First slide

Geometric progression

Question

Let $a, b$ and $c$ be distinct real numbers. If $a, b, c$ are in geometric progression and $a + b + c = x b,$ then $x$ lies in the set

Moderate

A

$(1, 3)$
B

$(- 1, 0) \cup (1, 2)$
C

$(- \infty, - 1) \cup (3, \infty)$
D

$(0, 1)$

Solution

$a + b + c = x b$
$\Rightarrow \frac{b}{r} + b + b r = x b$ , where $r$ is common
ratio of the G.P and $r \neq 1$
$\Rightarrow r + \frac{1}{r} = (x - 1)$
But $r + \frac{1}{r} < - 2$ or $r + \frac{1}{r} > 2$
$\begin{array}{l} ∴ x - 1 < - 2 or x - 1 > 2 \\ \Rightarrow x < - 1 or x > 3 \\ \Rightarrow x \in (- \infty, - 1) \cup (3, \infty) \end{array}$

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