{an}是等差数列则an=a1+d(n-1)
又a1=9d所以an=9d+d(n-1)=d(n+8)
ak是a1与a2k的等比中项
(ak)²=a1*a2k
[d(k+8)]²=9d*d(2k+8)
即(k+8)²=9(2k+8)
解得k=4或k=-2(舍)
所以k=4
选B
ak是a1与a2k的等比中项
ak^2=a1a2k
(a1+(k-1)d)^2=a1(a1+(2k-1)d)
(9d+kd-d)^2=9d(9d+2kd-d)
(8d+kd)^2=9d(8d+2kd)
(8+k)^2d^2=9d^2(8+2k)
(8+k)^2=9(8+2k)
64+16k+k^2=72+18k
k^2-2k-8=0
(k-4)(k+2)=0
k=4 k=-2(舍去,列数没有负数)
选B
谢谢