已知各项均不相等的等差数列{an}的前四项之和S4=14,且a1 a3 a7成等比数列

由题意可知 等差数列{an}的公差 d≠0
S4=a1+a2+a3+a4
=a1+a1+d+a1+2d+a1+3d
=4a1+6d=14
2a1+3d=7 (1)
a3²=a1a7
(a1+2d)²=a1*(a1+6d)
a1²+4a1d+4d²=a1²+6a1d
2a1d=4d²
a1=2d (2)
将(2)代入(1)中得：
4d+3d=7
d=1
a1=2
an=a1+(n-1)d=n+1
(2).记bn=1/an*a(n+1)=1/[(n+1)(n+2)]=1/(n+1)-1/(n+2)
Tn=b1+b2+...+bn
=1/2-1/3+1/3-1/4+...+1/(n+1)-1/(n+2)
=1/2-1/(n+2)=n/[2(n+2)]
Tn≤xa(n+1)=x(n+2)
n/[2(n+2)]≤x(n+2)
x≥n/[2(n+2)²]
对于n属于正整数恒成立
n/[2(n+2)²]=1/[2(n+4+4/n)]≤1/[2(4+2√n*4/n)]=1/[2(4+4)]=1/16
则 x最小值为1/16