Arithmetic and geometric progressions based question

Question

Q. Let a, b, c be positive integers such that b/a is an integer. If a, b, c are in geometric progression and the arithmetic mean of a, b, c is b+2, the value of (a²+a-14) /(a+1) is..?

Answer 1

Write $d=b/a$ so $a+b+c=3ad+6=a(1+d+d^2)$, i.e. $6=(1-d)^2a$. The only square dividing $6$ is $1$, so $a=6$. Hence $\frac{a^2+a-14}{a+1}=a-\frac{14}{a+1}=6-2=4$.