Mark the correct alternative in the following question:

As, $S_{n}=n^{2} q$

$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=n^{2} q$

$\Rightarrow 2 a+(n-1) d=\frac{n^{2} q \times 2}{n}$

$\Rightarrow 2 a+(n-1) d=2 n q$    ....(1)

Also, $S_{m}=m^{2} q$

$\Rightarrow \frac{m}{2}[2 a+(m-1) d]=m^{2} q$

$\Rightarrow 2 a+(m-1) d=\frac{m^{2} q \times 2}{m}$

$\Rightarrow 2 a+(m-1) d=2 m q \quad \ldots$ (ii)

Subtracting (i) from (ii), we get

$2 a+(n-1) d-2 a-(m-1) d=2 n q-2 m q$

$\Rightarrow(n-1-m+1) d=2 q(n-m)$

$\Rightarrow(n-m) d=2 q(n-m)$

$\Rightarrow d=\frac{2 q(n-m)}{(n-m)}$

$\Rightarrow d=2 q$

Substituting $d=2 q$ in (ii), we get

$2 a+(m-1) 2 q=2 m q$

$\Rightarrow 2 a+2 m q-2 q=2 m q$

$\Rightarrow 2 a=2 q$

$\Rightarrow a=q$

Now,

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

$=\frac{q}{2}[2 q+(q-1) 2 q]$

$=\frac{q}{2}\left[2 q+2 q^{2}-2 q\right]$

$=\frac{q}{2}\left[2 q^{2}\right]$

$=q^{3}$

Hence, the correct alternative is option (c).