In a geometric progression consisting of posi | vedclass

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Question

Let the series $a, a r, a r^{2}, \ldots \ldots \ldots$ are in geometric progression

given, $a=a r+a r^{2}$

$\Rightarrow 1=r+r^{2}$

$\Rightarr

Vedclass · Accepted Answer

$\frac{{\sqrt 5 - 1}}{2}$

Vedclass · Answer

Let the series $a, a r, a r^{2}, \ldots \ldots \ldots$ are in geometric progression

given, $a=a r+a r^{2}$

$\Rightarrow 1=r+r^{2}$

$\Rightarrow r^{2}+r-1=0$

$\Rightarrow r=\frac{-1+\sqrt{1-4 	imes-1}}{2}$

$\Rightarrow r=\frac{-1+\sqrt{5}}{2}$

$\Rightarrow r=\frac{\sqrt{5}-1}{2}$

[As terms of $G . P .$ are positive

$	herefore r 	ext { should be positive }]$

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals

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