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
(a) 0
(b) p − q
(c) p + q
(d) −(p + q)
In the given problem, we are given $S_{p}=q$ and $S_{q}=p$
We need to find $S_{p+q}$
Now, as we know,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
So,
$S_{p}=\frac{p}{2}[2 a+(p-1) d]$
$q=\frac{p}{2}[2 a+(p-1) d]$
$2 q=2 a p+p(p-1) d$ .........(1)
Similarly,
$S_{p}=\frac{q}{2}[2 a+(q-1) d]$
$p=\frac{q}{2}[2 a+(q-1) d]$
$2 p=2 a q+q(q-1) d$ ..............(2)
Subtracting (2) from (1), we get
$2 q-2 p=2 a p+[p(p-1) d]-2 a q-[q(q-1) d]$
$2 q-2 p=2 a(p-q)+[p(p-1)-q(q-1)] d$
$-2(p-q)=2 a(p-q)+\left[\left(p^{2}-q^{2}\right)-(p-q)\right]$
$-2=2 a+(p+q-1) d$ ..............(3)
Now,
$S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d]$
$S_{p+v}=\frac{(p+q)}{2}(-2)$$\ldots$ (Using 3)
$S_{p+\sqrt{q}}=-(p+q)$
Thus, $S_{p+q}=-(p+q)$
Hence, the correct option is (d).
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