Solution:
a) It follows by the recurrence relation that $x_{1}=2$, $x_{2}=1+2a$ and $x_{3}=1+a+2a^{2}$. Then $x_{1}+x_{3}=2x_{2} \Longleftrightarrow 3+a+2a^{2}=2(1+2a)$ with solutions $a=1$ and $a=\frac{1}{2}$. For $a=1$ we get $x_{n+1}=x_{n}+1$, i.e. the sequence is an arithmetic progression. For $a=\frac{1}{2}$ we see by induction on $n$ that $x_{n}=2$ for every $n$. Therefore $a=1$ is the only solution.

b) We prove by induction on $n$ that $x_{n+1}=1+a+\cdots+a^{n-1}+2a^{n}$, $n \geq 1$. For $a=1$ we have $x_{n}=n+1$, i.e. the sequence is not convergent. Let $a \neq 1$. Then
$$x_{n+1}=2a^{n}+\frac{1-a^{n}}{1-a}=a^{n}\left(2-\frac{1}{1-a}\right)+\frac{1}{1-a}$$
If $2-\frac{1}{1-a}=0$, i.e. $a=\frac{1}{2}$, we get $x_{n+1}=2$ for every $n$ and the sequence is convergent.
Since $\{a^{n}\}_{n=1}^{\infty}$ converges if and only if $|a|<1$ or $a=1$, we conclude that the given sequence is convergent for $a \in(-1,1)$ and its limit is equal to $\frac{1}{1-a}$ (since $\lim _{n \rightarrow \infty} a^{n}=0$ for $a \in(-1,1)$ ).