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If a, b and c be three distinct real numbers in G.P and $a+b+c = xb$ then $x$ cannot be:
Option 1)
-2
Option 2)
-3
Option 3)
4
Option 4)
2

Answers (1)

best_answer

General term of a GP -

$T_{n}= ar^{n-1}$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

from the concept we have learnt

Let the three terms be

$\frac{b}{r},b,br\rightarrow in \: \: GP$

given $a+b+c=xb$

$\Rightarrow \frac{b}{r}+b+br=xb$

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

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

from the AM-GM $r+\frac{1}{r}>2\: \: \: \: \: \: \: \: \: or\: \: \: \: \: \: \: \: \: \: \: \: \: \: r+\frac{1}{r}<-2$

So, $x-1>2\: \: \: \: \: \: \: \: \: \: \: \: \: or\: \: \: \: \: \: \: \: \: x-1<-2$

$\Rightarrow x>3\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or\: \: \: \: \: \: \: \: \: \: \: x<-1$

So, $x\neq 2$

Option 1)

-2

Option 2)

-3

Option 3)

4

Option 4)
2

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