limit of sequnce in real numbers For every tow real numbers  $a$ and $b$,with this condition that $0<a<b$ ,define   sequence ${x_{n}}$ to the following:
$x_{1}=a$  $x_{2}=b$ 
${x_{n}}=\frac{ x_{n-1}+ x_{n-2} }{2}$ (for $n>2$) 
then which of the following options is limit of  ${x_{n}}$  ?
1)b
2)$\frac{a+2b}{3}$
3)$\frac{a+b}{2}$
4)$\frac{3a+5b}{8}$
I think that option 1 is true because if we draw  ${x_{n}}$  on the axiom of real numbers then limit of ${x_{n}}$ is b but it is not true.

The characteristic equation for the recurrence is
$$y^2 - \dfrac{y}2 - \dfrac12 = 0 \implies (y-1)(y+1/2) = 0$$
This gives us
$$x_n = c_0 1^n + c_1 \left(-\dfrac12\right)^n$$
Hence, note that $\lim_{n \to \infty} x_n = c_0$. Setting $n=1$, we have $c_0 - \dfrac{c_1}2 = a$ and setting $n=2$, we have $c_0 + \dfrac{c_1}4 = b$. This gives us $c_0 = \dfrac{a+2b}3$.
Hence, we have that
$$\color{blue}{\boxed{\lim_{n \to \infty} x_n = \dfrac{a+2b}3}}$$

EDIT:
It cannot converge to $b$ because we have $x_3 = \dfrac{a+b}2$ and $x_4 = \dfrac{a+3b}4$. This means any further $x_n$ has to be in the interval $\left[\dfrac{a+b}2,\dfrac{a+3b}4\right]$ and clearly since $a<b$, we have $$a < \dfrac{a+b}2 < \dfrac{a+3b}4 < b$$
