Sequences And Series Ques 27
- Let $a, b, c$ be positive integers such that $b / a$ 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
(2014 Adv.)
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Solution: Plan
(i) If $a, b, c$ are in GP, then they can be taken $a s a, a r, a r^2$ where $r,(r \neq 0)$ is the common ratio.
(ii) Arithmetic mean of $x_1, x_2, \ldots, x_n=\frac{x_1+x_2+\ldots+x_n}{n}$
Let $a, b, c$ be $a, a r, a r^2$, where $r \in N$
$ \begin{array}{llrl} \text { Also, } & \frac{a+b+c}{3} =b+2 \\ \Rightarrow & a+a r+a r^2 =3(a r)+6 \\ \Rightarrow & a r^2-2 a r+a =6 \\ \Rightarrow & (r-1)^2 =\frac{6}{a} \end{array} $
Since, $6 / a$ must be perfect square and $a \in N$. So, $a$ can be $6$ only.
$\Rightarrow \quad r-1= \pm 1 \Rightarrow r=2$
and $\quad \frac{a^2+a-14}{a+1}=\frac{36+6-14}{7}=4$
BETA
