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 ................

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is B $4$

$\frac{b}{a}=\frac{c}{b}=\left(integer\right)$
$b^2=ac\Rightarrow c=\frac{b^2}{a}$
$\frac{a+b+c}{3}=b+2$
$a+b+c=3b+6\Rightarrow a-2b+c=6$
$a-2b+\frac{b^2}{a}=6\Rightarrow 1-\frac{2b}{a}+\frac{b^2}{a^2}=\frac{6}{a}$
${(\frac{b}{a}-1)}^2=\frac{6}{a}\Rightarrow a=6only$