Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1973 AHSME Problems/Problem 28"

(Solution to Problem 28)
 
m (typo)
Line 5: Line 5:
 
<math> \textbf{(A)}\ \text{which is a G.P} \qquad</math>
 
<math> \textbf{(A)}\ \text{which is a G.P} \qquad</math>
  
−
<math> \textbf{(B)}\ \text{whichi is an arithmetic progression (A.P)} \qquad</math>
+
<math> \textbf{(B)}\ \text{which is an arithmetic progression (A.P)} \qquad</math>
  
 
<math> \textbf{(C)}\ \text{in which the reciprocals of the terms form an A.P} \qquad</math>
 
<math> \textbf{(C)}\ \text{in which the reciprocals of the terms form an A.P} \qquad</math>

Revision as of 21:44, 11 December 2018

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

See Also

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
All AHSME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1973_AHSME_Problems/Problem_28&oldid=99388"