real analysis - About a recurrence sequence

I have a good understanding of the theorems of convergent and divergent sequence and I am able to apply them. But I really don't see how I can solve this thing. We have to use the theorem provided as well in order to solve the question.

Problem

Consider a real sequence $(a_n)$ defined through the formula
$$c_2 a_{n+2} + c_1a_{n+1} + x_0a_n = 0 \tag{1}$$
with $c_2, c_1, c_0 \in \Bbb{R}$. The characteristic polynomial of the linear recurrence $(1)$ is defined to be
$$p(x) = c_2 x^2 + c_1 x + c_0.$$
By means of the following theorem (do not prove it!)

Theorem If $p(x) = c_2(x - x_1)(x - x_2)$ where $x_1$ and $x_2$ are distinct nonzero complex numbers, then $(1)$ can be solved for $a_n$ and we have
$$a_n = B_1x_1^n + B_2x_2^n \quad \forall n \in \Bbb{N} \quad \text{with} \quad B_1, B_2 \in \Bbb{C} \tag{2}$$

1. find $B_1$ and $B_2$ for the sequence $(a_n)_{n\in\Bbb{N}}$ such that
   $$a_0 = a, \quad a_1 = b, \quad a_n = \frac{a_{n-1} + a_{n-2}}{2} \quad \forall n \ge 2$$
   with $a, b \in \Bbb{R}$.




2. Moreover, show that the above sequence converges and find its limit.

I tried to re organize the equation of

$$a_n=\frac{a_{n-1} +a_{n-2}}{2}$$
to
$$a_n-\frac{1}{2}a_{n-1}-\frac{1}{2}a_{n-2}=0$$
but then I didn't know how to take the choice for $x$ or how to proceed after that.

Answer

With your reworked equation we have that the characteristic polynomial is given by
$$p(x)=x^{2}-\frac{1}{2}x-\frac{1}{2}=(x-1)(x+\frac{1}{2}).$$
By the theorem this implies there exist $B_{1},B_{2}\in\mathbb{C}$ such that
$$a_{n}=B_{1}+B_{2}(-2)^{-n}.$$
In particular we have $B_{1}+B_{2}=a$ and $B_{1}-\frac{1}{2}B_{2}=b$. This gives us $B_{1}=\frac{1}{3}(a+2b)$ and $B_{2}=\frac{2}{3}(a-b)$.

Furthermore note that
$$\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}B_{1}+B_{2}(-2)^{-n}=B_{1}=\frac{1}{3}(a+2b).$$