2條F5 Math Question*

1.) common ratio=1.5
let a be the first term
a + 1.5a + 2.25a > 800
4.75a > 800
a > 168.42
if all 3 terms must be integers, then a = 172
so, the sum of the first 3 terms of a geometric series is 172+258+387=817

2.) 36-18+9-...=23又128分之125
therefore, first term = a = 36 and common ratio = R = -0.5
let n be the number of terms
[a(1-R)^n]/(1-R) = 3069/128
[36(1-(-0.5)^n)]/(1+0.5) = 3069/128
24[1-(-0.5)^n] = 3069/128
1-(-0.5)^n = 1023/1024
1-1023/1024 = (-0.5)^n
1/1024 = (-0.5)^n
(0.5)^10 = (-0.5)^n
n = 10
so, the first 10 terms of the geometric series = 23又128分之125

備註: (-0.5)^n = (-0.5)的n次
1023/1024 = 1024分之1023

2007-10-08 17:25:49 補充：
1.) this question is talking about geometric series, not Arithmetic Series~~

2007-10-08 19:20:29 補充：
2.) 其實可以check下答案~~first 10 terms: 36, -18, 9, -9/2, 9/4, -9/8, 9/16, -9/32, 9/64, -9/12836 - 18 9 - 9/2 9/4 - 9/8 9/16 - 9/32 9/64 - 9/128= 36 - 18 9 - 4.5 2.25 - 1.125 0.5625 - 0.28125 0.140625 - 0.0703125= 23.976562523又128分之125 = 23.9765625so, 一定係10個terms~

n th the sum of a geometric series = 1/2n[2a+(n-1)d]


1.a=4 d=2

so 1/2n[2(4)+(n-1)(2)] greader than 800
n^2 + 3n - 800 greader than 0
n greader than 26.824 or -29.824 greader than n(rej)
n=27


n th the sum of等級(呢個唔識eng)=[a(r^n-1)]/(r-1)

2. a=36 r=-1/2 because 36 * -1/2 =-18

so 23又128分之125=[(36)((-1/2)^n-1)]/((-1/2)-1)
23又128分之125=[(36)((-1/2)^n-1)]/(-3/2)
23又128分之125=(-24)[(-1/2)^n-1]
23又128分之125=(-24)(-1/2)^n+24
(-24)(-1/2)^n=-128分之1
(-1/2)^n=3072分之1
log(-1/2)^n=log3072分之1
(n+1)．log(1/2)=log3072分之1
n=10.5849625
n=11