Arithmetic?

Algebra Level 3

$\begin{array}{|c|c|c|c}\hline a_1 & a_2 & a_3 &\cdots \\ \hline b_1 & b_2 & b_3 & \cdots \\ \hline c_1& c_2 & c_3 & \cdots \\ \hline \vdots &\vdots&\vdots&\ddots\\ \end{array}$

Every row or column in the table is an arithmetic progression.

Is the diagonal $a_1,b_2,c_3,\ldots$ also in an arithmetic progression?

Try my set here .

Yes Not always

3 solutions

Poca Poca
May 27, 2018

The multiplication table is a counterexample. The differences withon rows/columns are constant, but since the numbers on the diagonal are the perfect squares the diffefences between them are increasing because they are sums of consecutive odd numbers.

Michael Mendrin
May 26, 2018

Here's one example of how it could be done where the diagonal is not in arithmetric progression

0	1	2	3
8	15	22	29
16	29	42	55
24	43	62	81

This is a doubly ruled surface that is not always a plane, i.e., this one would be a hyperbolic paraboloid. The general equation for this array is

$(a(n-x)+bx)(n-y)+(c(n-x)+dx)y +e$

where $n$ is the size of the square array, $x,y$ is the position of the numbers, and $a,b,c,d,e$ are arbitrary integers. If $a-b-c+d \ne 0$ , then the diagonal terms are not in arithmetic progression.

Zeeshan Ali
May 27, 2018

Here is an example that I reached to after a few hit-and-trials that show that diagonals are not always an AP;

	$d_{c_1}=0$	$d_{c_2}=1$	$d_{c_3}=2$	$d_{c_4}=3$
$d_{r_1}=3$	1	4	7	10
$d_{r_2}=4$	1	5	9	13
$d_{r_3}=5$	1	6	11	16
$d_{r_4}=6$	1	7	13	19

It was a fun fact how the differences in rows and columns formed an AP as well. (^ ^)

The diagonals in this case which are $1,5,11,19$ do not form an AP. However, the differences, $4,6,8,\dots$ , of these diagonals formed an AP.

Arithmetic?

3 solutions

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