若{an}为等差数列,a(p)=q,a(q)=p(p≠0),则a(p+q)=

问题描述:

若{an}为等差数列,a(p)=q,a(q)=p(p≠0),则a(p+q)=

ap=a1+(p-1)d=q
aq=a1+(q-1)d=p
相减
(p-1-q+1)d=q-p
(p-q)d=q-p
p≠q
所以d=-1
a(p+q)
=a1+(p+q-1)d
=a1+(p-1)d+qd
=ap+qd
=q+qd
=q(1+d)
=0