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: 
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 $$\displaystyle\lim\_{n\to\infty}x\_n=2+3\times\frac{1}{1+1/2}=4$$
You may have proved that somewhere. It comes from the usual technique to solving homogeneous linear recurrence relations. You assume $x\_n=ar^n$, plug that into the recurrence, get the characteristic polynomial, here $r^2=\frac 12(r+1)$ and solve for $r$. You get two roots, here $r=1, -\frac 12$ The general solution is then $x\_n=An^1+Bn^{-1/2}$ and you evaluate $A,B$ from the initial conditions. As the second term dies away to nothing as $n$ gets large, $x\_n \to A$ for $n$ large. You find $A=\frac 13(x\_1+2x\_2)$ when you do this.