{an},{√sn}都是等差数列,
∴2√S2=√S1+√S3,
即2√（2a1+d)=√a1+√（3a1+3d),
平方得4(2a1+d)=a1+3a1+3d+2√[a1(3a1+3d)],
4a1+d=2√[a1(3a1+3d)],
平方得16a1^2+8a1d+d^2=4a1(3a1+3d),
4a1^2-4a1d+d^2=0,
d=2a1.
又两数列公差相等,
∴√（4a1)-√a1=2a1,a1>0,
∴1=2√a1,a1=1/4.