Is there any way I can make an explicit formula for the sequance $a_n=x+ya_{n-1}$?

Question

Let $a_n$ be a sequence defined by recursion: $a_n=x+ya_{n-1}, a_1=k$. For example, if $(x,y)=(3,5)$, then the sequence would go
$$a=\{k,\space 3+5k,\space 3+5(3+5k),\space ...\}$$ Is there an explicit formula for $a_n$? If not, is there a way to tell if a number is a member of $a$?

Answer 1

Just another solution.

Considering $$a_n=x+y\,a_{n-1}$$ let $a_n=b_n+k$ and replace $$b_n+k=x +y\ b_{n-1}+k y$$ Let $k=x+k y \implies k=\frac x{1-y}$ (if $y \neq 1$) to make $$b_n=y \,b_{n-1}$$ which is simple.

Solve for $b_n$ and $a_n=b_n+\frac x{1-y}$

Answer 2

Hint: Let $b_n = a_{n+1}-a_n$. Then $b_n = y b_{n-1}$. Can you finish?