Evaluate Expression Involving Geometric Progression Terms

Given that $a, b, c$ are positive integers forming a geometric progression with common ratio $r$.

It is given that $\frac{b}{a}$ is an integer, so $r = \frac{b}{a}$ is a positive integer.

We can express $b$ and $c$ in terms of $a$ and $r$ as: $b = ar$ $c = ar^2$

According to the question, the arithmetic mean of $a, b, c$ is equal to $b + 2$: $\frac{a + b + c}{3} = b + 2$

Multiplying both sides by $3$: $a + b + c = 3b + 6$ $a - 2b + c = 6$

Substituting $b = ar$ and $c = ar^2$ into the equation: $a - 2ar + ar^2 = 6$ $a(r^2 - 2r + 1) = 6$ $a(r - 1)^2 = 6$

Since $a$ and $r$ are positive integers, $(r - 1)^2$ must be a perfect square that is a factor of $6$. The positive factors of $6$ are $1, 2, 3,$ and $6$. The only perfect square among these factors is $1$.

Therefore, we have: $(r - 1)^2 = 1$

Since $r$ is a positive integer: $r - 1 = 1 \implies r = 2$

Substituting $(r - 1)^2 = 1$ back into the equation $a(r - 1)^2 = 6$: $a(1) = 6 \implies a = 6$

Now, we calculate the value of the given expression $\frac{a^2 + a - 14}{a + 1}$: $\frac{a^2 + a - 14}{a + 1} = \frac{6^2 + 6 - 14}{6 + 1} = \frac{36 + 6 - 14}{7} = \frac{28}{7} = 4$