SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 5 & 6)


Question 5:
(a) Given a geometric progression with terms a, ar, ar2, ar3, … , arn-2, arn-1 and the sum of first n terms is Sn
Derive the formula for the sum of the first n terms, Sn for the geometric progression when |r| < 1. [3 marks]

(b) Hence, find the value of n for which the sum of the first 2n terms is 127/128 times of the sum of first n terms of a geometric progression which has a common ratio of -½. [4 marks]


Solution:
(a)

S n =a+ar+a r 2 +a r 3 +  +a r n2 +a r n1      (1) ( 1 )×r: r S n =ar+a r 2 +a r 3 +a r 4 +  +a r n1 +a r n      (2) ( 1 )( 2 ),  S n r S n =aa r n S n ( 1r )=a( 1 r n ) S n = a( 1 r n ) ( 1r ) , | r |<1


(b)

S 2n = 127 128 S n a( 1 ( 1 2 ) 2n ) 1( 1 2 ) = 127 128 ( a( 1 ( 1 2 ) n ) 1( 1 2 ) ) a ( 1 ( 1 2 ) 2n ) 3 2 = 127 128 ( a ( 1 ( 1 2 ) n ) 3 2 ) 128( 1 ( 1 2 ) 2n )=127( 1 ( 1 2 ) n ) 128128 ( 1 2 ) 2n =127127 ( 1 2 ) n Let  ( 1 2 ) n =x 128128 x 2 =127127x 128 x 2 127x1=0 ( 128x+1 )( x1 )=0 x= 1 128    or   x=1 ( 1 2 ) n = 1 128     ( 1 2 ) n = ( 1 2 ) 7 n=7 or    ( 1 2 ) n =1 ( 1 2 ) n = ( 1 2 ) 0 n=0 (not accepted)


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