In a geometric progression consisting of posi | vedclass

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(A) Let the terms of 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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the terms of 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:$a = ar + ar^2$Since $a > 0$,we can divide both sides by $a$:$1 = r + r^2$Rearranging the terms,we get 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)}$$r = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$Since the terms of the geometric progression are positive,the common ratio $r$ must be positive.Therefore,we take the positive root:$r = \frac{\sqrt{5} - 1}{2}$

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

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