Let a, b, c be positive integers such that fr | vedclass

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(D) Given that $a, b, c$ are in geometric progression,let $b = ar$ and $c = ar^2$,where $r$ is an integer since $\frac{b}{a} = r$ is an integer.The ar

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$4$

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(D) Given that $a, b, c$ are in geometric progression,let $b = ar$ and $c = ar^2$,where $r$ is an integer since $\frac{b}{a} = r$ is an integer.The arithmetic mean of $a, b, c$ is given by $\frac{a+b+c}{3} = b+2$.Substituting $b = ar$ and $c = ar^2$,we get $\frac{a + ar + ar^2}{3} = ar + 2$.Multiplying by $3$,we have $a + ar + ar^2 = 3ar + 6$,which simplifies to $a + ar^2 = 2ar + 6$.Rearranging the terms,we get $a(1 - 2r + r^2) = 6$,or $a(r-1)^2 = 6$.Since $a$ and $r$ are integers,$(r-1)^2$ must be a factor of $6$. The perfect square factors of $6$ are $1$. Thus,$(r-1)^2 = 1$,which implies $r-1 = 1$ (since $r$ must be positive for $a, b, c$ to be positive integers),so $r = 2$.Substituting $r = 2$ into $a(r-1)^2 = 6$,we get $a(1)^2 = 6$,so $a = 6$.Now,we calculate the value of $\frac{a^2+a-14}{a+1}$ for $a = 6$:$\frac{6^2 + 6 - 14}{6 + 1} = \frac{36 + 6 - 14}{7} = \frac{28}{7} = 4$.

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

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