1. (a) W1= 27 81 27 M1 1 2 1-r = r = 3 A1 (b) V2=9 V4=1 (A1) 2d =- 8 => d =- 4 V1= 13 (A1) -(2x 13 -4(N-1)) > 0 (accept equality) M1 -(30 -4N) > 0 N(15-2N) >0 N<7.5 (M1) N =7 A1 Note: 13 + 9 + 5 +1-3-7-11 >0 => N =7 or equivalent receives full marks. 2. (a) let the first three terms of the geometric sequence be given by u1, u1 r, ujr- : u1 = a + 2d, ujr = a + 3d and ujr = a + 6d (M1) a + 6d a+3d A1 = a+ 3d a+2d a2 + 8ad + 12d2 = a2 + 6ad + 9d2 A1 2a+3d = 0 a =2d AG [7] IB Questionbank Mathematics Higher Level 3rd edition 1
(b) 11= 2. u.r=32,(1,2 -9d) M1 r = 3 A1 geometric 4"term ujr3 = 27d A1 2 3 arithmetic 16" term a + 15d = - = d+ 15d 2 M1 27d = A1 2 Note: Accept alternative methods. [8] 3. P (six in first throw) = 1 6 25 1 P (six in third throw) = -X 36 x1 P (six in fifth throw) = 25 2 1 x- 36 6 1 25 1 25 1 x- + ... (M1) 36 2 P(A obtains first six) = 6' 36 6 6 -+-x-+ 25 recognizing that the common ratio is 1 36 P(A obtains first six) = 25 6 (by summing the infinite GP) M1 1- 36 6 = 11 4. METHOD 1 5(2a+9d) =60 (or 2a + 9d = 12) M1A1 10(2a + 19d) = 320 (or 2a + 19d = 32) A1 solve simultaneously to obtain M1 a =- 3, d = 2 (A1) (M1)(A1) (A1) A1 [7] A1 the 15th term is - 3 + 14 × 2 = 25 A1 Note: FT the final A1 on the values found in the penultimate line. IB Questionbank Mathematics Higher Level 3rd edition 2
METHOD 2 with an AP the mean of an even number of consecutive terms equals the mean of the middle terms (M1) @10 + @11 = 16 (or @10 + @11= 32) A1 2 a5 +@6 =6 (or a5 + a6 = 12) A1 2 @10-@5+@11-@6=20 5d + 5d = 20 M1 d= 2 and a =- 3 (or a5 = 5 or a10 = 15) A1 the 15th term is -3 + 14 × 2 =25 (or 5 + 10 × 2 =25 or 15 +5 ×2 =25) A1 Note: FT the final A1 on the values found in the penultimate line. [6] 5. METHOD 1 (a) un= Sn-Sn-1 (M1) 7" - a" 7"-1 - an-1 = ? A1 7" 7-1 (b) EITHER a A1 12 =1-2-1ª = "(1-4) M1 A1 common ratio = 7 a A1 OR n a Un = 1 - 7 n-1 -1+ 7 a M1 = (9) 1- . 4 ) 7 n-1 = 7ª(a) n-1 A1 11= 74, , common ratio =