So I came across a formula that looks like:  

$x\_n = \alpha x\_{n-1} + \beta$  

Since I don't have a strong mathematical background I didn't recognize it was an AGP and as I tried to express $x\_n$ with only the first term $x\_0$ I concluded that this was probably the formula:  

$x\_n = \alpha^nx\_0 + \frac{1-\alpha^n}{1-\alpha}\beta$  

(I don't know if it's correct though)


I wanted to know if it was correct so after a bit of searching I finally found <https://en.wikipedia.org/wiki/Arithmetico%E2%80%93geometric_sequence#cite_note-RHB118-1>  

However instead of bringing answers the article confused me:  

It states that an AGP is of the form: $u\_{n+1} = au\_{n} + b$, fair enough, but then it says that the first term of the sequence is $t\_1 = ab$ with $a$ the initial value of the arithmetic sequence and b the initial value of the geometric sequence ?  

1) is there a significance in changing from $u$ to $t$ ?  

2) Shouldn't it be $u\_1 = au\_0 + b$ ? How is this equivalent to $t\_1 = ab$ ?


In short I don't understand their notation, and I don't know how to apply their formula $t\_n = [a + (n-1)d]br^{n-1}$ to my problem ...



Explanation about arithmetico-geometric progression (AGP)
Let $a:=\alpha$ and $b:=\beta$. If $a=1$, then the 
equation 
$$x\_n=ax\_{n-1}+b \tag{1}$$
implies $x\_n=x\_0+bn$. 


If $a\ne1$, let $y\_n:=x\_n-c$, so that $x\_n=y\_n+c$. Substituting $y\_n+c$ for $x\_n$ in (1), we have $y\_n+c=ay\_{n-1}+ac+b$, which simplifies to $y\_n=ay\_{n-1}$ if $c=ac+b$, that is, if we let $c:=\frac b{1-a}$. Hence, $y\_n=a^n y\_0$, that is, 
$$x\_n=c+a^n(x\_0-c). 
$$


The sequence $(t\_n)$ in the linked Wikipedia article is the special case of the sequence $(x\_n)$ with $x\_1=ab$. 


