Let [tex]d[/tex] be the common difference between terms. Then the AP is
[tex]\left\{\underbrace{a,a+d,a+2d,\ldots,a+(p-1)d}_{\text{first }p\text{ terms}},\underbrace{a+pd,a+(p+1)d,\ldots,a+(p+q+1)d}_{\text{next }q\text{ terms}}\right\}[/tex]
Because the first [tex]p[/tex] terms add to zero, you have
[tex]\displaystyle\sum_{n=1}^p (a+(n-1)d)=(a-d)\sum_{n=1}^p1+d\sum_{n=1}^pn=(a-d)p+d\frac{p(p+1)}2=0[/tex]
Solving for [tex]d[/tex] yields [tex]d=\dfrac{2a}{1-p}[/tex].
Now, the sum of the next [tex]q[/tex] terms is
[tex]\displaystyle\sum_{n=p+1}^{p+q}(a+(n-1)d)=\sum_{n=p+1}^{p+q}\left(a+(n-1)\frac{2a}{1-p}\right)[/tex]
[tex]=\displaystyle\left(a-\frac{2a}{1-p}\right)\sum_{n=p+1}^{p+q}1+\frac{2a}{1-p}\sum_{n=p+1}^{p+q}n[/tex]
[tex]=\displaystyle\left(a-\frac{2a}{1-p}\right)q+\frac{2a}{1-p}\left(\frac{q(2p+q+1)}2\right)[/tex]
[tex]=\displaystyle\frac{aq(p+q)}{1-p}[/tex]
[tex]=\displaystyle-\frac{aq(p+q)}{p-1}[/tex]
as desired.
