已知等差数列{an}的公差d 不等于 0 ,它的前 n项和为Sn ,若Ss = 70 ,且a2...

1.
S5=5a1+10d=5(a1+2d)=70
a1+2d=14 a3=14
a7^2=a2×a22
(a3+4d)^2=(a3-d)(a3+19d)
a3=14代入,整理,得
d(d-4)=0 d=0(已知d不等于0,舍去)或d=4
a1=a3-2d=14-8=6
an=a1+(n-1)d=6+4(n-1)=4n+2
数列{an}的通项公式为an=4n+2
2.
Sn=na1+n(n-1)d/2=6n+4n(n-1)/2=2n(n+2)
1/Sn=1/[2n(n+2)]=(1/4)[1/n-1/(n+2)]
Tn=1/S1+1/S2+...+1/Sn=(1/4)[1-1/3+1/2-1/4+1/3-1/5+...+1/n-1/(n+2)]
=(1/4)[(1+1/2+1/3+...+1/n)-(1/3+1/4+...+1/(n+2))]
=(1/4)[1+1/2-1/(n+1)-1/(n+2)]
=(3/8)-(1/4)[1/(n+1)+1/(n+2)]
随n增大,1/(n+1)和1/(n+2)都递减,Tn递增,当n=1时,Tn取得最小值
Tmin=(3/8)-(1/4)(1/2+1/3)=(3/8)-(5/24)=1/6
当n->+无穷大时,1/(n+1)和1/(n+2)都大于0,且趋向于0,Tn3/8-0=3/8
综上,得1/6≤Tn最后貌似算错了？没错的。题目是正确的，过程也是正确的。