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

i) If a, b, c are in GP, then they can be taken as $a,ar,a{r}^2$ where $r,(r\ne 0)$ is the common ratio.
ii) Arithmetic mean of $x_1,x_2,\dots \dots x_n=\frac{x_1+x_2+\dots \dots +x_n}{n}$
Let a, b, c be $a,ar,a{r}^2$ where $rϵN$
Also, $\frac{a+b+c}{3}=b+2$
$\Rightarrow a+ar+a{r}^2=3\left(ar\right)+6$
$\Rightarrow a{r}^2-2ar+a=6$
$\Rightarrow (r-1{)}^2=\frac{6}{a}$
Since, $\frac{6}{a}$ must be perfect square and $aϵN$
So, a can be 6 only
$\Rightarrow r-1=\pm 1\Rightarrow r=2$
and $\frac{a^2+a-14}{a+1}=\frac{36+6-14}{7}=4$