Question

If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)

Answer 1

We know,
$S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

According to the question,
$$\begin{gathered} S_p=S_q \\ \Rightarrow \frac{p}{2}\left(2a+\left(p-1\right)d\right)=\frac{q}{2}\left(2a+\left(q-1\right)d\right) \\ \Rightarrow p\left(2a+\left(p-1\right)d\right)=q\left(2a+\left(q-1\right)d\right) \\ \Rightarrow 2ap+\left(p-1\right)pd=2aq+\left(q-1\right)qd \\ \Rightarrow 2ap+p^{2}d-pd=2aq+q^{2}d-qd \\ \Rightarrow 2ap-2aq=q^{2}d-qd-p^{2}d+pd \\ \Rightarrow 2a\left(p-q\right)=d\left(q^2-p^2\right)+d\left(p-q\right) \\ \Rightarrow 2a\left(p-q\right)=d\left(q+p\right)\left(q-p\right)+d\left(p-q\right) \\ \Rightarrow 2a\left(p-q\right)=d\left[\left(q+p\right)\left(q-p\right)+\left(p-q\right)\right] \\ \Rightarrow 2a\left(p-q\right)=d\left[-\left(q+p\right)\left(p-q\right)+\left(p-q\right)\right] \\ \Rightarrow 2a\left(p-q\right)=d\left(p-q\right)\left[1-q-p\right] \\ \Rightarrow 2a=d\left(1-q-p\right)\left(\because p\ne q\right) \\ \Rightarrow 2a=d\left(1-q-p\right)...\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} S_{p+q}=\left(\frac{p+q}{2}\right)\left(2a+\left(p+q-1\right)d\right) \\ =\left(\frac{p+q}{2}\right)\left(\left(1-q-p\right)d+\left(p+q-1\right)d\right)\left(\mathrm{from}\left(1\right)\right) \\ =\left(\frac{p+q}{2}\right)d\left(1-q-p+p+q-1\right) \\ =0 \end{gathered}$$

Hence, the sum of its first (p + q) terms is zero.