In an A.P., sum of first p terms is q and sum of first q terms is p. Sum of its p + q terms is?

AP is of form a,a+d,a+2d,...,a+(n-1)d,.. where a is the 1st term, a+(n-1)d is the nth term and d is the
common difference {t(n+1) - t(n)} between adjacent terms, n=1,2,3,... . d is non-zero. In general, the
sum, S(n), of the 1st n terms is (n/2)[2a + (n-1)d]. Here S(p) = (p/2)[2a + (p-1)d] = q and S(q) =
(q/2)[2a + (q-1)d] = p. S(p+q) = [(p+q)/2][2a + (p+q-1)d]. Now S(p) = ap + p(p-1)(d/2) = q..[1] and
S(q) = aq + q(q-1)(d/2) = p..[2]. {[1]+[2]}--> S(p)+S(q) = a(p+q) + [p(p-1) + q(q-1)](d/2) = (p+q). Now
p(p-1) + q(q-1) = p^2 + q^2 - (p+q) so S(p)+S(q) = a(p+q) + [p^2 + q^2 - (p+q)](d/2) = (p+q) or
2[S(p)+S(q)] = 2a(p+q) + [p^2+q^2-(p+q)]d = 2(p+q)..[3]. Now 2S(p+q) = 2a(p+q) + [(p+q)(p+q-1)]d
= 2a(p+q) + [p^2+q^2+2pq - (p+q)]d..[4]. {[4]-[3]}--> 2[S(p+q) - {S(p)+S(q)}] = 2pqd, ie., S(p+q) =
pqd + (p+q).