The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{ } \forall \text{ }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.
Best Answer
$(a^2+ac+c^2)-(a^2+ab+b^2)=a(c-b)+c^2-b^2=(c-b)(a+b+c)\\(b^2+bc+c^2)-(a^2+ac+c^2)=b^2-a^2+(b-a)c=(b-a)(a+b+c)$
Are they equal?