S_{n} = qn^{2} + pn

**Taking n = 1, we get**

S_{1} = q(1)^{2} + p(1)

⇒ S_{1} = q + p

⇒ a_{1} = q + p

**Taking n = 2, we get**

S_{2} = q(2)^{2} + p(2)

⇒ S_{2} = 4q + 2p

∴ a_{2} = S_{2} – S_{1} = 4q + 2p – q - p = 3q + p

**Taking n = 3, we get**

S_{3} = q(3)^{2} + p(3)

⇒ S_{3} = 9q + 3p

∴ a_{3} = S_{3} – S_{2} = 9q + 3p – 4q – 2p = 5q + p

So, a = q + p,

d = a_{2} – a_{1} = 3q + p – (q + p) = 3q + p – q – p = 2q

**Hence, the common difference is 2q.**