RI APGP C7B Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ________________________________________________________________ ____________________________________ Tutorial 7B: Arithmetic Progression and Geometric Progression Page 1 of 2 Tutorial 7B: Arithmetic Progression and Geometric Progression Section A (Basic Questions) 1 In the sequence 1.0, 1.1, 1.2, … , 99.9, 100.0, each number after the first is 0.1 greater than the preceding number. Find (a) how many numbers there are in the sequence, (b) the sum of all the numbers in the sequence. Solutions (a) Sequence is an AP. Let the first term be a and the common difference be d. Hence, a = 1.0 and d = 0.1 Let the number of terms be n. nth term : (1 )nua n d  Hence, 100.0 = 1.0 + (n – 1)(0.1)  n = 991 (b) Required sum of AP, 2(1 )2 n nSa n d  = 991 2(1.0) (991 1)(0.1) 50045.52   2 The first term of a geometric progression is 10 and its sum to infinity is 15. Find (a) the third term of the progression. (b) the sum of the first 5 terms of the progression. Solutions (a) Let the first term be a and the common ratio be r. Hence, a = 10. Given that the sum to infinity = 15. Therefore, 151 a r   10 151 r 

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ____________________________________ Tutorial 7B: Arithmetic Progression and Geometric Progression Page 2 of 2  1 3r  Hence, the third term of the progression = 31ar  = 2 11 0(10) 39   (b) Sum of first 5 terms = (1 ) 1 nar r   = 5 5 110 1 3 1 121015 11 38 11 3          . 3 The ninth term of an arithmetic progression is 43 and the sum of the first 15 terms is 570. It is given that the sum of the first n terms is greater than 2265. Find the least possible value of n. Solutions Let the first term be a and the common difference be d. Given that the ninth term = 43. Therefore, (9 1) 43ad  84 3ad … (1) Given that the sum of the first 15 terms = 570. Therefore, 15 2 (15 1) 5702 ad  73 8ad … (2) Solving (1) and (2) : 3a and 5d  Given 2265nS  . Therefore, 2 ( 1) 22652 n an d  6 5( 1) 4530nn  25 4530 0nn  (5 151)( 30) 0nn  30n or 30.2n Hence, least n = 31. -30 30.2