Congruent sets of an arithmetic sequence and a geometric sequence

Solution 1:

We have three cases.

1. $$2aq=a+aq^2$$ or $$q=1,$$ which is impossible because $d\neq0$.

2. $$2aq^2=a+aq$$ or $$2q^2-q-1=0,$$ which gives $$q=-\frac{1}{2}.$$

3. $$2a=aq+aq^2.$$ Can you end it now?

Solution 2:

Surely $q=1$ would mean that $d=0$, which is precluded by the conditions of your problem.

Have you considered that $M=N$ means only that the elements are the same, but does not imply anything about in which ordering. What you have proven is that the ordering you've attempted doesn't work.

Let's try adifferent ordering. Due to $d\ne 0$, $M$ and $N$ both consist of three different numbers, and $a$ is common in both. Thus, you have only two possibilities for the ordering. One is the one you've examined ($a+d=aq, a+2d=aq^2$), and the other is: $a+d=aq^2, a+2d=aq$. Now, applying the same logic as you did, but to this ordering, you get: $$a(2q^2-q-1)=0$$ which, after cancelling $a$ ($a\ne 0$) and solving for $q$ gives you the solutions $q=1$ or $q=-1/2$. The first solution is forbidden, so we are left with $q=-1/2$ (answer D).

One example is $4, 1, -2$ (arithmetic progression) vs. $4,-2,1$ (geometric progression).