Answer
a) Recursive formula is $a_{n+1}=\sqrt {6+a_n}$
b) Converges to $3$
Work Step by Step
a) We can see that our recursion formula is $a_{n+1}=\sqrt {6+a_n}$
b) Let us consider that $a_{n+1} \approx a_n =l$
For a converging series, we have: $\lim\limits_{n \to \infty}a_{n+1}= \lim\limits_{n \to \infty} a_n=x$
Now, $\lim\limits_{n \to \infty}a_{n+1}= \lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt {6+a_n}$
So, $x=\sqrt {6+x}$
After squaring, we have: $x^2=6+x \\ \implies x^2-x-6=0 \\ \implies (x-3)(x+2) =0 \\ \implies x=-2, 3$
But the limit does not attain a negative value for a square root of a number sequence, so $x=-2$ must be neglected. So, our sequence converges to $3$.
