Question

等差数列{an}中,当m≠2001时,有a2001=m,am=2001,若p∈N*,且p>am,试比较am+p与0大小关

Answer 1

a(n)=a+(n-1)d

a(2001)=m ==>a+2000d=m

a(m)=2001 ==>a+(m-1)d=2001

两式相减得 (2001-m)d=(m-2001), 所以d=-1

a(m+p)=a(m)+(p-1)d=2001+(p-1)d=2001-(p-1)=2002-p

又p>am=2001, 所以p>=2002, a(m+p)=2002-p<=0.