Given that the first term and the common ratio of a geometric series are 486 and 1/3 respectively and that the sum to term of this series is 728 8/9 , find
(a)the value of n,
(b)the n th term of the given geometric series,
answers:(a)8 (b)2/9 pls.
geometric series??
2007-06-01 7:40 pm
回答 (2)
2007-06-01 9:07 pm
✔ 最佳答案
a)a = 486,R = 1/3,S(n) = 6560/9
S(n) = a(1 - R^n)/(1 - R)
6560/9 = 486(1 - (1/3)^n) / (1 - 1/3)
1 - (1/3)^n = 6560/6561
(1/3)^n = 1/6561
3^n = 6561
n = 8
b)
by G(n) = aR^(n - 1)
G(n) = 486(1/3)^(8 - 1)
G(n) = 486/2187 = 2/9
參考: eason mensa
2007-06-01 8:08 pm
a) Let the the first term be a and the sum be s, ratio r.
s = a [1-r^(n+1)]/(1-r)
r^(n+1)=1-s(1-r)/a
r=1/3, s = 728 8/9, a = 486
1/3^(n+1)=1/6561
ln(/3^(n+1))=ln(1/6561)
-(n+1)ln3=-ln(6561)
n+1=ln(6561)/ln(3)
= ln(3^8)/ln3
=8 ln3/ln3
=8
n=7
b) nth term = 486*1/3^7=2 X 3^5/3^7 = 2/9
s = a [1-r^(n+1)]/(1-r)
r^(n+1)=1-s(1-r)/a
r=1/3, s = 728 8/9, a = 486
1/3^(n+1)=1/6561
ln(/3^(n+1))=ln(1/6561)
-(n+1)ln3=-ln(6561)
n+1=ln(6561)/ln(3)
= ln(3^8)/ln3
=8 ln3/ln3
=8
n=7
b) nth term = 486*1/3^7=2 X 3^5/3^7 = 2/9
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