a,b,c are three distinct real numbers and they are in G P If a+b+c = xb, then prove that - Maths - Complex Numbers and Quadratic Equations

Question

a,b,c are three distinct real numbers and they are in G P If
a+b+c = xb, then prove that x<-1 or x>3

Santosh Kumar · Accepted Answer

$a + b + c = x b \Rightarrow x b = a + b + c \Rightarrow x = \frac{a + b + c}{b} = 1 + (\frac{a + c}{b}) = 1 + (\frac{a}{b} + \frac{c}{b}) ...... (I)$

Let us consider $\frac{a}{b} and \frac{c}{b}$

when, both ratio $\frac{a}{b} and \frac{c}{b}$ is (+)ve.

A.M > G.M

$\frac{\frac{a}{b} + \frac{c}{b}}{2} \geq \sqrt{\frac{a}{b} + \frac{c}{b}} \frac{a}{b} + \frac{c}{b} \geq 2 \sqrt{(\frac{a c}{b^{2}})} \to (1) \geq 2 (∵ b^{2} = a c) ∴ x \geq 1 + 2 [using equation (1.)] \Rightarrow x \geq 3 when both \frac{a}{b} and \frac{c}{b} is (-) v e .$

Matrix a × value of $\frac{a}{b} + \frac{c}{b}$ is –2.

For example, let us take a G.P series 1, –2, 4.

Hence, a = 1, b = –2 and c = 4.

$\frac{a}{b} + \frac{c}{b} = \frac{1}{- 2} + \frac{4}{- 2} = \frac{- 1}{2} - 2 = \frac{- 5}{2}$

let us take another G.P series 6, –12, 24.

$\frac{a}{b} + \frac{c}{b} = \frac{6}{- 12} + \frac{24}{- 12} = \frac{- 1}{2} - 2 = \frac{- 5}{2} Hence, \frac{a}{b} + \frac{c}{b} < - 2 Thus, x = 1 + (\frac{a}{b} + \frac{c}{b}) < 1 - 2 \Rightarrow x < - 1 ∴ x \geq 3 or x < - 1.$

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a,b,c are three distinct real numbers and they are in G.P. If a+b+c = xb, then prove that x<-1 or x>3.