Use the condition that $a,b,c$ are in geometric progression

If $a,b,c$ are in G.P., then $b^2=ac.$

Also, it is given that $\dfrac{b}{a}$ is an integer. Let $\frac{b}{a}=r,$ where $r$ is a positive integer.

Then $b=ar.$ Since $a,b,c$ are in G.P., the common ratio is $r$, so $c=ar^2.$

Use the arithmetic mean condition

The arithmetic mean of $a,b,c$ is $b+2$, so $\frac{a+b+c}{3}=b+2.$

Substitute $b=ar$ and $c=ar^2$: $\frac{a+ar+ar^2}{3}=ar+2.$

Multiply by $3$: $a(1+r+r^2)=3ar+6.$

Rearranging, $a(r^2-2r+1)=6.$

Since $r^2-2r+1=(r-1)^2,$ we get $a(r-1)^2=6.$

Use positivity and integrality

Here $a$ is a positive integer and $(r-1)^2$ is a perfect square. So $(r-1)^2$ must be a positive divisor of $6$ that is also a perfect square.

Perfect square divisors of $6$ are only $1$.

Hence, $(r-1)^2=1 \implies r-1=\pm 1.$

Since $r$ is a positive integer, possibilities are:

• $r=2$
• $r=0$ (not allowed, since $r$ must be positive)

So $r=2.$

Then $a\cdot 1=6 \implies a=6.$

Evaluate the required expression

We need $\frac{a^2+a-14}{a+1}.$

Substitute $a=6$: \frac{6^2+6-14}{6+1}=rac{36+6-14}{7}=rac{28}{7}=4.

Final answer

$\boxed{4}$

Comparison with stored correct answer

Stored correct answer = $4$.

Our derived answer also equals $4$, so they agree.