A magic square of order $N$ is an $N\times N$ matrix with positive integral entries such that the elements of every row, every collumn and the two diagonals all add up to the same number.
If a magic square is filled with numbers in AP starting with $a \in \mathbb{N}$ and common difference $d \in \mathbb{N}$ what is the value of this common sum?
I am thinking that it is just a sum of the AP given? I mean $$\frac{N}{2}[2a+(N-1)d]$$ Am I right?
Let $s$ be the common sum. Then each row sums to $s$, and there are $N$ rows, so the sum of all the numbers in the square is $Ns$. That is,
$$\frac{N^2}2\Big(2a+(N^2-1)d\Big)=Ns\;,$$
and $$s=\frac{N}2\Big(2a+(N^2-1)d\Big)\;.$$
(Note that you miscounted the number of terms in the sequence when you wrote down your expression for the sum.)