Let $a_{n},b_{n}$ be two geometric sequences with ratios $q_{a},q_{b}$ such that $a_{1},b_{1}\neq0$ and also $a_{1}\neq b_{1}$. If $c_{n}=b_{n}-a_{n}$ is a geometric sequence with ratio $q_{c}$ then $q_{a}=q_{b}=q_{c}$
Attempt:
If $c_{n}$ is a geomtric sequence with ratio $q_{c}$ then: \begin{align*} & q_{c}=\frac{c_{n+1}}{c_{n}}=\frac{a_{n+1}-b_{n+1}}{a_{n}-b_{n}}=\frac{a_{n}q_{a}-b_{n}q_{b}}{a_{n}-b_{n}}\\ \Rightarrow\quad & q_{c}\left(a_{n}-b_{n}\right)=a_{n}q_{a}-b_{n}q_{b}\\ \Rightarrow\quad & a_{n}\left(q_{c}-q_{a}\right)=b_{n}\left(q_{c}-q_{b}\right) \end{align*} I thought this last equation would help me, but still im stuck.
EDIT: In the question it was stated that $q_b = 2$ but I thought it doesn't change anything, so I just used $q_b$ instead.
After your last step, you get that the ratio $\frac{a_n}{b_n}$ is constant for all $n$, and you know that both $a$ and $b$ are geometric, so you can get : $$\forall n\in\mathbb{N}, \left(\frac{q_a}{q_b}\right)^n \text{ is constant }$$
And that is only possible if $q_a=q_b$, and from that you can deduce that $q_a=q_b=q_c$