Question

If the sum of p terms of an arithmetic progression is same as the sum of q terms show that sum of its (p+q) terms is zero

Answer 1

Sp=q⇒p2{2a+(p−1)d}=q.

∴pa+p(p−1)d2=q,or,

pa+(p2−p)d2=q..........(1).

Sq=p⇒q2{2a+(q−1)d}=p.

∴qa+q(q−1)d2=p,or,

qa+(q2−q)d2=p...........(2).

∴(1)−(2)⇒(p−q)a+{(p2−q2)−(p−q)}d2=q−p.

÷ing by (p−q)≠0,a+{(p+q)−1}d2=−1...(⋆)

Therefore, the Desired Sum, Sp+q

=p+q2{2a+(p+q−1)d}

=(p+q){a+(p+q−1)d2}

=(p+q)(-1)...[because, (star)] =-(p+q)=(p+q)(−1)...[∵,(⋆)]=−(p+q).

This proves the Result.

Next, to find S_(p-q),Sp−q, we have to find a, &, da,&,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.