(1/2)已知(an)是各项不同的正数的等差数列,lga1.lga2.lga4成等差数列,又bn=1/a2^n.n=1.2.3.证明bn为

问题描述:

(1/2)已知(an)是各项不同的正数的等差数列,lga1.lga2.lga4成等差数列,又bn=1/a2^n.n=1.2.3.证明bn为

lga1+lga4=2lga2
即a1*a4=a2^2
a1*(a1+3d)=(a1+d)^2
a1^2+3a1d=a1^2+2a1d+d^2
a1d=d^2
(an)是各项不同的正数的等差数列所以a1>0,d>0
所以a1=d
a2=a1+d=d+d=2d
bn=1/a2^n.=1/(2d)^n