All terms of a geometric progression are posi | vedclass

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(D) Let the geometric progression be $a, ar, ar^2, ar^3, \dots$ where $a > 0$ and $r > 0$.According to the problem,each term is equal to the sum of th

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the geometric progression be $a, ar, ar^2, ar^3, \dots$ where $a > 0$ and $r > 0$.According to the problem,each term is equal to the sum of the two terms following it:$ar^{n-1} = ar^n + ar^{n+1}$Dividing both sides by $ar^{n-1}$ (since $a 
eq 0$ and $r 
eq 0$):$1 = r + r^2$Rearranging the terms gives the quadratic equation:$r^2 + r - 1 = 0$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 all terms are positive,the common ratio $r$ must be positive.Therefore,we take the positive root:$r = \frac{\sqrt{5} - 1}{2} = \frac{1}{2}(\sqrt{5} - 1)$

All terms of a geometric progression are positive. If each term is equal to the sum of the two terms following it,then the common ratio of this progression is:

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