If a, b and c be three distinct numbers in G. | vedclass

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Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$a + ar + a{r^2} = xar$

since $a 
e 0\,\,$    so $\frac{{{r^2}r + 1}}{r} = x$;    $1 + r + \frac{1}{r} = x$

$r + \frac{1}{r} \in ( - \infty , - 2] \cup [2,\infty )\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x \in ( - \infty , - 1] \cup [3,\infty )$

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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