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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)

Sn=a+ar+ar2+ar3+  +arn2+arn1     (1)(1)×r: rSn=ar+ar2+ar3+ar4+  +arn1+arn     (2)(1)(2), SnrSn=aarnSn(1r)=a(1rn)Sn=a(1rn)(1r), |r|<1


(b)

S2n=127128Sna(1(12)2n)1(12)=127128(a(1(12)n)1(12))a(1(12)2n)32=127128(a(1(12)n)32)128(1(12)2n)=127(1(12)n)128128(12)2n=127127(12)nLet (12)n=x128128x2=127127x128x2127x1=0(128x+1)(x1)=0x=1128   or   x=1(12)n=1128   (12)n=(12)7n=7or   (12)n=1(12)n=(12)0n=0 (not accepted)


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