What is the general term of $a_{n+1}=\frac{2a_n-1}{5a_n-1} \ , \ \ a_1=1$? I've struggled to solve this exercise

$$a_{n+1}=\frac{2a_n-1}{5a_n-1}\ , \ \ a_1=1$$
$$b_{n+1}=(5a_n-1)b_n \ , \ \ b_1=1$$
Find $b_{\ 40}$ .

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I thought 'taking inverse' will be useful, but not yet... :-(
$\color{red}{01.}$ How can I find $b_{\ 40}$ ?
$\color{red}{02.}$ How can I find the general term of $a_n$, $b_n$ ?
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Thank you for your attention to this matter.
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 A: since
$$b_{n+1}=(5a_{n}-1)b_{n}\Longrightarrow a_{n}=\dfrac{1}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)$$
take in
$$a_{n+1}=\dfrac{2a_{n}-1}{5a_{n}-1}\Longrightarrow \dfrac{1}{5}\left(\dfrac{b_{n+2}}{b_{n+1}}+1\right)=\dfrac{\frac{2}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)-1}{\frac{b_{n+1}}{b_{n}}}$$
$$\Longrightarrow \dfrac{b_{n+2}+b_{n+1}}{5b_{n+1}}=\dfrac{\frac{2}{5}b_{n+1}+\frac{2}{5}b_{n}-b_{n}}{b_{n+1}}$$
$$\Longrightarrow b_{n+2}+b_{n+1}=2b_{n+1}-3b_{n}$$
$$\Longrightarrow b_{n+2}=b_{n+1}-3b_{n},b_{1}=1,b_{2}=(5a_{1}-1)b_{1}=4$$
The corresponding characteristic equation is
$$r^2=r-3\Longrightarrow r=\dfrac{1\pm\sqrt{11}i}{2}$$
so
$$b_{n}=A\left(\dfrac{1+\sqrt{11}i}{2}\right)^n+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^n$$
since $b_{1}=1,b_{2}=4$
so we have
$$\begin{cases}
A\left(\dfrac{1+\sqrt{11}i}{2}\right)+B\left(\dfrac{1-\sqrt{11}i}{2}\right)=1\\
A\left(\dfrac{1+\sqrt{11}i}{2}\right)^2+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^2=4
\end{cases}$$
$$\Longrightarrow \begin{cases}
3A+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^2=\dfrac{1-\sqrt{11}i}{2}\\
A\left(\dfrac{1+\sqrt{11}i}{2}\right)^2+3B=\dfrac{1+\sqrt{11}i}{2}
\end{cases}\Longrightarrow \begin{cases}
A=\dfrac{7+\sqrt{11}i}{\sqrt{11}i-11}\\
B=\dfrac{\sqrt{11}i-7}{11+\sqrt{11}i}
\end{cases}$$
so
$$b_{n}=\dfrac{7+\sqrt{11}i}{\sqrt{11}i-11}\cdot\left(\dfrac{1+\sqrt{11}i}{2}\right)^n+\dfrac{\sqrt{11}i-7}{11+\sqrt{11}i}\cdot\left(\dfrac{1-\sqrt{11}i}{2}\right)^n$$
A: here is a partial answer:
the eigenvalues of $A = \pmatrix{2 & -1\\5&-1}$ are $\lambda_{1,2} = \dfrac{1}{2} \pm i\dfrac{\sqrt{11}}{2}=\sqrt 3(\cos t \pm i\sin t)$ and the corresponding eigenvectors are 
$u_{1,2}=\pmatrix{2\\3\mp i\sqrt{11}}$ 
let $a, b, c$ be complex numbers such that 
$z = au_1 + bu_2 = a \pmatrix{2\\3 -i\sqrt{11}} + b \pmatrix{2\\3 + i\sqrt{11}} = 2c\pmatrix{1\\1}$
we need $a + b = c, 3a + 3b +i\sqrt{11}(b-a)=2c$  that means 
$i\sqrt{11}(a-b) = c, a + b =  c$ we can choose 
$c = 2\sqrt{11}, a = \sqrt{11} - i, b = \sqrt{11} + i$
$\begin{align}
A^k z &= A^k(au_1 +bu_2) = a\lambda_1^ku_1+b \lambda_2^k u_2 \\
& =3^{k/3} \{ a(\cos kt + i \sin kt)u_1 +  b(\cos kt - i \sin kt)u_2  \}\\
& =3^{k/3} \pmatrix{2(a+b)\cos kt + 2(a-b)i \sin kt)\\
(a+b)(3\cos kt + \sqrt{11} \sin kt+i(a-b)(3\sin kt -\sqrt{11}\cos kt)} \\
& = 3^{k/3}  \pmatrix{4(\sqrt{11}\cos kt +  \sin kt)\\
2\sqrt{11}(3\cos kt + \sqrt{11} \sin kt)+2(3\sin kt -\sqrt{11}\cos kt) } \\
&=3^{k/3}  \pmatrix{4(\sqrt{11}\cos kt +  \sin kt)\\
4\sqrt{11}\cos kt + 28 \sin kt)} \\
\end{align}$
so $$a_k = \dfrac{\sqrt{11}\cos kt +  \sin kt}{\sqrt{11}\cos kt + 7 \sin kt}  $$
