If the sum of n terms of an A.P. is (pn + qn2),

It is known that, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

According to the given condition,

$\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]=\mathrm{pn}+\mathrm{qn}^{2}$

$\Rightarrow \frac{\mathrm{n}}{2}[2 \mathrm{a}+\mathrm{nd}-\mathrm{d}]=\mathrm{pn}+\mathrm{qn}^{2}$

$\Rightarrow \mathrm{na}+\mathrm{n}^{2} \frac{\mathrm{d}}{2}-\mathrm{n} \cdot \frac{\mathrm{d}}{2}=\mathrm{pn}+\mathrm{qn}^{2}$

Comparing the coefficients of n2 on both sides, we obtain

$\frac{\mathrm{d}}{2}=\mathrm{q}$

$\therefore d=2 q$

Thus, the common difference of the A.P. is $2 q$.