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8. Sequences and Series
hard
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
