www.1862.net > 已知A3=3/2,S3=9/2.求A1与q.

已知A3=3/2,S3=9/2.求A1与q.

最佳答案 a3=a1q^2=3/2 S3=a1+a1q+a1q^2=9/2 q^2/(1+q+q^2)=1/3 2q^2-q-1=0 (2q+1)(q-1)=0 q=-1/2, a1=(3/2)/q^2=6 或:q=1, a1=3/2

S3=(a1+a3)*3/2=9/2 所以 a1+a3=3 所以 a3=3-a1=3/2 所以 公差 d=(a3-a1)/2=0 答:a3=3/2, d=0 *这个数列是常数列,每项都相等,所以也可看作是公比 q=1 的等比数列

q等于负二分之一,a等于六

a1*q^2=a3=3/2 a1(1-q^3)/(1-q)=S3=9/2 粮食相除q^2(1-q)/(1-q^3)=1/3 3q^2(1-q)=(1-q^3) 2q^3-3q^2+1=0 2q^2(q-1)-(q-1)(q+1)=0,而q≠1 2q^2-q-1=0 (2q+1)(q-1)=0,而q≠1 2q+1=0,q=-1/2 a1=a3/q^2=6 故a1=6 ,q=-1/2

(1)解:设数列{an}的公差为d,则∵S3=9,且a5是a3和a8的等比中项,∴3a1+3d=9(a1+4d)2=(a1+2d)(a1+7d)∵d≠0,∴d=1∴a1=2∴an=n+1;(2)证明:∵1anan+1=1(n+1)(n+2)=1n+1?1n+2∴Tn=12?13+13?14+…+1n+1?1n+2=12?1n+2=n2(n+2)∵Tn≤λan+1对任意的n∈N*...

a3/q² + a3/q + a3 = S3 3/2(1/q² + 1/q + 1) = 9/2 1/q² + 1/q + 1 = 3 1/q² + 1/q - 2 = 0 (1/q - 1)(1/q + 2) = 0 q = 1 或 q = -1/2 当 q = 1 时 ,a1 = 3/2 当 q = -1/2 时 ,a1 = 6

a3=a1*q^2=3/2 s3=a1*(1+q+q^2)=9/2 两个相除 1+q+q^2=3q^2 1+q=2q^2 q=1或者-1/2 若q=1 a1=3/2, 若q=-1/2 a1=6

网站地图

All rights reserved Powered by www.1862.net

copyright ©right 2010-2021。
www.1862.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com