在等差数列{an}中,公差d≠0,a1,a2,a4,成等比数列,已知数列a1,a3,ak,ak1,ak2…akn也成等比数列,求数列{kn}的通项kn

a1,a2,a4,成等比数列a2^2=a1a4(a1+d)^2=a1(a1+3d)解得a1=dan=a1+(n-1)d=nd数列a1,a3,ak,ak1,ak2…akn也成等比数列公比q=a3/a1=3d/d=3ak1=3^2a3=27d,k1=27akn/ak(n-1)=d*kn/d*k(n-1)=kn/k(n-1)=3kn是以k1=27为首项,3...