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.!
