Question

If $a_1=1$ and $a_{n+1}=\frac{4+3a_n}{3+2a_n},n\ge 1$ and if $\underset{n\to \infty }{lim}{a}_n=n$, then the value of a is?

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $2$
• D. None of these

Answer 1

The correct option is A $\sqrt{2}$

We have, $a_{n+1}=\frac{4+3a_n}{3+2a_n}$
$\Rightarrow \underset{n\to \infty }{lim}{a}_{n+1}=\underset{n\to \infty }{lim}\frac{4+3a_n}{3+2a_n}$
$\Rightarrow a=\frac{4+3a}{3+2a}\Rightarrow 2a^2=4$
$\Rightarrow a=\sqrt{2}$
and $a\ne -\sqrt{2}$ because each $a_n>0$.
Therefore, $\underset{n\to \infty }{lim}{a}_n=a>0$.