Q:

Could its generalised version be reduced & evaluated in terms of the declared primative constants alone?

Let $a_0$ and $b_0$ be non-equal positive integers. Let's define the recursions
$$a_n = (a_{n-1} + b_{n-1})/2 $$
$$b_n = (a_n \cdot b_{n-1})^{1/2}$$
Can the above two terms be reduced in terms of $a_0$ & $b_0$ alone? If yes, could $\mathbf{b_∞}$ be evaluated ?

A:

This is the problem of Arithmatic-Geometric Mean of Gauss.
We'll proceed through the following claims.
$\bullet~$ $\textbf{Claim (a):}~$ The sequences $\{ a_{n} \}_{n \geqslant 0}$ and $\{b_{n} \}_{n \geqslant 0}$ always follow the criterion $$ a_{n} \geqslant b_{n} \quad \text{for all } n \in \mathbb{N} $$
$\bullet~$ Proof:$~$ The sequences $\{a_{n} \}$ and $\{ b_{n}\}$ follow the recursions
$$ a_{n} = \frac{a_{n - 1} + b_{n - 1}}{2} ~\text{ and }~ b_{n} = \sqrt{a_{n - 1}b_{n - 1}} \quad \text{for any } n \in \mathbb{N} $$
As for any $n$ $\in$ $\mathbb{N}$, according to problem $~a_{n}, b_{n} > 0,$ Then by A.M-G.M inequality we have that
\begin{align*}
&\frac{a_{n -1} + b_{n - 1}}{2} \geqslant \sqrt{a_{n - 1}b_{n - 1}}\\
\implies & a_{n} \geqslant b_{n} \quad \text{for any } n \in \mathbb{N}
\end{align*}
This inequality can be strictened to $~a_{n} > b_{n}~$ for all $~n \in \mathbb{N}$
Then for the initial terms, we also have that $0 < b_{0} = b < a_{0} = a~$ [Let]
$\bullet~$ Claim (b): The sequence $\{a_{n}\}$ is a decreasing sequence and the sequence $\{b_{n} \}$ is an increasing sequence.
$\bullet~$ Proof: For the sequence $\{a_{n} \}$ we  have that
\begin{align*}
&a_{n} - a_{n - 1} = \frac{b_{n - 1} - a_{n -1}}{2}\\
\implies & a_{n} - a_{n-1} \leqslant 0 \quad [\text{as } a_{n} \geqslant b_{n} \text{ for any } n \in \mathbb{N}]\\
\implies & a_{n} \leqslant a_{n -1} \quad \text{for any } n \in \mathbb{N}
\end{align*}
This can also be strictened to $~a_{n} < a_{n - 1}$. Hence we have that $\{ a_{n} \}_{n \geqslant 0}$ is a decreasing sequence.
Again for $\{ b_{n}\}$, similarly we have
\begin{align*}
&\frac{b_{n}}{b_{n -1}} = \sqrt{\frac{a_{n -1}}{b_{n -1}}} \geqslant 1\\
\implies& \frac{b_{n}}{b_{n - 1}} \geqslant 1\\
\implies& b_{n} \geqslant b_{n -1} \quad \text{for any } n \in \mathbb{N} 
\end{align*}
This can also be strictened to $~b_{n} > b_{n - 1}$. Hence we have that $\{ b_{n} \}_{n \geqslant 0}$ is an increasing sequence.
$\bullet~$ Claim (c): The sequences $\{ a_{n} \}_{n \geqslant 0}$ and $\{ b_{n} \}_{n \geqslant 0}$ converge and converge at the same limit, i.e.,
$$ \lim_{n \to \infty} a_{n} = \lim_{n \to \infty} b_{n} = g $$
$\bullet~$ Proof: The sequence $\{ a_{n} \}$ is decreasing and is bounded below by $0$. Hence by MCT, the sequence is convergent. i.e.,
$$ \lim_{n \to \infty} a_{n} = L \quad [\text{ Say }] $$
Also observe that the sequence $\{a_{n}\}$ has a lower bound and $b_{n} < a_{n} $ for all $n \in \mathbb{N}$.
Hence $\{ b_{n}\}_{n \geqslant 0}$ is bounded above, thus by MCT, convergent. i.e.,
$$ \lim_{n \to \infty} b_{n} = L' \quad [\text{ Say }] $$
Now from the recursions, we have that
$$ a_{n} = \frac{a_{n - 1} + b_{n - 1}}{2} \implies \lim_{n \to \infty} a_{n} = \frac{\lim\limits_{n \to \infty}a_{n - 1} + \lim\limits_{n \to \infty}b_{n - 1}}{2} \implies 2L = L + L' \implies L = L' $$
Hence we are done!
$$ \lim_{n \to \infty} a_{n} = \lim_{n \to \infty} b_{n} = g $$
Where $g$ is the Arithmatic Geometric Mean of Gauss

