If $a^2 + b^2+16c^2=2(3ab+6bc + 4ac)$ , where $a,b,c$ are non zero numbers. Then $a,b,c$ are in __________?
1. Harmonic progression 2. Geometric progression 3. Arithmetic progression 4. None of these
My attempt:
$ a \rightarrow 4a'$
$ b \rightarrow 4b'$
$a'^2 +b'^2 +c^2 - 6a'b' -3b'c -2a'c=0$
Ok, this looks nicer than the original thing but still the solution doesn't seem in sight. Further, I started wondering, would the relationship between numbers be preserved under transformations to the equation?
Try $a=b=4$ and $c=\frac{5+\sqrt{41}}{2}.$
We got: non one of them.