已知数列{an}是等差数列,bn=（0.5）^an,已知b1+b2+b3=21/8,b1*b2*b3=1/8

Question

已知数列{an}是等差数列,bn=（0.5）^an,已知b1+b2+b3=21/8,b1*b2*b3=1/8

Answer 1

∵bn=(1/2)^an
∴b1b2b3=(1/2)^(a1+a2+a3)=1/8
∴a1+a2+a3=3
又∵(an)是等差数列
∴a1+a3=2a2
∴3a2=3
a2=1
∴b2=(1/2)^1=1/2
又∵ b1+b2+b3=21/8,b1b2b3=1/8,
∴b1+b3=17/8,b1b3=1/4
b1和b3是方程8x²-17x+2=0的两个根
(8x-1)(x-2)=0
∴x=1/8或x=2
当b1=1/8 b3=2时
a1=3 a3=-1
a3=a1+2d
-1=3+2d
2d=-4
d=-2
an=a1+(n-1)d
=3-2(n-1)
=5-2n
当b1=2 b3=1/8时,
a1=-1 a3=3
a3=a1+2d
3=-1+2d
2d=4
d=2
an=a1+(n-1)d
=-1+2(n-1)
=2n-3
综上所述,an的通向公式为：an=5-2n或an=2n-3