Arithmetic problem involving summation of terms

Question

This is the question which i am referring to

If in an AP , $S_p=S_q $ then prove that $S_{p+q} = 0 $ where $S_n$ denotes the sum of the first $n$ terms.

My try: Here I used the given condition and solved further as follows

$S_p=S_q $

$\frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d] $

$2ap+pd(p-1)-2aq-qd(q-1)=0$

$2a(p-q) +d(p^2-q^2-p+q)=0$

$2a(p-q)+d(p-q)(p+q-1)=0$

$p-q[2a+d(p+q-1)]=0$

$\frac{p-q}{2}[2a+d(p+q-1)]=0$

this is not what is is required to prove instead we have to prove

$\frac{p+q}{2}[2a+d(p+q-1)]=0$

Please mention where i am wrong.

Answer 1

You did nothing wrong.

As you did, we have $$(p-q)(2a+d(p+q-1))=0.$$ Note that this is equivalent to $$p-q=0\ \ \ \text{or}\ \ \ 2a+d(p+q-1)=0.$$

The latter follows the claim. (I think we need to have $p\not =q$ in the first place.)