等比数列a1a4=-512,a1+a4=-56,且公比q为整数,求a10
问题描述：
等比数列a1a4=-512,a1+a4=-56,且公比q为整数,求a10
答
a(n)=aq^(n-1),n=1,2,...a(1)a(4)=-512,a(1)+a(4)=-56.由韦达定理,a(1)和a(4)为方程x^2+56x-512=0的2个根.0=x^2+56x-512=(x+64)(x-8).若a(1)=-64=a,a(4)=8=aq^3,则,q=-1/2.与q为整数矛盾.所以,只能a(1)=8=a,a(4)=-64...