设等差数列第一项为a1,公差d,
ak1=a1,ak2=a1+4d,ak3=a1+16d;
因为等比,ak1*ak3=ak2*ak2,
解得,a1=2d,
ak1=2d,ak2=6d,ak3=18d,公比q=3;
得:akn=2d*3^(n-1);
kn=2*3^(n-1)-1;
设等差数列第一项为a1,公差d,
ak1=a1,ak2=a1+4d,ak3=a1+16d;
因为等比,ak1*ak3=ak2*ak2,
解得,a1=2d,
ak1=2d,ak2=6d,ak3=18d,公比q=3;
得:akn=2d*3^(n-1);
kn=2*3^(n-1)-1;