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+ … +arn−2+arn−1 … … … … (1)(1)×r: rSn=ar+ar2+ar3+ar4+ … +arn−1+arn … … … … (2)(1)−(2), Sn−rSn=a−arnSn(1−r)=a(1−rn)Sn=a(1−rn)(1−r), |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)128−128(−12)2n=127−127(−12)nLet (−12)n=x128−128x2=127−127x128x2−127x−1=0(128x+1)(x−1)=0x=−1128 or x=1(−12)n=−1128 (−12)n=(−12)7n=7or (−12)n=1(−12)n=(−12)0n=0 (not accepted)
(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+ … +arn−2+arn−1 … … … … (1)(1)×r: rSn=ar+ar2+ar3+ar4+ … +arn−1+arn … … … … (2)(1)−(2), Sn−rSn=a−arnSn(1−r)=a(1−rn)Sn=a(1−rn)(1−r), |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)128−128(−12)2n=127−127(−12)nLet (−12)n=x128−128x2=127−127x128x2−127x−1=0(128x+1)(x−1)=0x=−1128 or x=1(−12)n=−1128 (−12)n=(−12)7n=7or (−12)n=1(−12)n=(−12)0n=0 (not accepted)