It's Somewhat Tricky!

Given the fact that $a,b,c$ are in GP, let $b=ar$ and $c=ar^2$ , such that $r\in Z^{+}$ .

$\dfrac{a+ar+ar^2}{3}=ar+2\\\Rightarrow a-2ar+ar^2=6\\\Rightarrow a(r-1)^2=6$

Now we know that no divisor of $6$ is perfect square except $1$ , so $(r-1)=1$ or $r=2$ . It gives $a=6$ .

Substituting $a=6$ , we get $\dfrac{6^2+6-14}{6+1}=\dfrac{28}{7}=4$

Let us say that $a=x,b=xr,c=xr^2$ .Thus, $\cfrac{b}{a}=r$ .Now,the arithmetic mean of $a,b,c$ $=\dfrac{a+b+c}{3}\\ =\dfrac{x+xr+xr^2}{3}$ .Which is given to be equal to $xr+2$ .Thus, $\dfrac{x+xr+xr^2}{3}=xr+2\\ \Rightarrow x+xr+xr^2=3xr+6$ .Dividing by $x$ gives, $1+r+r^2=3r+\dfrac{6}{x}\\ \Rightarrow r^2-2r+(1-\dfrac{6}{x})=0$ .This is a quadratic equation whose roots are $\dfrac{2\pm\sqrt{4-4+\dfrac{24}{x}}}{2}$ .Now,for $r$ to be an integer, $\dfrac{24}{x}=k^2$ .This condition is satisfied by $x=6$ and $x=24$ but for the latter, $r$ is not an integer.Thus, $\boxed{x=6=a}$ .

Using a time saver and 'IIT like' method,

Answer must be an Integer

Putting Values and checking,

Value of expression becomes an integer on a=6

Since, other values of b and c can be found for this value of a, hence a=6 must give an answer.

Now, why it's only on a=6, lets see.

Breaking the given expression in two parts,

I get a-14/(a+1)

again which is positive integer for a=6 and 13,

for a=13 b comes out to be non integral value . Hence it is not allowed.

Let $a= \frac{b}{r}$ and $c=br$ where r is the common ratio. Now we have $\frac{b}{r} +b + br = 3(b+2)$ . Now we have $b(1+r+r^2) = 3r(b+2)$ or $b(1+r^2 - 2r) = 6r$ . Hence $b(r-1)^2 = 6r$ . Since $b= ar$ , we can rewrite the equation to $a(r-1)^2 = 6$ . Bearing in mind that a and r have to be positive integers, the only possible value for r is 2. Now we have a=6.

Evidently, the required expression can be simplified to $a - \frac{14}{a+1} = 6 -\frac{14}{7} = 6-2 = \boxed{4}$