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 ia zero $(p\ne q)$.

Answer 1

We know,

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

According to the question,

$S_p=S_q$

$\Rightarrow \frac{p}{2}(2a+(p-1\left)d\right)=\frac{q}{2}(2a+(q-1\left)d\right)$

$\Rightarrow p(2a+(p-1\left)d\right)=q(2a+(q-1\left)d\right)$

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

$\Rightarrow 2ap+p^{2}d-pd=2aq+q^{2}d-qd$

$\Rightarrow 2ap-2aq=q^{2}d-qd-p^{2}d+pd$

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

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

$\Rightarrow 2a(p-q)=d\left[\right(q+p\left)\right(q-p)+(p-q\left)\right]$

$\Rightarrow 2a(p-q)=d[-(q+p\left)\right(q-p)+(p-q\left)\right]$

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

$\Rightarrow 2a=d(1-q-p)(\because p\ne q)$

$\Rightarrow 2a=d(1-q-p)...\left(1\right)$

Now,

$S_{p+q}=\left(\frac{p+q}{2}\right)(2a+(p+q-1\left)d\right)$

$=\left(\frac{p+q}{2}\right)\left(\right(1-q-pd+(p+q-1)d\left)\right)\left(from\right(1\left)\right)$

$=\left(\frac{p+q}{2}\right)d(1-q-p+p+q-1)$

$=0$

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