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 $a,b,c$ is $b+2$, then the value of $\frac{a^2+a-14}{a+1}$ is

Answer 1

We know that,
$\frac{a+b+c}{3}=b+2\Rightarrow a+c=2(b+3)$
We know that $a,b,c$ are in $G.P.$
Let the common ratio be $r\in I$
$\Rightarrow a+a{r}^2=2(ar+3)\Rightarrow a=\frac{6}{1+r^2-2r}\Rightarrow a=\frac{6}{(r-1{)}^2}$
As both $a$ and $r$, will be integer so only possible solution will be,
$r=2\Rightarrow a=6$
Now,
$\frac{a^2+a-14}{a+1}=\frac{36+6-14}{7}=4$