Question

For a series in A.P $S_p=q,S_q=p$ find $S_{p+q}=$?

Answer 1

Formula,

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

Consider sum of first p terms,

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

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

$2ap+p^{2}d-pd=2q$.......(1)

Consider sum of first q terms,

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

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

$2aq+q^{2}d-qd=2p$........(2)

(1)-(2) gives,

$2ap+p^{2}d-pd-2aq-q^{2}d+qd=2q-2p$

$2a(p-q)+(p-q)(p+q)d-d(p-q)=-2(p-q)$

$2a+(p+q)d-d=-2$

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

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

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