If the sum of p terms of an A.P. is $q$ and the sum of $q$ terms is $p,$ then the sum of $p+q$ terms will be

The correct option is D: $-(p+q)$

We know that the sum of ${}^{\prime }{n}^{\prime }$ terms of an A.P is given by $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Sum of ${}^{\prime }{p}^{\prime }$ terms is given

$S_p=q$

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

$\Rightarrow 2ap+(p-1)pd=2q....\left(i\right)$

And, Sum of ${}^{\prime }{q}^{\prime }$ terms is given as

$S_q=p$

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

$\Rightarrow 2aq+(q-1)qd=2p....\left(ii\right)$

on Subtracting eq.$\left(ii\right)$ from eq.$\left(i\right),$ we get

$2ap+(p-1)pd-[2aq+(q-1\left)qd\right]=2q-2p$

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

$\Rightarrow 2a(p-q)+p^{2}d-pd-q^{2}d+qd=-2(p-q)$

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

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

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

$\Rightarrow 2a+(p+q)d-d=-2$ [cancelling $(p-q)$ from both sides]

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

Now, we have to fing sum of $(p+q)$ terms

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

$=\frac{(p+q)}{2}[-2]$ [Using $eq.\left(iii\right)$]

$=-(p+q)$