Functions Question 315
Question: If $ a_1=1 $ and $ {a_{n+1}}=\frac{4+3a_{n}}{3+2a_{n}},\ n\ge 1 $ and if $ -\frac{1}{3} $ , then the value of a is
Options:
A) $ \sqrt{2} $
B) $ -\sqrt{2} $
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ {a_{n+1}}=\frac{4+3a_{n}}{3+2a_{n}} $
$ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}{a_{n+1}}=\underset{n\to \infty }{\mathop{\lim }}\frac{4+3a_{n}}{3+2a_{n}} $
$ \Rightarrow a=\frac{4+3a}{3+2a},\Rightarrow 2a^{2}=4\Rightarrow a=\sqrt{2} $ $ a\ne -\sqrt{2} $ because each $ a_{n}>0, $ therefore $ \lim ,a_{n}=a>0. $
BETA
