A geometric progression consists of positive | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the geometric progression be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.Given that each term is the sum of the next two terms,we have $T_n = T_{n+1} + T_{n+2}$.Substituting the general term formula $T_n = ar^{n-1}$,we get $ar^{n-1} = ar^n + ar^{n+1}$.Dividing both sides by $ar^{n-1}$ (since $a 
eq 0$ and $r 
eq 0$),we obtain $1 = r + r^2$.Rearranging the terms gives the quadratic equation $r^2 + r - 1 = 0$.Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since the terms are positive,the common ratio $r$ must be positive.Therefore,$r = \frac{\sqrt{5} - 1}{2}$.

$A$ geometric progression consists of positive terms. If each term is equal to the sum of the next two terms,what is the common ratio of the progression?

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