3.APGP _ Series Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T1 Solutions (Sequencs and Series) 1(a) First term =1, common difference = d 5 10 20,,S S S form a GP .  10 20 5 10 SS SS        2 10 5 202 9 2 4 . 2 192 2 2 d d d           2 2 9 2 4 . 2 19d d d     25 10 0dd   2d or d = 0 (rejected as AP is increasing)       1 2 2 22 2 2 2 2 1100 2 1 1 100 2 ( 1) 100 0 1 100 2 99 0 1 100 n n S S n n nn n nn n             Since 2 2 99 0nn   as discriminant < 0 and coefficient of n2 is +ve.   2 1 100 0 ( 1 10)( 1 10) 0 ( 11)( 9) 0 9 n nn nn n              least n is 10. (b) nth month Outstanding amt owed at the start of the month (in hundreds) Outstanding amt owed at the end of the nth month (in hundreds) 1 34 34 2 (34)2 (34)2 3 (34)22 34(2)2-70 4   22 34(2) 70 334(2) 70(2) 70 5   32 34(2) 70(2) 70 4234(2) 70(2) 70(2) 70   n 1334(2) 70(2) .......... 70(2) 70nn    Total amount of money owed at the end of nth month

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T2 = 100( 1334(2) 70(2) .......... 70(2) 70nn    ) 2 1 70((2) 1)100 34(2) 21 n n        1100 70 2 n To be free from debt, 170 2 0n 2 140 ln140 7.129ln 2 n n     Least n = 8 Earliest month to be free from debt = Jan 2014 2(i) Length of film in the 1st layer = 42 Length of film in the 2nd layer = (42 + 2x) Length of film in the 3rd layer = (42 + 4x) Length of film in the nth layer = [42 + 2x(n −1)] = 124 2x (n −1) = 82 ( 1) 41xn   (shown) 2(ii) nS = 16766 mm, a = 42 mm, l = 124 mm Using ()2 n nS a l 2 2(16766π) 20242π 124π nSn al   Using (i), x(n – 1) = 41, 41 0.203981x n  x = 0.204 (to 3 s.f.) 2(iii) The sum of the first 10 terms = k (the sum of the first 5 terms) 10 5(1 ) (1 ) 11 a r a r krr   10 51 (1 )r k r   5 5 5(1 )(1 ) (1 )r r k r   

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T3 Since 51, 1 0rr   Therefore, 51kr (shown) 2(iv) Since k =33 and a = 50, 5 5 1 33 32 2 r r r    Suppose the roll of film is cut into n pieces. The sum of the first n terms  total length of the film 50(2 1) 5258021 2 1 1051.6 2 1052.6 ln 2 ln1052.6 10.03974 n n n n n        the largest possible value of n is 10. Hence the largest number of pieces is 10 pieces. 3 22 21 n nU ar    . (i) Given that n n nU U U U U     