JEE MAIN - Mathematics (2007 - No. 19)
In a geometric progression consisting of positive terms, each term equals the sum of the next two terns. Then the common ratio of its progression is equals
$${\sqrt 5 }$$
$$\,{1 \over 2}\left( {\sqrt 5 - 1} \right)$$
$${1 \over 2}\left( {1 - \sqrt 5 } \right)$$
$${1 \over 2}\sqrt 5 $$.
Explanation
Let the series $$a,ar,$$ $$a{r^2},........$$ are in geometric progression.
given, $$a = ar + a{r^2}$$
$$ \Rightarrow 1 = r + {r^2}$$
$$ \Rightarrow {r^2} + r - 1 = 0$$
$$ \Rightarrow r = {{ - 1 \mp \sqrt {1 - 4 \times - 1} } \over 2}$$
$$ \Rightarrow r = {{ - 1 \pm \sqrt 5 } \over 2}$$
$$ \Rightarrow r = {{\sqrt 5 - 1} \over 2}$$
[ As terms of $$G.P.$$ are positive
$$\therefore$$ $$r$$ should be positive]
given, $$a = ar + a{r^2}$$
$$ \Rightarrow 1 = r + {r^2}$$
$$ \Rightarrow {r^2} + r - 1 = 0$$
$$ \Rightarrow r = {{ - 1 \mp \sqrt {1 - 4 \times - 1} } \over 2}$$
$$ \Rightarrow r = {{ - 1 \pm \sqrt 5 } \over 2}$$
$$ \Rightarrow r = {{\sqrt 5 - 1} \over 2}$$
[ As terms of $$G.P.$$ are positive
$$\therefore$$ $$r$$ should be positive]
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