If a1=1 and an+1=4+3an3+2an,n≥1 and if limn→∞ an=a, then the value of a, is

A
$\sqrt{2}$
B
$- \sqrt{2}$
C
2
D
none of these

Solution:

We have,

$\begin{array}{l} a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}} \\ \Rightarrow lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{4 + 3 a_{n}}{3 + 2 a_{n}} \\ \Rightarrow a = \frac{4 + 3 a}{3 + 2 a}, where a = lim_{x \to \infty} a_{n} \\ \Rightarrow 2 a^{2} = 4 \Rightarrow a = \pm \sqrt{2} \end{array}$

But, $a \neq - \sqrt{2} because each a_{n} > 0 for all n$ .

Hence $a = \sqrt{2 .}$

If $a_{1} = 1 and a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}, n \geq 1 and if lim_{n \to \infty} a_{n} = a$ , then the value of $a$ , is

Solution:

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