In the Usual Notation, we have,
#S_p=q rArr p/2{2a+(p-1)d}=q.#
#:. pa+p(p-1)d/2=q, or,#
#pa+(p^2-p)d/2=q..........(1).#
#S_q=p rArr q/2{2a+(q-1)d}=p.#
#:. qa+q(q-1)d/2=p, or,#
#qa+(q^2-q)d/2=p...........(2).#
#:. (1)-(2) rArr (p-q)a+{(p^2-q^2)-(p-q)}d/2=q-p.#
#-:"ing by "(p-q)!=0, a+{(p+q)-1}d/2=-1...(star)#
#"Therefore, the Desired Sum, "S_(p+q)#
#=(p+q)/2{2a+(p+q-1)d}#
#=(p+q){a+(p+q-1)d/2}#
#=(p+q)(-1)...[because, (star)] =-(p+q)#.
This proves the Result.
Next, to find #S_(p-q),# we have to find #a, &, d# from #(1), &, (2).
Modifying #(1)" by "-:"ing it by "p," we get, "a+(p-1)d/2=q/p...(1').#
#(star)-(1) rArr q(d/2)=-1-q/p=-(p+q)/p.#
#:. d=-(2(p+q))/(qp).#
Then, by #(1'), a=q/p-(p-1)d/2=q/p-(p-1){-(p+q)/(pq)}#
#=1/(pq){q^2+(p-1)(p+q)}=1/(pq)(q^2+p^2+pq-p-q).#
Thus, with, #a=1/(pq)(q^2+p^2+qp-q-p), &, d=-2/(pq)(q+p),#
#S_(p-q)=(p-q)/2{2a+(p-q-1)d}#
#=(p-q)/2[2/(pq)(q^2+p^2+qp-q-p)+(p-q-1){-2/(pq)(q+p)}]#
#=((p-q)/2)(2/(pq)){q^2+p^2+qp-q-p+(1+q-p)(q+p)}#
#=(p-q)/(pq){q^2+p^2+qp-1(q+p)+1(q+p)+(q^2-p^2)}#
#=(p-q)/(pq)(2q^2+qp)#
#:. S_(p-q)={(p+2q)(p-q)}/p.#
Enjoy Maths.!
