高一等比数列题!

来源:百度知道 编辑:UC知道 时间:2024/07/07 08:01:24
1.四个数-3,a,b,27中,前三个数成等差数列,后三个数成等比数列,则a-b的值为多少?

2.在等比数列{an}中,a7乘a11=6,a4+a14=5,则a20/a10的值为多少?

要过程,谢谢!

1. a=(b-3)/2, b^2=a*27=27(b-3)/2
2b^2-27b+81=0, (2b-9)(b-9)=0, b=9/2 or b=9

a=3/4 or a=3,

a-b=3/4-9/2=-15/4, or a-b=3-9=-6

2. a7*a11=a4^2*q^10=6, a4+a14=a4(1+q^10)=5, a4^2(1+q^10)^2=25

(1+q^10)^2/q^10 = 25/6,令q^10=a

6a^2-13a+6=0, (2a-3)(3a-2)=0, a=3/2 or a=2/3

a20/a10=q^10=a = 2/3 or 3/2

2a=b-3,b*b=27a.2b*b-27b+51=0,得a=3或0.75,b=9或4.5,∴a-b=-6或-3.75
2.a7*a11=a4*a14=6,又a4+a14=5,得a4=2,a14=3或a4=3,a14=2。
a20/a10=q的十次方=a14/a4=2/3或3/2

2a=b-3
b^2=27a
b^2=27(b-3)/2
2b^2-27b+81=0
(2b-9)(b-9)=0
b=9/2或b=9
a=3/4或a=3
a-b=3/4-9/2=-15/4或a-b=3-9=-6

a7*a11=a9^2=(a4q^3)(a4q^7
(a4)^2q^10=6
q^10=6/(a4)^2...........1
a4+a14=a4(1+q^10)=5
q^10=(5/a4)-1...........2
将1式代入2式
6/(a4)^2=(5/a4)-1
整理得
(a4)^2-5a4+6=0
(a4-2)(a4-3)=0
a4=2或a4=3
q^10=6/(a4)^2
q^10=6/2^2=3/2或q^10=6/3^2=2/3