If a, b and c be three distinct numbers in G.P. and a + b + c = xb then x can not be

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8. Sequences and Series

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If $a, b$ and $c$ be three distinct numbers in $G.P.$ and $a + b + c = xb$ then $x$ can not be

A

$-2$

B

$-3$

C

$4$

D

$2$

(JEE MAIN-2019)

Solution

$a + ar + a{r^2} = xar$

since $a \ne 0\,\,$ so $\frac{{{r^2}r + 1}}{r} = x$; $1 + r + \frac{1}{r} = x$

$r + \frac{1}{r} \in ( – \infty , – 2] \cup [2,\infty )\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x \in ( – \infty , – 1] \cup [3,\infty )$

Standard 11

Mathematics

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