#a_n# denotes the #n^(th)# term of the A.P.
Let, #d# be the common difference of the A.P., and, let #S_n#
be the sum of its first #n# terms.
Then, we know that,
#a_n=a_1+(n-1)d, and, S_n=n/2{2a_1+(n-1)d}......(ast).#
We are given that, for #p,q in NN; pltq,#
#a_(p+1) + a_(p+2) +a_(p+3)+...+a_q=0............(star).#
Adding #{a_1+a_2+...+a_p}# on both sides of this eqn., we get,
#{a_1+a_2+...+a_p}+{a_(p+1) + a_(p+2) +a_(p+3)+...+a_q},#
#={a_1+a_2+...+a_p}+{0}.........[because, (star)], i.e., #
# S_q=S_p.#
#q/cancel2[2a_1+(q-1)d]=p/cancel2[2a_1+(p-1)d]......[because, (ast)].#
#:. 2qa_1+q(q-1)d-{2pa_1+p(p-1)d}=0.#
#:. 2a_1(q-p)+d{q^2-q-(p^2-p)}=0.#
#:. 2a_1(q-p)+d{q^2-p^2-q+p}=0.#
#:. 2a_1(q-p)+d{(q-p)(q+p)-1(q-p)}=0.#
#:. (q-p)[2a_1+d(q+p-1)]=0.#
#:. q=p," which is impossible as "qltp" (given); or, "[2a_1+d(q+p-1)]=0.#
#:. [2a_1+d(q+p-1)]=0.#
# rArr S_(p+q)=(p+q)/2[2a_1+d(q+p-1)]=0.#
Enjoy Maths.!
