finding a coefficient to obtain a geometric sequence I have the following:
$$u_0=2 \\
u_{n+1} = \frac{2}{3}(u_n)-n-\frac{8}{3}\\
v_n=u_n + x\cdot n -1$$
I need to find the value of $x$ in order for $v_n$ to be a geometric sequence, which mean that I need to compute the following :
$$\frac{v_{n+1}}{v_n}$$
edit:
thanks for the hints but when i solve the equation it give 2 roots

The hard part is to show that $$ u_{n} = \left(\frac{2}{3}\right)^n - 3 \, n + 1.$$
The remaining part is
\begin{align}
v_{n} &= u_{n} + x \, n - 1 \\
&= \left(\frac{2}{3}\right)^n + (x - 3) \, n.
\end{align}
To show the form of $u_{n}$ compute the first few terms to develop a pattern and then check:
\begin{align}
f_{n} &= \frac{2}{3} \, u_{n} - n - \frac{8}{3} \\
&= \frac{2}{3} \, \left(\frac{2}{3}\right)^n - 2 \, n + \frac{2}{3} - n - \frac{8}{3} \\
&= \left(\frac{2}{3}\right)^{n+1} - 3 \, (n+1) + 1 \\
&= u_{n+1}. 
\end{align}
With these components then the desired answer can quickly be obtained.
