A combined arithmetic and geometric sequence question Here is a question I am currently struggling with -

The first, the tenth and the twentieth terms of an increasing arithmetic sequence are also consecutive terms in an increasing geometric sequence. Find the common ratio of the geometric sequence.

Here's what I've done so far -
$u_1=v_1$
$u_{10}=v_2$
$u_{20}=v_3$
We know that,
$\displaystyle \frac{v_2}{v_1}=\frac{v_3}{v_2}$
and,
$u_1=u_1$
$u_{10}=u_1+9d$
$u_{20}=u_1+19d$
Therefore,
$\displaystyle \frac{u_1+9d}{u_1}=\frac{u_1+19d}{u_1+9d}$
Upon simplifying -
$(u_1+9d)^{2}=u_1(u_1+19d)$
$\therefore \displaystyle u_1=81d$
Now, what do I do after this?
 A: You have then$$\frac{v_2}{v_1}=\frac{u_{10}}{u_1}=\frac{90d}{81d}=\frac{10}9\quad\text{and}\quad\frac{v_3}{v_2}=\frac{u_{20}}{u_{10}}=\frac{100d}{90d}=\frac{10}9.$$Therefore, the answer is $\frac{10}9$.
Note that there is really no need to do what comes after “and”. It's just to double check things.
A: We could also set up the relations between the terms of the geometric sequence in this way:
$$ v_2 \ \ = \ \ r·u_1 \ \ = \ \  u_1 \ + \ 9·d  \ \ \ , \ \ \ v_3 \ \ = \ \ r^2·u_1 \ \ = \ \ u_1  \ + \ 19·d  $$
$$ \Rightarrow \ \ (r \ - \ 1)·u_1 \ \ = \ \ 9·d \ \ \ , \ \ \ (r^2 \ - \ 1)·u_1 \ \ = \ \ 19·d $$
$$ \Rightarrow \ \ \frac{(r^2 \ - \ 1)·u_1}{(r \ - \ 1)·u_1} \ \ = \ \  r \ + \ 1  \ \ = \ \ \frac{19·d}{9·d} \ \ \Rightarrow \ \ r \ \ = \ \ \frac{19}{9} \ - \ 1 \ \ = \ \ \frac{10}{9} \ \ . $$
This is not the description of a unique sequence:  the problem statement applies equally well to
•  $ \ \ \ d \ = \ \frac19 \ \ \rightarrow \ \ u_1 \ = \ 9 \ \ , \ \ u_{10} \ = \ 10 \ \ , \ \ u_{20} \ = \ \frac{100}{9} \ \ \ $ or to
•  $ \ \ \ d \ = \ -3 \ \ \rightarrow \ \ u_1 \ = \ -243 \ \ , \ \ u_{10} \ = \ -270 \ \ , \ \ u_{20} \ = \ -300 \ \ . $
In this approach, $ \ u_1 \ $ and $ \ d \ $ are "eliminated in one stroke", which is not important here since we are not asked to determine them.  (The fact that this happens is the "cause" of the sequence not being unique.)  Once we know $ \ r \ \ , $ we can find the relation between these two quantities by inserting it into, for instance, $ \ \left(\frac{10}{9} \ - \ 1 \right)·u_1 \ \ = \ \ 9·d  \ \Rightarrow \ u_1 \ = \ 81·d \ \ , \ $ as you observed.
