In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to

Solution:

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,

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

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

$2 p=2 a q+q(q-1) d$...............(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,

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

$S_{p+0}=\frac{(p+q)}{2}(-2)$ $\ldots$ (Using 3)

$S_{p+q}=-(p+q)$

Thus, $S_{p+q}=-(p+q)$

Hence, the correct option is (b).