Question

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,...

Answer 1

因为 lga1 lga2 lga4 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+3d a2 =a1+d (d>0,a1>0)

(a1+d)^2 = a1*(a1+3d) 故 a1d = d*d 即 a1 = d

an = nd

bn = 1/2nd bn-1 = 1/2(n-1)d bn+1 = 1/2(n+1)d

bn*bn = bn-1*bn+1 又 b1=1/2d

即 bn 为等比数列