an=a1+(n-1)d
am+an=a1+(m-1)d+a1+(n-1)d
=2a1+(m+n-2)d
同理,
ap+aq=2a1+(p-q-2)d
2a1+(m+n-2)d=2a1+(p-q-2)d
(m+n)d-2d=(p-q)-2d
m+n=p-q
所以,当m+n=p+q时,一定有am+an=ap+aq