The first term of an A.P.is a,

Given first term is ' $\mathrm{a}$ ' and sum of first $\mathrm{p}$ terms is $\mathrm{S}_{p}=0$

Now we have to find the sum of next q terms

Therefore, total terms are $p+q$

Hence, sum of all terms minus the sum of the first $p$ terms will give the sum of

next q terms

But sum of first $p$ terms is 0 hence sum of next $q$ terms will be the same as sum

of all terms

So, we have to find sum of $p+q$ terms

Sum of $n$ terms of $A P$ is given by $S_{n}=\left(\frac{n}{2}\right)(2 a+(n-1) d)$

Where $\mathrm{a}$ is first term and $\mathrm{d}$ is the common difference

Using the given hint we get

$\Rightarrow$ required sum $=\mathrm{S}_{p+q}-\mathrm{S}_{p}$

Using $\mathrm{S}_{\mathrm{n}}$ formula

$\Rightarrow$ required sum $=\frac{p+q}{2}(2 a+(p+q-1) d)$

Now we have to find $d$

We use the given $\mathrm{S}_{p}=0$ to find $\mathrm{d}$

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

$\Rightarrow 0=2 a+(p-1) d$

$\Rightarrow d=-\frac{2 a}{p-1}$

Put this value of $d$ in 1 we get

$\Rightarrow$ required sum $=\frac{p+q}{2}\left(2 a+(p+q-1)\left(-\frac{2 a}{p-1}\right)\right)$

Taking LCM and simplifying we get

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

On computing we get

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

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

$=\mathrm{a}(\mathrm{p}+\mathrm{q})\left(1+(-1)-\frac{\mathrm{q}}{\mathrm{p}-1}\right)$

$=-\frac{a(p+q) q}{p-1}$

Hence proved.