question in 0580/43 May/June 2025

20 These are the first five terms of a sequence. 48 24 12 6 3 (a) Find the next term. … [1] (b) Find the nth term. … [2]

3 marks

Mark scheme: 20(a) 1 3 1 1.5 or 1 or 2 2 20(b) n 2 n + k  1   1  96   oe final answer M1 for answer 96   oe  2   2   1  n or for 96   oe seen  2 

This question in 0580/41 Oct/Nov 2025

10 Find the nth term of each sequence. (a) 17, 9, 1, – 7, -15, f … [2] (b) 3, 12, 27, 48, 75, f … [2]

4 marks

Mark scheme: 10(a) 25 – 8n oe final answer 2 B1 for answer k – 8n (any k) oe or 25 – jn (j ≠ 0) or for correct answer seen then spoilt 10(b) 3n2 oe final answer 2 B1 for quadratic or for second difference = 6 (at least two, with none incorrect) or for correct answer seen then spoilt

This question in 0580/42 Oct/Nov 2025