Let a and d be the first term and the common difference of the A.P. respectively.
Thus, Sp=p2[2a+(p−1)d]
and Sq=q2[2a+(q−1)d]
Now according to the given condition,
p2[2a+(p−1)d]=q2[2a+(q−1)d]⇒p[2a+(p−1)d]=q[2a+(q−1)d]⇒2ap+pd(p−1)=2aq+qd(q−1)⇒2a(p−q)+d[p(p−1)−q(q−1)]=0⇒2a(p−q)+d[p2−p−q2+q]=0⇒2a(p−q)+d[(p−q)(p+q)−(p−q)]=0⇒2a(p−q)+d[(p−q)(p+q−1)]=0⇒2a+d(p+q−1)=0⇒d=−2ap+q−1∴Sp+q=p+q2[2a+(p+q−1).d]=p+q2[2a+(p+q−1)(−2ap+q−1)]=0