a1=4dak=a1+(k-1)d=(k+3)da2k=a1+(2k-1)d=(2k+3)d因为ak是a1与a2k的等比中项所以[(k+3)d]^2=4d*(2k+3)dd≠0所以(k+3)^2=4(2k+3)即k^2+6k+9=8k+12即k^2-2k-3=0(k+1)(k-3)=0所以k=3或k=-1(负值舍去)所以k=3
