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)
-（p+q）
