Question

In an $A.P$.$S_p=q,S_q=p$. Where $S_r$ denotes the sum of the first $r$ terns. Then $S_{p+q}=$

• A. $0$
• B. $-(p+q)$
• C. $p+q$
• D. $pq$

Answer 1

The correct option is B $-(p+q)$

Sp = q, sq = p, find $S_{p+q}=?$

First term = a, Common difference = d

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

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

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

$2ap+p(p-1)d=2q\times q$

$2aq+q(q-1)d=2p\times p$

$\Rightarrow 2apq+pq(p-1)d=2q^2$

$2apq+qp(q-1)d=2p^2$

_____________________

$pqd(p-1-q+1)=2(q^2-p^2)$

$\Rightarrow qpd=-2(p+q)$

$d=\frac{-2(p+q)}{pq}...\left(3\right)$

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

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

$S_{p+q}=\frac{p+q}{2}\times \frac{2q}{p}+\frac{(p+q)q}{2}\times \frac{-2(p+q)}{qp}$

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

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

So, option (B) is correct.