an是各项为不同正数的等差数列,又lga1.lga2.lga4成等差数列,切bn=1/a(2^n).若bn的前三项和为7/24,则a1=
问题描述:
an是各项为不同正数的等差数列,又lga1.lga2.lga4成等差数列,切bn=1/a(2^n).若bn的前三项和为7/24,则a1=
答
lga1,lga2,lga4成等差数列 a2^2=a1*a4 => a1=d =>an=a1n
bn=1/a(2^n)=1/[a12^n]=1/a1*(1/2)^n {bn}为等比数列
{bn}的前三项和为7/24,1/2a1+1/4a1+1/8a1=7/24 => a1=d =3