Congruent sets of an arithmetic sequence and a geometric sequence

Question

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) $-2$

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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?

Answer 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?

Answer 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).

Answer 3

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 $$ \ \{ \ a \ \ , \ \ a \ - \ \frac34 a \ = \ \frac14 a \ = \ aq^2 \ \ , \ \ a \ - \ 2·\frac34 a \ = \ -\frac12 a \ = \ aq \ \} \ \ . $$

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[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.]