Official Solution: A geometric sequence is one in which the ratio of any term after the first to the preceding term is a constant. If the letters \(a\), \(b\), \(c\), \(d\) represent a geometric sequence in normal alphabetical order, which of the following must also represent a geometric sequence for all values of \(k\)?
I. \(dk\), \(ck\), \(bk\), \(ak\)
II. \(a + k\), \(b + 2k\), \(c + 3k\), \(d + 4k\)
III. \(ak^4\), \(bk^3\), \(ck^2\), \(dk\)
A. I only
B. I and II only
C. II and III only
D. I and III only
E. I, II, and III
Our first task is to ensure that we understand the definition of a geometric sequence. Let's use the sequence given to us: \(a\), \(b\), \(c\), \(d\). We are told that the ratio of any term after the first to the preceding term is a constant. In other words, \(\frac{b}{a} = \text{some constant}\), which is the same constant for the other ratios (\(\frac{c}{b}\) and \(\frac{d}{c}\)). Let's name that constant \(r\). Thus, we have the following:
\(\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r\)
By a series of substitutions, we can rewrite the sequence in terms of just \(a\) and \(r\):
\(a\), \(b\), \(c\), \(d\) is the same as \(a\), \(ar\), \(ar^2\), \(ar^3\)
Rewriting the sequence this way highlights the role of the constant ratio \(r\). That is, to move forward in the sequence one step, we just multiply by a constant factor \(r\). Rewriting also lets us substitute into the alternative sequences and watch what happens.
I. \(dk\), \(ck\), \(bk\), \(ak\)
This sequence is the same as \(ar^3k\), \(ar^2k\), \(ark\), \(ak\). To move forward in the sequence, we divide by \(r\). This is the same thing as multiplying by \(\frac{1}{r}\). Since this factor is constant throughout the sequence, the sequence is geometric.
II. \(a + k\), \(b + 2k\), \(c + 3k\), \(d + 4k\)
This sequence is the same as \(a + k\), \(ar + 2k\), \(ar^2 + 3k\), \(ar^3 + 4k\). To move forward in this sequence, we cannot simply multiply by a constant expression. The presence of the plus sign means that we will not have a constant ratio between successive terms, and this sequence is not geometric.
III. \(ak^4\), \(bk^3\), \(ck^2\), \(dk\)
This sequence is the same as \(ak^4\), \(ark^3\), \(ar^2k^2\), \(ar^3k\). To move forward in the sequence, we multiply by \(r\) and divide by \(k\). In other words, we multiply by \(\frac{r}{k}\), which is a constant factor. This sequence is therefore geometric.
Only sequences I and III are geometric.
Answer: D
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