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Question
If a, b, c are three distinct real numbers in G.P. and a + b + c = xb, then prove that either x< −1 or x > 3.
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Solution
\[\text { Let r be the common ratio of the given G . P } . \]
\[ \therefore b = \text { ar and } c = a r^2 \]
\[\text { Now }, a + b + c = bx\]
\[ \Rightarrow a + ar + a r^2 = arx\]
\[ \Rightarrow r^2 + \left( 1 - x \right)r + 1 = 0\]
\[ \text { r is always a real number } . \]
\[ \therefore D \geq 0\]
\[ \Rightarrow \left( 1 - x \right)^2 - 4 \geq 0\]
\[ \Rightarrow x^2 - 2x - 3 \geq 0\]
\[ \Rightarrow \left( x - 3 \right)\left( x + 1 \right) \geq 0\]
\[ \Rightarrow x > 3 \text { or }x < - 1 \text { and } x \neq 3 \text { or } - 1 \left[ \because \text { a, b and c are distinct real numbers } \right]\]
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