Infinite series

From the given equation the sequence may considered as : $a_n=p \lambda^n +q$

Putting this in the equation $(p \lambda^n+q) +(p \lambda^{n+1} +q)=2(p \lambda^{n+2} +q)$

$1+ \lambda=2 \lambda^2$

Solving this equation we get $\lambda=1, -1/2$ .First solution will not be acceptable because the sequence will never be generated.

So, $\lambda=-\dfrac{1}{2}$ .

Sequence will be $a_n=p(-\dfrac{1}{2})^n + q)$

The given info $a_1=5$ and $a_2=8$ will be used to find two unknown constants.

$a_1=-\dfrac{p}{2} +q =5$ and $\dfrac{p}{4} + q=8$ .

On solving this $p=4\,;\,q=7$

The final sequence will become $a_n=4 (-\dfrac{1}{2})^n + 7$

Now $\displaystyle \lim_{n \to \infty} a_n=7$

Relevant wiki: Linear Recurrence Relations - With Repeated Roots

From $a_n + a_{n+1} = 2a_{n+2}$ we have the characteristic equation as follows:

$\begin{aligned} 2r^2 - r-1 & = 0 \\ (2r+1)(r-1) & = 0 \\ \implies a_n & = c_1(1)^n + c_2\left(-\frac 12\right)^n \end{aligned}$

$\begin{cases} a_1 = 5: & \implies c_1 - \dfrac {c_2}2 = 5 & ...(1) \\ a_2 = 8: & \implies c_1 + \dfrac {c_2}4 = 8 & ...(2) \end{cases}$

$\begin{aligned} (2)-(1): \quad \frac {c_2}4 + \frac {c_2}2 & = 8-5 \\ \frac 34 c_2 & = 3 \\ \implies c_2 & = 4 \end{aligned}$

$\begin{aligned} (1): \quad c_1 - \frac 42 & = 5 \\ \implies c_1 & = 7 \end{aligned}$

$\begin{aligned} \implies a_n & = 7 + 4\left(-\frac 12\right)^n \\ \implies \lim_{n \to \infty} a_n & = \boxed{7} \end{aligned}$