Question: ![enter image description here](https://i.stack.imgur.com/sOBaz.jpg)
I cannot understand how the n term tends to  

1/3((x1)+2(x2))

Answer: 
How is $x\_3$ defined? It is defined as the arithmetic mean between $x\_1$ and $x\_2$. And what is $x\_4$? $x\_4$ is the arithmetic mean of $x\_3$ and $x\_2$.

My suggestion: draw a line segment and call $x\_1$ to the left ending, $x\_2$ to the right one. $x\_3$ is right in the middle of the segment. $x\_4$ in the middle, between $x\_3$ and $x\_2$. $x\_5$ in the middle between $x\_3$ and $x\_4$. And so on and so forth. How do you get to $x\_3$?

$$x\_3 = 5 - 1.5$$

because 1.5 is half the distance between $x\_1$ and $x\_2$.
How do you get to $x\_4$? $x\_5$? $x\_6$?

$$x\_4 = x\_3 + 0.75 = 5 - 1.5 + 0.75\\
x\_5 = x\_4 - 0.375 = 5 - 1.5 + 0.75 - 0.375\\
x\_6 = x\_5 + 0.1875 = 5 - 1.5 + 0.75 - 0.375 + 0.1875$$

if you keep doing this, you should be able to convince yourself that $x\_n$ has an expression that starts with $5$ and then an alternating sum. Well, if you factor $1.5$ out, you get:

$$x\_n = 5 + 1.5(-1 + 0.5 - 0.25 + 0.125 - \cdots )$$

But that alternating sum is 

$$\sum\_{i=0}^n (-1)^{i+1}\frac1{2^i}$$

which, when $n \to \infty$, yields $-\frac23$. Plugging in, we get

$$n \to \infty\ \Longrightarrow\ 5 - 1.5\cdot(\frac23) = 4$$

The attached formula can be derived if you think separately of the positive and negative terms of the summation above for $x\_n$.
We have $x\_1=2$ , $x\_2=5$ and $$x\_n=\frac12\left(x\_{n-1}+x\_{n-2}\right)$$
$$x\_n-x\_{n-1}=\left(\frac{-1}2\right)\left(x\_{n-1}-x\_{n-2}\right)$$
\begin{align}
 x\_2-x\_1 &=\left(\frac{-1}{2}\right)^03 \\
 x\_3-x\_2&=\left(\frac{-1}{2}\right)^13 \\
 x\_4-x\_3&=\left(\frac{-1}{2}\right)^23\\
x\_5-x\_4&=\left(\frac{-1}{2}\right)^33\\
&\cdots\cdots\cdots\cdots\\
&\cdots\cdots\cdots\cdots\\
x\_n-x\_{n-1}&=\left(\frac{-1}{2}\right)^{n-2}3\\
\end{align}
on adding we have $x\_n-x\_1=3\left(1+(-1/2)+(-1/2)^2+\cdots(-1/2)^{n-2}\right)$
Now $$\displaystyle\lim\_{n\to\infty}x\_n=2+3\times\frac{1}{1+1/2}=4$$