數學 AP & GP
回答 (2)
2011-05-09 6:15 pm
✔ 最佳答案
在 一GP中，已知S3 =9/2，S6= 14/3，求首項 a 1 與 公比qSol
S3=a1+a1q+a1q^2
=a1(1+q+q^2)
=a1(1－q^3)/(1－q)
=9/2
S6=a1+a1q+a1q^2+a1q^3+a1q^4+a1q^5
=a1(1+q+q^2+q^3+q^4+q^5)
=a1(1－q^6)/(1－q)
=14/3
So
S6/S3=(1－q6)/(1－q^3)=(14/3)/(9/2)
1+q^3=28/27
q^3=1/27
q=1/3(另2根為虛根)
S3=a1(1+1/3+1/9)=9/ 2
a 1(13/9)=9/ 2
a1=81/26
2011-05-10 2:35 am
S6/S3=(1－q6)/(1－q^3)=(14/3)/(9/2)
1+q^3=28/27
唔明呢步...有冇過程.....thx~
1+q^3=28/27
唔明呢步...有冇過程.....thx~
收錄日期: 2021-05-02 10:42:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110508000051KK01274