设{an}是公差不为零的等差数列,Sn为其前n项和,满足(a2)^2+(a3)^2=(a4)^2+(a5)^2,S7=7

1) 设首项为a1,公差为d,
a2^2+a3^2=a4^2+a5^2
(a5+a3)*(a5-a3)+(a4+a2)(a4-a2)=0
由于 a5-a3=a4-a2=2d≠0,所以 a2+a3+a4+a5=0
则 4a1+10d=0 (1)
7a1+21d=7 (2)
解得 a1=-5,d=2
所以 an=2n-7
Sn=(a1+an)*n/2=(2n-12)n/2=n(n-6)
2) am*a(m+1)/a(m+2)
=(2m-7)(2m-5)/(2m-3)
=(4m^2-24m+35)/(2m-3)
=2m-9+8/(2m-3)
所以 2m-3能整除8.
2m-3=-1或1
m=1或2
当m=1时,a1*a2/a3=15=2*11-7=a11
当m=2时,a2*a3/a4=3=2*5-7=a5