Given A.P. has first term(a) = (x – y)²,

number of terms(n) = n and

Common difference(d) = x² + y² – (x – y)² = x² + y² – x² + y² + 2xy = 2xy

So, Sum of A.P. = S_n = n[2a + (n – 1)d] / 2

= n[2(x – y)²+(n – 1)(2xy)]/2

= n[(x – y)²+(n – 1)(xy)]

Hence, the sum of the first n terms of A.P. is n[(x – y)²+(n – 1)(xy)].