1973 AHSME Problems/Problem 28

Problem

If $a$ , $b$ , and $c$ are in geometric progression (G.P.) with $1 < a < b < c$ and $n>1$ is an integer, then $\log_an$ , $\log_bn$ , $\log_cn$ form a sequence

$\textbf{(A)}\ \text{which is a G.P} \qquad$

$\textbf{(B)}\ \text{which is an arithmetic progression (A.P)} \qquad$

$\textbf{(C)}\ \text{in which the reciprocals of the terms form an A.P} \qquad$

$\textbf{(D)}\ \text{in which the second and third terms are the }n\text{th powers of the first and second respectively} \qquad$

$\textbf{(E)}\ \text{none of these}$

Solution

Using the change of base formula, the three logarithmic terms can be written as $\frac{\log n}{\log a}, \frac{\log n}{\log b}, \frac{\log n}{\log c}$ Since $a$ , $b$ , and $c$ are members of a geometric sequence, $b = ar$ and $c = ar^2$ . That means the three logarithmic terms can be rewritten as $\frac{\log n}{\log a}, \frac{\log n}{\log ar}, \frac{\log n}{\log ar^2}$ $\frac{\log n}{\log a}, \frac{\log n}{\log a + \log r}, \frac{\log n}{\log a + 2 \log r}$ Note that if we take the reciprocals of each term, the next term can be derived from the previous term by adding $\tfrac{\log r}{\log n}$ , so the answer is $\boxed{\textbf{(C)}}$ .

1973 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
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1973 AHSME Problems/Problem 28

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