In a geometric progression consisting of positive terms, each term equals the sum of the next two terms.…

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals
  1. $\frac{1}{2}(1-\sqrt{5})$
  2. $\frac{1}{2} \sqrt{5}$
  3. $\sqrt{5}$
  4. $\frac{1}{2}(\sqrt{5}-1)$

Solution

Given $a r^{n-1}=a r^n+a r^{n+1}$ $\begin{aligned} & \Rightarrow 1=r+r^2 \\ & \therefore r=\frac{\sqrt{5}-1}{2} . \end{aligned}$

Asked in: JEE Main 2007

Practice more Sequences and Series questions on Aicharya