由公式Sn=na1+n(n-1)d/2有
Sp=pa1+p(p-1)d/2=q.(1)
Sq=qa1+q(q-1)d/2=p.(2)
(1)-(2)得(p-q)a1+(p+q-1)(p-q)d/2=q-p
∵p≠q
∴p-q≠0
∴a1+(p+q-1)d/2=-1
∴S(p+q)=(p+q)a1+(p+q)(p+q-1)d/2=(p+q)[a1+(p+q-1)d/2]=-(p+q)