Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesG.P.Easy2 minPYQ_2019

MathematicsEasystatement

If $a, b$ and $c$ be three distinct real numbers in G.P. and $a + b + c = x b,$ then $x$ cannot be:

Options:

Answer:

Solution:

Given $a, b, c$ are in G.P.

Let common ratio of G.P. be $r$ , then $b = a r, c = a r^{2}$

Given equation can be written as

$a + a r + a r^{2} = x . a . r$

$\Rightarrow x = r + \frac{1}{r} + 1$

As we know $r > 0,$ then $r + \frac{1}{r} \geq 2 \Rightarrow x \geq 3$

$r < 0,$ then $r + \frac{1}{r} \leq - 2 \Rightarrow x \leq - 1$

$\Rightarrow x \in (- \infty, - 1] \cup [3, \infty)$

Hence, $x \neq 2 .$

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:G.P.

⏱ 2mℹ️ Source: PYQ_2019

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