1973 AHSME Problems/Problem 28


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

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

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

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

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

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


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 $\boxed{\textbf{(C)}}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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