MCQ
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$1$
- B$2$
- C$3$
- ✓$4$
Also $\frac{a+a r+a r^2}{3}=a r+2 \Rightarrow a+a r^2=2 a r+6$
$\Rightarrow \quad a ( r -1)^2=6 \quad \Rightarrow \quad r$ must be 2 and $a =6$.
Thus $\frac{a^2+a-14}{a+1}=\frac{36+6-14}{7}=4$ Ans.
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