calculus - Given two real numbers $a$ and $b$ such that $a

Let $a$ and $b$ be two given real numbers such that $a < b$, and let $\left\{x_n\right\}$ and $\left\{ y_n \right\}$ be the sequences defined as follows:
Let us choose $x_1$ and $y_1$ such that $$a < x_1 < b, \qquad a < y_1 < b$$ arbitrarily, and then let
$$x_2 = \frac{a+x_1}{2}, \qquad y_2 = \frac{y_1 + b}{2},$$
$$x_3 = \frac{a + x_1 + x_2 }{3}, \qquad y_3 = \frac{ y_1 + y_2 + b}{3},$$
and so on

$$ x_n = \frac{a+ x_1 + \cdots + x_{n-1} }{n}, \qquad y_n = \frac{ y_1 + \cdots + y_{n-1} + b}{n} $$
for $n= 3, 4, 5, \ldots$. Then what can we say about the convergence of these sequences?

To generalize this problem a little further, let $\left\{ r_n \right\}$ be a given sequence of positive real numbers, and let us now define
$$x_2 = \frac{r_1 x_1 + a r_2}{r_1 + r_2}, \qquad y_2 = \frac{r_1 y_1 + r_2 b}{r_1 + r_2},$$
$$x_3 = \frac{r_1 x_1 + r_2 x_2 + r_3 a }{r_1 + r_2 + r_3}, \qquad y_3 = \frac{ r_1 y_1 + r_2 y_2 + r_3 b}{r_1 + r_2 + r_3 },$$
and so on
$$ x_n = \frac{r_1 x_1 + \cdots + r_{n-1} x_{n-1} + r_n a }{r_1 + \cdots + r_n }, \qquad y_n = \frac{ r_1 y_1 + \cdots + y_{n-1} + r_n b}{r_1 + \cdots + r_n} $$
for $n= 3, 4, 5, \ldots$. Then what can we say about the convergence of these sequences?

What if we proceed as follows?

Let $r_0 > 0$ be given, and let
$$x_2 = \frac{r_0a+ r_1 x_1}{r_0 + r_1 }, \qquad y_2 = \frac{ r_1 y_1 + r_0 b}{r_1 + r_0},$$
$$x_3 = \frac{r_0 a + r_1 x_1 + r_2 x_2 }{r_0 + r_1 + r_2}, \qquad y_3 = \frac{ r_2 y_2 + r_1 y_1 + r_0 b}{r_2 + r_1 + r_0},$$
and so on
$$ x_n = \frac{r_0 a + r_1 x_1 + \cdots + r_{n-1} x_{n-1} }{r_0 + \cdots + r_{n-1} }, \qquad y_n = \frac{r_{n-1} y_{n-1} + \cdots + r_1 y_1 + r_0 b}{r_{n-1} + \cdots + r_0} $$
for $n= 3, 4, 5, \ldots$. What can we say about the convergence of these sequences now?

I can handle the situation only if we have only unit weights and only average of two terms is involved at a time, but I simply have no idea of what happens in this case!!

So, I would be really grateful for a detailed answer!