Why the linear system?

An arithmetic progression means that

a1 can be written a

a2 can be written a + d

a3 can be written a + 2d

a4 can be written a + 3d

a5 can be written a + 4d

a1 can be written a + 5d

Sub these in. Solve the system as normal and you'll find that you get lots of nice cancelling.

with solution -1,2 (independent of a1 to a6)

If they form an arithmetic progression, then a2 = a1 + d and a3 = a1 + 2d, etc

Then, we can substitute all the progression elements in terms of a1

. a1x + (a1+d)y = a1+2d or x = (a1+2d-a1y-dy)/a1

(a1+3d)x + (a1+4d)y = a1+5d

Substituting

(a1+3d)(a1+2d-a1y-dy)/a1 + a1y + 4dy = a1 + 5d

(a1+3d)(a1+2d-a1y-dy) + a1^2y + 4a1dy = a1^2 + 5a1d For clarity, I'm just going to use a

a^2 + 2ad - a^2y - ady + 3ad + 6d^2 - 3ady -3d^2y + a^2 y + 4ady = a^2 + 5ad

a^2 -a^2 - a^2y + a^2y + 2ad + 3ad -5ad - ady -3ady + 4ady + 6d^2 - 3d^2y =0 which is

6d^2 - 3d^2y = 0 factor out d^2

6 - 3y = 0 At this point, all the information pointing to the arithmetic progression has been simplified away. Thus, it does not matter which number you choose for a1 or what arithmetic difference you use. You will always end up with the same answer.

I don't know.