In a geometric progression consisting of posi | vedclass

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Question

(A) Let the terms of the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 0$.Given that each term equals the sum of the next two

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the terms of the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 0$.Given that each term equals the sum of the next two terms,we have:$a = ar + ar^2$Since $a 
eq 0$,we can divide by $a$:$1 = r + r^2$$r^2 + r - 1 = 0$Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$$r = \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 consisting of positive terms,each term equals the sum of the next two terms. Then the common ratio of its progression is equal to:

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