A) \[\sqrt{2}\]
B) \[-\sqrt{2}\]
C) 2
D) None of these
Correct Answer: A
Solution :
We have \[{{a}_{n+1}}=\frac{4+3{{a}_{n}}}{3+2{{a}_{n}}}\] \[\Rightarrow \,\,\underset{n\to \infty }{\mathop{\lim }}\,\,{{a}_{n+1}}=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{4+3{{a}_{n}}}{3+2{{a}_{n}}}\] \[\Rightarrow \,\,a=\frac{4+3a}{3+2a}\,\Rightarrow 2{{a}^{2}}=4\,\,\Rightarrow \,\,a=\sqrt{2}\] \[a\ne -\sqrt{2}\] because each \[{{a}_{n}}>0,\] therefore \[\lim \,{{a}_{n}}=a>0.\]
You need to login to perform this action.
You will be redirected in
3 sec
