$\begin{array}{1 1}0 \\ p+q \\ q \\p\end{array} $

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- Sum of first $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[2a+(n-1)d]$

Given that the sum of first $p$ terms=sum of first $q$ terms of an $A.P.$

$\Rightarrow\:S_p=S_q$

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

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

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

$\Rightarrow\:2a(p-q)+d[(p^2-q^2)-(p-q)]=0$

$p^2-q^2=(p-q)(p+q)$

$\Rightarrow\:(p-q)[2a+d(p+q-1)]=0$..........(A)

But we know that sum of first $p+q$ terms= $S_{p+q}=\large\frac{p+q}{2}$$[2a+(p+q-1)d]$

$\therefore\:$from (A) $S{p+q}=0$

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