What is $4{(\frac ab )}^2 + \frac ba - a + 1$ 
What is $4{(\frac ab )}^2 + \frac ba - a + 1$ when $a, b, c$ are natural number and form geometry sequence with $b/a$ is an integer.
the average of a,b,c is b + 1

$a=a, b = ar, c=ar^2 = b^2 /a $ when $ r = b/a$. Also $c/b=b/a \rightarrow ac = b^2$
$a+b+c = 3b + 3 \rightarrow a+b+ b^2/a = 3b + 3$

So $r=b/a\in \mathbb{N}$ and thus $b=ar$ and $c=ar^2$. Now we have $$a-2ar+ar^2=3$$
so $$a(1-r)^2 = 3\implies a= 3\;\;\wedge\;\; 1-r = \pm 1\implies r= 2$$
So $b=6$ and now you can finish.
