Sn = qn2 + pn
Taking n = 1, we get
S1 = q(1)2 + p(1)
⇒ S1 = q + p
⇒ a1 = q + p
Taking n = 2, we get
S2 = q(2)2 + p(2)
⇒ S2 = 4q + 2p
∴ a2 = S2 – S1 = 4q + 2p – q - p = 3q + p
Taking n = 3, we get
S3 = q(3)2 + p(3)
⇒ S3 = 9q + 3p
∴ a3 = S3 – S2 = 9q + 3p – 4q – 2p = 5q + p
So, a = q + p,
d = a2 – a1 = 3q + p – (q + p) = 3q + p – q – p = 2q
Hence, the common difference is 2q.