In a geometric progression with positive term | vedclass

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(D) Let the geometric progression be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.Given that each term is equal to the sum of the next two terms,we

Vedclass · Accepted Answer

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

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(D) Let the geometric progression be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.Given that each term is equal to the sum of the next two terms,we have:$a_n = a_{n+1} + a_{n+2}$$ar^{n-1} = ar^n + ar^{n+1}$Dividing by $ar^{n-1}$ (since $a, r 
eq 0$):$1 = r + r^2$$r^2 + r - 1 = 0$Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$Since the terms are positive,the common ratio $r$ must be positive.Therefore,$r = \frac{\sqrt{5} - 1}{2}$.

In a geometric progression with positive terms,if each term is equal to the sum of the next two terms,then the common ratio of the progression is = .......

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