JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 4)
Let a, b, c be positive integers such that $${b \over a}$$ 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 $${{{a^2} + a - 14} \over {a + 1}}$$ is
Answer
4
Explanation
Let $a=a, b=a r$ and $c=a r^2$, where $r$ is integer since $${b \over a}$$ is an integer.
According to the question, we have
$\frac{a+b+c}{3}=b+2$ [$$ \because $$ $($ A.M. $)=(b+2)$]
$$ \Rightarrow $$ $$ \frac{a+a r+a r^2}{3}=a r+2 $$
$\begin{aligned} & a+a r+a r^2=3 a r+6 \\\\ &\Rightarrow a r^2-2 r+a=6\end{aligned}$
$\Rightarrow \underbrace{r^2-2 r+1}_{\text {integer }}=\underbrace{\frac{6}{a}}_{\text {integer }}$
$\Rightarrow(r-1)^2=\frac{6}{a}$
If $a=1,2,3,4,5,6$, then it is not a perfect square and integer.
Therefore, the only possibility is that $a=6$. Thus,
$\frac{a^2+a-14}{a+1}=\frac{36+6-14}{6+1}=\frac{284}{7}=4$
According to the question, we have
$\frac{a+b+c}{3}=b+2$ [$$ \because $$ $($ A.M. $)=(b+2)$]
$$ \Rightarrow $$ $$ \frac{a+a r+a r^2}{3}=a r+2 $$
$\begin{aligned} & a+a r+a r^2=3 a r+6 \\\\ &\Rightarrow a r^2-2 r+a=6\end{aligned}$
$\Rightarrow \underbrace{r^2-2 r+1}_{\text {integer }}=\underbrace{\frac{6}{a}}_{\text {integer }}$
$\Rightarrow(r-1)^2=\frac{6}{a}$
If $a=1,2,3,4,5,6$, then it is not a perfect square and integer.
Therefore, the only possibility is that $a=6$. Thus,
$\frac{a^2+a-14}{a+1}=\frac{36+6-14}{6+1}=\frac{284}{7}=4$
Comments (0)
