Congruent sets of an arithmetic sequence and a geometric sequence Suppose we have a $a,d,$ and $q$ such that $a \neq 0, d \neq 0.$ Then, let $M = \{a, a + d, a + 2d\}$ and $N = \{a, aq, aq^2\}.$ Given that $M = N,$ find the value of $q.$
(A) $\frac12$
(B) $\frac13$
(C) $-\frac14$
(D) $-\frac12$
(E) $-2$

I immediately thought about setting $a + d = aq$ and $a + 2d = aq^2.$ I then proceeded to do $aq^2 - aq = d,$ and substitute in for $d,$ which gave me $a + (aq^2 - aq) = aq.$ Simplifying then gave me $aq^2 - 2aq + a = 0,$ and dividing by $a$ gave me $q^2 - 2q + 1 = 0,$ which should mean that $q = 1.$ However, that's not an answer choice. What should I do instead?

One could also work with the difference between terms, rather than eliminating it.  The two possible orderings of the elements in the set are
$ \ \{ a \ , \ aq \ = \ a + d \ , \ aq^2 \ = \ a + 2d \}  \ $ or $ \ \{ a \ , \ aq \ = \ a + 2d \ , \ aq^2 \ = \ a + d \}  \ \ . $  [Initially, it "feels like" the second arrangement should be unreasonable...]
For the first arrangement, subtracting the first term from  the second and third produces $ \ a·(q - 1) \ = \ d \ \ , \ \ a·(q^2 - 1) \ = \ 2d \ \ . $  Taking $ \ a \neq 0 \ \ $ (otherwise, set $ \ N \ $ would only contain zeroes),   dividing the latter equation here by the former yields
$$ \frac{q^2 \ - \ 1}{q \ - \ 1} \ \ = \ \ q \ + \ 1 \ \ = \ \ \frac{2d}{d} \ \ = \ \ 2 \ \ \Rightarrow \ \ q \ = \ 1 \ \ \Rightarrow \ \ q \ - \ 1 \ \ = \ \ \frac{d}{a} \ \ = \ \ 0 \ \ . $$
This just gives us the "constant" sequence of terms you were concerned about (and which the stated conditions exclude).  But in fact, $ \ q = 1 \ $ isn't even permissible for this ratio.  So it turns out that the first arrangement of elements is the "incorrect one".
The seemingly unreasonable ordering leads to
$$ \ a·(q^2 - 1) \ = \ d \ \ , \ \ a·(q - 1) \ = \ 2d  $$
$$ \Rightarrow \ \ \frac{q^2 \ - \ 1}{q \ - \ 1} \ \ = \ \ q \ + \ 1 \ \ = \ \ \frac{d}{2d} \ \ = \ \ \frac12 \ \ \Rightarrow \ \ q \ = \ -\frac12 $$ $$ \Rightarrow \ \ q^2 \ - \ 1 \ \ = \ \ -\frac34 \ \ = \ \ \frac{d}{a} \ \ \Rightarrow \ \ d \ \ = \ \ -\frac34 · a \ \  . $$
We note that this leaves $ \ a \ $ unspecified, so there are an infinite number of such sequences possible.  The elements of the sets are thus
$$ \ \{ \ a \ \ , \ \  a \ - \ \frac34 a \ = \ \frac14 a \ = \ aq^2 \ \ , \ \ a \ - \ 2·\frac34 a \ = \ -\frac12 a \ = \ aq \ \} \ \ .  $$
$$ \ \ $$
[I also had an argument in which we could avoid the question of sequence ordering by looking at the sum of the elements, which gives us
$$ a·( 1 \ + \ q \ + \ q^2) \ \ = \ \ a \ + \ (a + d) \ + \ (a + 2d) \ \ = \ \ 3a \ + \ 3d $$ $$ \Rightarrow \ \ q^2 \ + \ q \ - \ \left(2 \ + \ 3·\frac{d}{a}\right) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ q \ \ = \ \ -\frac12 \ \pm \ \frac{\sqrt{1 \ + \ 4·\left(2 \ + \ 3·\frac{d}{a}\right)}}{2} \ \ = \ \ -\frac12 \ \pm \ \frac{\sqrt{ \ 9  +  \left(12·\frac{d}{a} \right)}}{2} $$ $$  = \ \ -\frac12 \ \pm \ \frac{3·\sqrt{ \ 1  +  \left(\frac{4d}{3a} \right)}}{2} \ \ .   $$
Setting the discriminant equal to zero gives us $ \ q \ = \ -\frac12 \ $ and $ \ 4·d \ = \ -3·a \ \ $ as above, but I didn't really see a satisfying explanation for doing so, other than "tidiness" of the result.]
