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Question

Let $$a, b, c$$ be positive integers such that $$\displaystyle \frac{b}{a}$$ 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 $$\displaystyle \dfrac{a^2+a-14}{a+1}$$ is ................


A
3
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B
4
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C
5
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D
6
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Solution

The correct option is B $$4$$
$$\frac{b}{a} = \frac{c}{b} = (integer)$$
$$b^2 = ac \Rightarrow c = \frac{b^2}{a}$$
$$\frac{a+b+c}{3} = b +2$$
$$a+b+c = 3b + 6  \Rightarrow a - 2b + c= 6$$
$$a-2b + \frac{b^2}{a} = 6 \Rightarrow 1 - \frac{2b}{a} + \frac{b^2}{a^2} = \frac{6}{a}$$
$$\displaystyle \left ( \frac{b}{a}- 1 \right )^2 = \frac{6}{a}  \Rightarrow a = 6 \ only$$

Mathematics

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