Question

Let a, b, c be positive integers such that $\frac{b}{a}$ is an integer. If a, b, c are in geometric progression and the arithmetic mean of $a,b,c\text{is}b+2$, then the value of $\frac{a^2+a-14}{a+1}$ is

Answer 1

$a,ar,a{r}^2,0>1r$ is integer
$\frac{a+b+c}{3}=b+2$
$a+ar+a^{r}2=3(ar+2)$
$a+ar+a{r}^2=3ar+6$
$a{r}^2–2ar+a–6=0$
$a(r–1{)}^2=6$
$a=6,r=2$
So $\frac{a^2+a-14}{a+1}=\frac{6^2+6-14}{6+1}=\frac{28}{7}=4$