<s>[INST] 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.
 [/INST] Well,  If $a,b,c$ form an arithmetic progression then there is a $d$ so that $a=b-d$ and $c=b+d$.
And to show that $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ form an arithmetic progression we have to show there is a $k$ so that
$k = (a^2+ac+c^2) - (a^2+ab+b^2)$ and $k= (b^2+bc+c^2) -(a^2+ac+c^2)$ or in other words that
$(a^2+ac+c^2) - (a^2+ab+b^2)= (b^2+bc+c^2) -(a^2+ac+c^2)$.  So if we replace $a=b-d$ and $c = b+d$ we must show
$((b-d)^2+(b-d)(b+d)+(b+d)^2) - ((b-d)^2+(b-d)b+b^2)= (b^2+b(b+d)+(b+d)^2) -((b-d)^2+(b-d)(b+d)+(b+d)^2)$
Can we?
$((b-d)^2+(b-d)(b+d)+(b+d)^2) - ((b-d)^2+(b-d)b+b^2)=$
$((b-d)(b+d)+(b+d)^2) - ((b-d)b+b^2)=$
$((b-d)d+(b+d)^2) - (b^2)=$
$(b-d)d + 2bd+d^2=$
$3bd $.
And
$(b^2+b(b+d)+(b+d)^2) -((b-d)^2+(b-d)(b+d)+(b+d)^2)=$
$(b^2+b(b+d)) -((b-d)^2+(b-d)(b+d))=$
$(b^2) -((b-d)^2-d(b+d))=$
$(b^2) -(b^2-2bd-bd)=$
$3bd$.
Yep, they are equal all right.
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To take your idea of $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$ into account:
We know $b = \frac{a+c}2$ and we need to prove
$a^2+ac+c^2= \frac {(a^2+ab+b^2)+(b^2+bc+c^2)}2$ given that $b = \frac {a+c}2$
So
$\frac {(a^2+ab+b^2)+(b^2+bc+c^2)}2=$
$\frac {a^2 + c^2 + b(a+c) + 2b^2}2=$
$\frac {a^2 +c^2 + \frac {a+c}2(a+c) + 2(\frac {a+c}2)^2}2=$
$\frac {a^2 + c^2 + \frac {(a+c)^2}2 + 2\frac {(a+c)^2}4}2=$
$\frac {a^2 + c^2 + (a+c)^2}2=$
$\frac {a^2 + c^2 + a^2 + 2ac + c^2}2 =$
$\frac {2a^2 + 2ac + 2c^2}2 =$
$a^2 + ac + c^2$.
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