A) \[\frac{\sqrt{5}-1}{2}\]
B) \[\frac{1-\sqrt{5}}{2}\]
C) 1
D) \[2\sqrt{5}\]
Correct Answer: A
Solution :
Under condition \[{{T}_{n}}={{T}_{n+1}}+{{T}_{n+2}}\] \[\Rightarrow a{{r}^{n-1}}=a{{r}^{n}}+a{{r}^{n+1}}\Rightarrow {{r}^{n-1}}={{r}^{n}}(1+r)\] \[\Rightarrow \] \[{{r}^{2}}+r-1=0\Rightarrow r=\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}\] Since, each term is \[+ve\]. Hence common ratio is \[\frac{\sqrt{5}-1}{2}\].
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