Given two real numbers $a$ and $b$ such that $aLet $a$ 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! 
 A: This problem is in fact much easier than it seems at first.  For $n \geq 2$, we have $$x_n = \dfrac{r_0a+r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1}}{r_0+r_1+\cdots+r_{n-1}} \\\Rightarrow r_0a+r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1} = (r_0+r_1+\cdots+r_{n-1})x_n.$$
Thus $$x_{n+1} = \dfrac{r_0a+r_1x_1+r_2x_2+\cdots+r_{n}x_{n}}{r_0+r_1+\cdots+r_{n}} = \dfrac{(r_0+r_1+\cdots+r_{n-1})x_n+r_nx_n}{r_0+r_1+\cdots+r_n} = x_n.$$
Thus $$\frac{r_0a+r_1x_1}{r_0+r_1} = x_2=x_3=\cdots.$$  An analogous result holds for $\{y_i\}$.

If instead you use the formula $$x_n = \dfrac{r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1}+r_na}{r_1+r_2+\cdots+r_n},$$ the solution needs to be modified a bit.  We have $$r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1} = (r_1+r_2+\cdots+r_n)x_n-r_na.$$
Then $$x_{n+1}=\dfrac{r_1x_1+r_2x_2+\cdots+r_{n}x_{n}+r_{n+1}a}{r_1+r_2+\cdots+r_{n+1}}=\dfrac{(r_1+r_2+\cdots+r_n)x_n-r_na+r_nx_n+r_{n+1}a}{r_1+r_2+\cdots+r_{n+1}}\\=x_n-\dfrac{r_{n+1}x_n+r_na-r_nx_n-r_{n+1}a}{r_1+r_2+\cdots+r_{n+1}}=x_n-\dfrac{(r_{n+1}-r_n)(x_n-a)}{r_1+r_2+\cdots+r_{n+1}}.$$
I'm not completely sure where to proceed from here but if I solve it I will come back and edit my solution in.
