kwrl.net
当前位置:首页 >> 在等比数列{An}中,已知A3=3/2,s3=9/2,求A1与q求解... >>

在等比数列{An}中,已知A3=3/2,s3=9/2,求A1与q求解...

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

s3=a3(1/q²+1/q+1)=3/2(1+q+q²)/q²=9/2 得:1+q+q²=3q² 2q²-q-1=0 (2q+1)(q-1)=0 q=-1/2, 1 当q=-1/2时, a1=a3/q²=3/2/(-1/2)²=6 当q=1时, a1=a3=3/2

∵a3=a1q=3/2 S3=a1(1一q^3)/(1一q)解一下

解: S3=9/2 a3/q²+a3/q+a3=9/2 a3=3/2代入,整理,得 2q²-q-1=0 (q-1)(2q+1)=0 q=1或q=-½ q=1时,a1=a3=3/2 q=-½时,a1=a3/q²=(3/2)/(-½)²=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

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

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

解: S3=9/2 a3/q²+a3/q+a3=9/2 a3=3/2代入,整理,得 2q²-q-1=0 (q-1)(2q+1)=0 q=1或q=-½ q=1时,a1=a3=3/2 q=-½时,a1=a3/q²=(3/2)/(-½)²=6

a3=a1*q²=3/2 S3=a1+a2+a3=a1+a1*q+3/2=9/2 即求一个二元方程组: {a1*q²=3/2 {a1+a1*q=3 2式乘以q, a1*q+a1*q²=3q, a1q=3q-3/2 由1式,得a1q=3/(2q) 所以:3/(2q)=3q-3/2 1/q=2q-1 2q²-q-1=0 q=-1/2或1 q=-1/2时a1=6 q=1...

依题意, 当q=1,很明显,a1 = a2 =a3 = 3/2,s3=9/2, 满足要求; 当q 不等于1 时, a3= a1*q^2 = 3/2 , s3 = a1*(1-q^3)/(1-q) = 9/2; 两式相除消掉a1 ,得到 q^2(1-q)/(1-q^3)=q^2(1-q)/(1-q)(1+q+q^2)=q^2/(1+q+q^2)1/3,则 化简后为 2q^2-q-1=0,...

网站首页 | 网站地图
All rights reserved Powered by www.kwrl.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com