DHS Quiz 1_Sequences and Series_Maclaurin Suggested Solutions

Name: Index Number: Class: DUNMAN HIGH SCHOOL Quiz 1 Year 5 MATHEMATICS (Higher 2) 9758 22 August 2024 Additional Material: Printed Answer Booklet 40 minutes READ THESE INSTRUCTIONS FIRST Answer all the questions. Write your answers on the Printed Answer Booklet. Follow the instructions on the front cover of the answer booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you must present the mathematical steps using mathematical notations and not calculator commands. You must show all necessary working clearly. The number of marks is given in brackets [ ] at the end of each question or part question.

1 The first three terms of an infinite series are 16, x, 9. (i) Find the value(s) of x, if the series is (a) geometric, (b) arithmetic. [3] (ii) If all the terms in part (i)(a) are positive, calculate its sum to infinity. [2] (iii) Given that the sum of the first n terms in part (i)(b) is less than 64,− find the least value of n. [3] Qn Suggested Solution Comments / Common Mistakes 1(i) (a) 9 16 x x= 2 144x = 12x= (b) 16 9xx− = − 12.5x= • Formulate equation using common ratio or consecutive terms of GP • Formulate equation using common difference of AP • Students should recognise that information in part (ii) may not necessarily apply to part (i). For part (i)(a), students may make the wrong assumption that x or r (common ratio) must be positive hence missed out on one solution. • Note that the question asked for values of x and not d or r. (ii) 16 6431 4 S == − • Students should apply first term 1 ratioS = − formula for GP sum • Students may mistook 1 2 39, , 12,U U x U= = = hence taking the common ratio 4 3r = instead (then sum to infinity won’t exist) 2 16 9 16 9 12 a ar x ar x = = = =  = OR

(iii) Let the number of terms be n.  2(16) ( 1)( 3.5) 642 n n+ − − − [32 3.5 3.5] 128nn − + − 23.5 35.5 128 0nn− −  2.82 or 12.96nn−  OR From GC, least n = 13 • Students may apply wrong AP sum formula or wrong common difference • Note exact answers are not required hence students can use GC to solve the inequality graphically or by table • Students are reminded that the least value of n = 13 is obtained from the solution to the inequality n > 12.96, and not just derived by rounding up n = 12.96 to the nearest integer. Total marks: 8 n nS + 64 12 25 13 1−

2 (i) Find 1 1 2 rn r=   in terms of n. [2] A sequence