Question
Let the three non-zero real numbers $x(y - z),y(z - x),z(y - x)$ form a geometric progression in that order, with common ratio ' $r$ ', then the possible value $(s)$ of $\lbrack r\rbrack$ is equal to ([.] denotes greatest integer function)
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Solution
Sol. [Key Concept: Three reals in G.P.: use the common ratio to test each relation.]
$\Rightarrow \ $ Let $x(y - z) = a,y(z - x) = b,z(y - x) = C$
$$\Rightarrow \ a + b = C $$and $\ b^{2} = ac = a(a + b)$
$$\Rightarrow \ \frac{b^{2}}{a^{2}} - \frac{b}{a} - 1 = 0 \Rightarrow r = \frac{b}{a} = \frac{1 \pm \sqrt{5}}{2}$$
✘ (a) $\lbrack r\rbrack \neq - 2$
✔ (b) $r = \frac{1 - \sqrt{5}}{2} \approx - 0.618,\ \lbrack r\rbrack = - 1$
✘ (c) $\lbrack r\rbrack \neq 0$
✔ (d) $r = \frac{1 + \sqrt{5}}{2} \approx 1.618,\ \lbrack r\rbrack = 1$
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