Let the first term of the AP be a and the common difference be d

Given: S_m = m²p and S_n = n²p

To prove: S_p = p³

According to the problem

⇒2a + (m - 1)d = 2mp
⇒2a + (n - 1)d = 2np
Subtracting the equations we get,

(m - n)d = 2p(m - n)

Now m is not equal to n

So d = 2p
Substituting in 1^st equation we get

2a + (m - 1)(2p) = 2mp
⇒a = mp - mp + p = p

Hence proved.