设{an}为等差数列,Sn为数列{an}的前n项的和,S10=310,S20=1220,Tn为数列{Sn/n}的前n项的

s20=a1+a2+.+a10+a11+a12+.+a20
=s10+a1+10d+a2+10d+.+a10+10d
=s10+s10+100d
=2s10+100d
2s10+100d=s20
2*310+100d=1220
100d=600
d=6
s10=(a1+a10)*10/2
=5(a1+a1+9d)
=5(2a1+9*6)
=10a1+270
10a1+270=310
10a1=40
a1=4
an=a1+(n-1)d
=4+6(n-1)
=6n-2
sn=(a1+an)n/2
=(4+6n-2)n/2
=n(3n+1)
sn/n
=n(3n+1)/n
=3n+1
Tn=3*1+1+3*2+1+.+3n+1
=3(1+2+.+n)+n
=3n(n+1)/2+n
=(3n^2+3n+2n)/2
=(3n^2+5n)/2

有前n项和公式可以算出a1=4，d=6
Sn=3n^2+n
Sn/n=3n+1
Tn=3/2n^2+5/2n

S10=10a1+45d=310
S20=20a1+190d=1220
解得：a1=4、d=6。
Sn=na1+n(n-1)d/2=4n+3n^2-3n=4n^2-n
Sn/n=4n-1是等差数列，首项为S1/1=3。
Tn=n(3+4n-1)/2=n(2n+1)