首项a1=2,公差d=2
ak=a1+(k-1)d=2k
S(k+2)=(k+2)(a1+a(k+2))/2
=(k+2)(a1+a1+(k+2-1)d)/2
=(k+2)(a1+k+1)
=(k+2)(k+3)
a1,ak,S(k+2)成等比
ak^2=a1*S(k+2)
即
4k^2=2*(k+2)(k+3)
k^2-5k-6=0
(k-6)(k+1)=0
k=6或k=-1
由于an递增,所以k=6
an=2n