Common sum in magic square

Question

A magic square of size $N,N ≥ 2$, is an $N ×N$ matrix with integer entries such that the sums of the entries of each row, each column and the two diagonals are all equal. If the entries of the magic square are made up of integers in arithmetic progression with first term a and common difference d, what is the value of this common sum?

My attempt for this question is Suppose i took $n^2$ numbers as $ a, a+d,a+2d,...a+(n^2-1)d$ and now i want to put these numbers in n×n square matrix such that sum of each row and sum of each column and sum of each diagonal is same. So i want to know what is value of that common sum?

Answer 1

You have to just sum the series
$$a,a+d,a+2d,\cdots,a+(N^2-1)d$$

$$S=\frac{N^2}{2}\cdot(2a+(N^2-1)d)$$

Sum of all columns is same Thus sum of one column is $$S=\frac{N}{2}\cdot(2a+(N^2-1)d)$$

Answer 2

You can add up all the entries in the square.

$$a+(a+d)+ \cdots + a+(n^2-1)d = S.$$

If the sum of the entries in each column is $C$ then you have

$$NC = S.$$