If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Let a and d be the first term and the common difference of the A.P. respectively.
Here,
\begin{align} S_p = \frac{p}{2}\left[2a+(p-1)d\right]\end{align}
\begin{align} S_q = \frac{q}{2}\left[2a+(q-1)d\right]\end{align}
According to the given condition,
\begin{align} \frac{p}{2}\left[2a+(p-1)d\right]=\frac{q}{2}\left[2a+(q-1)d\right]\end{align}
\begin{align} ⇒p\left[2a+(p-1)d\right]=q\left[2a+(q-1)d\right]\end{align}
\begin{align} ⇒2ap + pd(p-1)=2aq+qd(q-1)\end{align}
\begin{align} ⇒2a(p-q) +d[p(p-1)-q(q-1)]=0\end{align}
\begin{align} ⇒2a(p-q) +d[p^2 -p -q^2 +q]=0\end{align}
\begin{align} ⇒2a(p-q) +d[(p-q)(p+q)-(p-q)]=0\end{align}
\begin{align} ⇒2a(p-q) +d[(p-q)(p+q-1)]=0\end{align}
\begin{align} ⇒2a +d(p+q-1)=0\end{align}
\begin{align} ⇒d=\frac{-2a}{p+q-1} \;\;\;\;...(1)\end{align}
\begin{align} \therefore S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).d\right] \end{align}
\begin{align} ⇒ S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).\left(\frac{-2a}{p+q-1}\right)\right] \;\;\;\; [From (1)] \end{align}
\begin{align}=\frac{p+q}{2}\left[2a-2a\right]\end{align}
\begin{align}=0\end{align}
Thus, the sum of the first (p + q) terms of the A.P. is 0.
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