For each given geometric sequence with 1st term a(1), common ratio r, nth term a(n) and sum of the first n terms is S(n), find the value(s) indicated in brackets.
1. a(1)=1/2,a(n)=64,S(n)=255/2; (n and r)
2. a=3,r=2 ;(the sum from the 7th to 10th term)
geometric sequence
2011-07-25 11:38 pm
回答 (3)
2011-07-26 12:17 am
2011-07-26 12:41 am
2011-07-26 12:19 am
1)
a(n)=a(r)^(n-1)
64=(1/2)(r)^(n-1)
128= r^(n-1)
128=(r^n)/(r)
r^n=128r--------------------(1)
S(n)=[1/2(1-r^n)]/(1-r)
255/2=(1/2)[(1-r^n)/(1-r)]
255=(1-128r)/(1-r)
255-255r=1-128r
-127r= -254
r=2
Sub r=2 into (1)
r^n=128r
2^n=128x2
2^n=256
log (2^n)=log 256
n log 2 =log 256
n = (log 256)/(log 2)
n = 8
Ans: n=8,r=2
2)
S(10)=[3(1-2^10)]/(1-2)
S(10)=3069
S(6)=[3(1-2^6)]/(1-2)
S(6)= 189
the sum from the 7th to 10th term
=S(10) - S(6)
=3069-189
=2880
Ans: the sum from the 7th to 10th term = 2880
a(n)=a(r)^(n-1)
64=(1/2)(r)^(n-1)
128= r^(n-1)
128=(r^n)/(r)
r^n=128r--------------------(1)
S(n)=[1/2(1-r^n)]/(1-r)
255/2=(1/2)[(1-r^n)/(1-r)]
255=(1-128r)/(1-r)
255-255r=1-128r
-127r= -254
r=2
Sub r=2 into (1)
r^n=128r
2^n=128x2
2^n=256
log (2^n)=log 256
n log 2 =log 256
n = (log 256)/(log 2)
n = 8
Ans: n=8,r=2
2)
S(10)=[3(1-2^10)]/(1-2)
S(10)=3069
S(6)=[3(1-2^6)]/(1-2)
S(6)= 189
the sum from the 7th to 10th term
=S(10) - S(6)
=3069-189
=2880
Ans: the sum from the 7th to 10th term = 2880
參考: 自己
收錄日期: 2021-04-16 13:08:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110725000051KK00580