(I)设数列的首项为a 1,则
∵S 5=35,且a 2,a 7,a 22成等比数列
∴
5 a 1 +10d=35
( a 1 +6d ) 2 =( a 1 +d)( a 1 +21d)
∵d≠0,∴d=2,a 1=3
∴a n=3+(n-1)×2=2n+1;
(II)S n=
n(3+2n+1)
2 =n(n+2)
∴
1
S n =
1
n(n+2) =
1
2 (
1
n -
1
n+2 )
∴T n=
1
2 (1-
1
3 +
1
2 -
1
4 +
1
3 -
1
5 +…+
1
n -
1
n+2 ) =
1
2 (1+
1
2 -
1
n+1 -
1
n+2 ) =
3
4 -
2n+3
2(n+1)(n+2)