Question

If the sum of $p$ terms of an $A.S$ is $q$ and the sum of $q$ terms is $p$, show that the sum of (p+q) terms is $-(p+q)$.

Answer 1

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

Given that,

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

$⟹2a+(p-1)d=\frac{2q}{p}$..........(i)

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

$⟹2a+(q-1)d=\frac{2p}{q}$.............(ii)

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

$⟹(p-q)d=\frac{2(q^2-p^2)}{pq}=\frac{-2(p-q)(p+q)}{pq}$

$\therefore d=\frac{-2(p+q)}{pq}$

From $\left(i\right)⟹2a+(p-1)\left\{\frac{-2(p+q)}{pq}\right\}=\frac{2q}{p}$

$⟹a=\frac{q}{p}+\frac{(p-1)(p+q)}{pq}$

$\therefore a=\frac{q^2+(p-1)(p+q)}{pq}$

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

$S_{p+q}=\left(\frac{p+q}{2}\right)[\frac{2[q^2+(p-1\left)\right(p+q\left)\right]}{pq}-\frac{2(p+q-1)(p+q)}{pq}]$

$S_{p+q}=(p+q)\left[\frac{q^2+(p+q)(p-1-p-q+1)}{pq}\right]$

$=(p+q)\left[\frac{q^2+(p+q)(-q)}{pq}\right]$

$=(p+q)\left[\frac{q-p-q}{p}\right]$

$=(p+q)\left[\frac{-p}{p}\right]$

$=-(p+q)$

Hence proved.