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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?

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up vote 2 down vote accepted

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.)

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I dont understand why you have written $(N^2-1)$, as total number of entries $N^2$ thatswhy? –  Bunuelian Trick Oct 3 '12 at 8:14
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@Flute: Yes: the arithmetic progression has $N^2$ terms, so if $a$ is the first term and $d$ the common difference, the last term is $a+(N^2-1)d$. –  Brian M. Scott Oct 3 '12 at 8:22
    
thank you Brian –  Bunuelian Trick Oct 3 '12 at 8:23
    
@Flute: You’re welcome. –  Brian M. Scott Oct 3 '12 at 8:28
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