已知an是不同正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...
问题描述:
已知an是不同正数等差数列,lga1、lga2、lga4成等差数列,又bn=1/a2n,n=1,2,3,...
(1)证明{bn}为等比数列
(2)如果数列{bn}的前3项的和等于7/24,求数列{an}的首项a1和公差d
答
1.因为lga1 lga2lga4成等差数列故2lga2= lga1+lga4即 a2^2 =a1*a4 又 an 是不同正数的等差数列故a4 = a1+3da2 =a1+d (d>0,a1>0)(a1+d)^2 = a1*(a1+3d)故a1d = d*d 即 a1 = d an = nd bn = 1...