Question

设正项数列{an}前n项和是sn,若{an},{根号下sn}都是等差数列,且等差相等,则a1等于?

Answer 1

{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.