Prove that this is an arithmetic progression.

Question

If a, b, c is an arithmetic progression prove that $$a^2-bc, b^2-ac, c^2-ab$$ is an arithmetic progression. Well firstly I stated that: b-a=c-b, then I showed that $$b^2-ac-a^2+bc=c^2-b^2-ab+ac$$ What should I do next?

Answer 1

We know that $b=a+d, c=a+2d$ for some $d\in\mathbb R$.

Substitute that in your equality

$$b^2-ac-a^2+bc=c^2-b^2-ab+ac$$

and you get the identity $3d^2+3ad=3d^2+3ad$.

Thus the equality always holds and by the definition of arithmetic progression, since the differences between the consecutive terms are the same, the terms form an arithmetic progression.

Answer 2

Let $A=a^2-bc; B=b^2-ac; C=c^2-ab$

If you have really proved the equation you state, then you can rewrite it $B-A=C-B$ - there is a common difference, so $A,B,C$ is an arithmetic progression. Note that your equation factorises as $(b-a)(a+b+c)=(c-b)(a+b+c)$.

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It is sometimes useful to write a three term arithmetic progression as $b-d, b, b+d$ - the symmetry sometimes makes things obvious. Here it would give directly:

$A=(b-d)^2-b(b+d)=d^2-3bd$

$B=b^2-(b-d)(b+d)=d^2$

$C=(b+d)^2-b(b-d)=d^2+3bd$