sequences and series - Arithmetic and geometric progression question (too long for title)

Good evening,
I've been struggling with this one for a while now.

The sum of an increasing geometrical progression is equal to 65. If we substract 1 from the smallest number there, and 19 from the biggest, the three numbers are going to form an arithmetical progression. Find these numbers.

I've gone trough many methods and none of them worked, so I't would be nice if you could explain this step by step. Not to look like someone just hunting for answers, I'll provide what I know:
The geometric progression, consisted of members b1, b2, b3.
The sum of it is equal to 65. b1q = b2; b1q^2=b3

The aritmetic progression, consisted of a1, a2, a3.
a1 = b1 - 1; a2 = b2 ; a3 = b3 - 19.
a1 + d = a2, a1 + 2d = a3
I'm sort of new to this site, so sorry if I didnt provide something, like tags, because these type of things are called progressions here and there no such things in tag list. Im in 12th grade which is equal to senior year in USA as far as I know if that matters. thank you!

Answer

You can develop a system of two equations in two unknowns as follows. Represent the terms of the geometric progression as $c$, $rc$, and $r^2c$. Since you are told that they sum to $65$, you know
$$c + rc + r^2c = 65\tag{1}$$
Now the terms of the arithmetic progression must be $c-1$, $rc$, and $r^2c-19$. Since this is an arithmetic progression, the difference between any two successive terms is the same; thus
$$(rc)-(c-1) = (r^2c - 19)-(rc)\tag{2}$$
Equations (1) and (2) provide your system of equations. Cleaning them up a little bit, you have the system

$$\left\{\begin{array}{l}r^2c + rc + c = 65 \\ r^2c - 2rc + c = 20
\end{array}\right.$$
to solve for $r$ and $c$, which will then give you the terms of the progressions.

Subtracting the equations in this system gives you immediately that $3rc=45$. Since necessarily $r\neq 0$ (why?), you can substitute $c=15/r$ in (say) the first equation and clean it up to get
$$3r^2 - 10r + 3 = 0$$
This factors as $$(r-3)(3r-1)=0$$
so the solutions are $r=3$ (which gives $c=5$) and $r=1/3$ (which gives $c=45$). Since the geometric progression is known to be increasing, we eliminate the solution $r=1/3$ (it produces a decreasing geometric progression). This means the geometric progression is
$$\boxed{5,15,45}$$ and the arithmetic progression is $$\boxed{4,15,26}.$$