Show that three numbers form an arithmetic progression 
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.

Since $2b=a+c$
$$a^2+ab+b^2+b^2+bc+c^2=a^2+b(a+c)+2b^2+c^2=a^2+\frac{(a+c)^2}{2}+2\times \frac{(a+c)^2}{4}+c^2=a^2+(a+c)^2+c^2=2(a^2+ac+c^2)$$
