Mark the correct alternative in the following question:

As, $a_{p}=q$

$\Rightarrow a+(p-1) d=q \quad \ldots \ldots(\mathrm{i})$

Also, $a_{(p+q)}=0$

$\Rightarrow a+(p+q-1) d=0 \quad \ldots$. (ii)

Subtracting (i) from (ii), we get

$a+(p+q-1) d-a-(p-1) d=0-q$

$\Rightarrow(p+q-1-p+1) d=-q$

$\Rightarrow q d=-q$

$\Rightarrow d=\frac{-q}{q}$

$\Rightarrow d=-1$

Substituting $d=-1$ in (i), we get

$a+(p-1) \times(-1)=q$

$\Rightarrow a-p+1=q$

$\Rightarrow a=p+q-1$

Now,

$a_{q}=a+(q-1) d$

$=p+q-1+(q-1) \times(-1)$

$=p+q-1-q+1$

$=p$

Hence, the correct alternative is option (b).