设等差数列{an}公差为d.
a3b3=a3/S3=a3/(3a1+3d)=a3/[3(a1+d)]=a3/(3a2)=1/2
a3=(3/2)a2=a2+d
d=a2/2
S5+S3
=5a1+10d+3a1+3d
=5(a2-d)+10d+3(a2-d)+3d
=8a2+5d=8a2+(5/2)a2
=(21/2)a2
=21
a2=2
d=a2/2=2/2=1
a1=a2-d=2-1=1
an=1+n-1=n
Sn=1+2+...+n=n(n+1)/2
bn=2/[n(n+1)]=2[1/n -1/(n+1)]