设数列{an}的前n项和为Sn，an与2的算术平均值等于Sn与2的几何平均值，求an

[a(n) + 2]/2 = [S(n)*2]^(1/2), S(n)>=0, a(n)>=-2.
[a(n) + 2]^2 = 8S(n), [a(1)+2]^2 = 8S(1) = 8a(1), [a(1)]^2 + 4a(1) + 4 = 8a(1), a(1) = 2.
[a(n+1)+2]^2 = 8S(n+1),
8a(n+1)=8S(n+1)-8S(n)=[a(n+1)+2]^2-[a(n)+2]^2=[a(n+1)]^2 + 4a(n+1) + 4 - [a(n)+2]^2,
[a(n+1)-2]^2 = [a(n)+2]^2,
|a(n+1)-2| = a(n)+2,
若a(n+1)>=2,则a(n+1)-2=a(n)+2, a(n+1)=a(n)+4,{a(n)}是首项为a(1)=2,公差为4的等差数列。a(n)=2 + 4(n-1), S(n)=2n+2n(n-1)=2n^2,满足题意。
若-2<=a(n+1)<=2,则2-a(n+1)=a(n)+2,a(n+1)=-a(n), a(2)=-a(1)=-2,a(n+2)=-a(n+1)=a(n).a(2n+1)=a(2n-1)=...=a(1)=2, a(2n+2)=a(2n)=...=a(2)=-2.
n=2m时，a(n)=-2,S(n)=0;n=2m-1时，a(n)=2,S(n)=2.满足题意。

综合，有
a(n) = 2+4(n-1)=4n-2,n=1,2,...
或
a(2n-1)=2,a(2n)=-2,n=1,2,...

an=4n-2