数列{an}为等差数列，Sn为其前n项和，已知S7=7,S15=75,Tn为数列{Sn/n}的前n项和，求Tn

S7=7,S15=75
7*(a1+a7)/2=7
a1+a7=2
a1+a7=2a4=2
a4=1

15*(a1+a15)=75
a1+a15=5
a1+a15=2a8=5
a8=5/2

a4+4d=a8
1+4d=5/2
4d=3/2
d=3/8

a1=a4-3d=1-9/8=-1/8
an=-1/8+3(n-1)/8=(3n-4)/8
Sn=n*(-1/8+(3n-4)/8)/2
Sn/n=(3n-5)/16=3n/16-5/16

Tn=3(1+2+..+n)/16-5n/16=3*n(n+1)/32-5n/16=3n^2/32-7n/32

S7=7 S15=75

a4=S7/7=1 a8=S15/15=5

d=1 an=n-3 a1=-2

Sn=n(a1+an)/2

设Bn=Sn/n=(a1+an)/2=a1+d(n-1)/2
=-2+(n-1)/2=n/2-5/2 为等差数列B1=-2

Tn=n(B1+Bn)/2=n(n/2-5/2-2)/2=(n^2)/2-9/4n