1. Tardigrade
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  4. If a1 = 1 and an+1 = (4+3an/3+2an), n ge 1 and if displaystyle limn→∞ an = n, then the value of a is

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

Limits and Derivatives

Solution:

We have $a_{n+1} = \frac{4+3a_{n}}{3+2a_{n}} \Rightarrow \displaystyle\lim_{n\to\infty} a_{n+1} = \lim _{n\to \infty } \frac{4+3a_{n}}{3+2a_{n}}$
$\Rightarrow \quad a = \frac{4+3a_{n}}{3+2a_{n}} \Rightarrow 2a^{2} = 4 \Rightarrow a = \sqrt{2}$
$a \ne \sqrt{2}$ because each $a_{n} > 0$, therefore $lim \,a_{n} = a > 0.$