RI APGP C7B Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________________________________________ Additional Practice Questions for Chapter 7B: Arithmetic Progression and Geometric Progression Page 1 of 20 Additional Practice Questions for Chapter 7B: Arithmetic Progression and Geometric Progression 1 JJC/Prelim 9233/2006/01/Q1 The sum of the first 100 terms of an arithmeti c progression is 15050; the first, third and eleventh terms of this progression are three co nsecutive terms of a geometric progression. Find the first term, a and the non-zero common difference, d, of the arithmetic progression. [5] Solutions 100 100 2 99 150502Sa d  2 99 301ad --------- (1) Also, 2311 31 1 1 31 uu uu uuu     2 21 0ad a a d   246da d -------- (2) Since 0d  3 2da Substitute into (1), 32 99 301 2aa   2, 3.ad  2 Jack starts working in a company with an annual salary of $16 000 in the first year. He will receive an annual salary increase of 4% each year. Assuming that he works in the company till his retirement, find (i) the amount he will earn in his 25th year, [2] (ii) the total amount that he will earn over the 25-year period. [2] (iii) the minimum number of years he has to work for his total earnings to exceed $1 000 000. [4] Solutions (i) 25 1 25 (16000)(1.04) 41012.87u  (nearest cent) He will earn $41 012.87 in his 25th year. (ii) 25 25 16000(1.04 1) 666334.531.04 1S   (nearest cent) He earns $666 334.53 over the 25-year period.

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 7B: Arithmetic Progression and Geometric Progression Page 2 of 20 (iii) 16000(1.04 1) 10000001.04 1 n   1.04 1 2.5 1.04 3.5 ln1.04 ln 3.5 ln1.04 ln 3.5 ln 3.5 ln1.04 31.9 n n n n n n       Minimum number of years he has to work = 32 3 VJC Promo 9758/2020/Q7 An infinite geometric series has first term a and common ratio r, where 0r  . The third term is 36 and the sum to infinity is 243. (i) Find the value of a and r. [3] An arithmetic series has first term 1 and common difference d. The sum of the first 6 terms of the arithmetic series is equal to the sum of the first 3 terms of the geometric series. (ii) Find the value of d. [3] (iii) Find the least value of n for which the thn term of the arithmetic series is more than the sum of the first 2n terms of the geometric series. [3] Solutions (i) Let nu be the thn term of the geometric series. 1n nua r  3 2 36 36 (1) u ar     243 2431 243 1 (2) S a r ar       Substitute (2) into (1),   2243 1 36rr 23243 243 36 0rr   From GC, 12 (reject 0) or 33rr   2Common ratio, . 3r 

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 7B: Arithmetic Progression and Geometric Progression Page 3 of 20 Substitute 2 3r  into (2), 2243 1 813a   First term, 81.a  (ii)     3 281 1 36 21 6 1 22 1 3 3 2 5 171 11 d d d        (iii)    2 2 2 281 1 3 11 1 1 21 3 211 1 1 1 2 4 3 1 3 2243 1 11 10 03 n n n n n n                  2 2Let f 243 1 11 10 3 n nn      n  f n 22 11 > 0 23 0.000002 < 0 The least value of n is 23.

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 7B: Arithmetic Progression and Geometric Progression Page 4 of 20 4 YIJC Prelim 9758/2023/01/Q5 (a) An infinite geometric progression has first term a and common ratio r, where a and r are non-zero. The sum of all the terms after the nth term of the progression is equal to twice the nth term. Show that the sum to infinity of the progression is three times the first term. [3] (b) The positive integers, starting at 1, are grouped into sets, as follows. { 1 }, {2,3}, {4,5,6}, ... (i) Find, in terms of r, the first integer and the last integer in the rth set. [3] (ii) Prove that the sum of the integers in the rth set is   21 1.2 rr  [2] Solutions (a) Sum of the all the terms after the nth term  1 11 1 n n n araSS rr ar r        Given 2nnSS u  , therefore 121 2(1 ) 2 3 n nar arr rr r    Hence 3 (Shown)21 1 3 aaSa r    (b)(i) Total number of integers in the first (r1)th brackets is   111 2 3 . . . (1 ) 1 (1 ) 22 rrrrr      Hence, first integer in the rth bracket   21 2122 rr rr   Last integer in the rth bracket

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 7B: Arithmetic Progression and Geometric Progression Page 5 of 20 2 2 2 2 (1 )2 22 2 2 2 rr r rr r rr      Alternative method: Last integer in the rth bracket = First integer in the (r+1)th bracket minus 1 =    21 1122 rr rr     (b)(ii) There are r integers in the rth bracket. First integer in the rth bracket = 2 2 2 rr Last integer in the rth bracket = 2 2 rr Sum of all the integers in the rth bracket = 22 2 22 2 22 2 2 2 rr r r r r r          =  21 12 rr  (Shown) 5 9758/2017/02/Q2 An arithmetic progression has first term 3. The sum of the first 13 terms of the progression is 156. (i) Find the common difference. [2] A geometric progression has first term 3 and common ratio r. The sum of the first 13 terms of the progression is 156. (ii) Show that 13 52 51 0rr   . Show that the common ratio cannot be 1 even though 1r  is a root of this equation. Find the possible values of the common ratio. [4] (iii) It is given that the common ratio of the geometric progression is positive, and that the nth term of this geometric progre ssion is more than 100 times the nth term of the arithmetic progression. Write down an inequality, and hence find the smallest possible value of n. [3]

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ _______________________________________________________________ Additional Practice Questions for Chapter 7B: Arithmetic Progression and Geometric Progression Page 6 of 20 Solutions (i) Let a and d be the first term and common difference of the AP. Given that 3a and 13 156S  ,   13 2 3 13 1 1562 1.5 d d    (ii) Let a and r be the first term and common ratio of the GP. Given that 3a and 13 ' 156S  ,   13 13 13 31 1561 15 2 1 52 51 0 (shown) r r rr rr       When 1r  , 13 52 51 0rr   .  1r  is a root of the equation. However, if 1r  , all the terms of the GP will be 3 and 13 3(13) 39 156S   .  1r  . Using GC, 1.45107r or 1.21002r 1.45 (3