Difference between revisions of "1959 AHSME Problems/Problem 33"
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Thus, the fourth term of the arithmetic progression is <math>\tfrac{1}{12}</math>. | Thus, the fourth term of the arithmetic progression is <math>\tfrac{1}{12}</math>. | ||
− | So, fourth term of the harmonic progression is <math>12</math>. | + | So, the fourth term of the harmonic progression is <math>12</math>. |
Now, we can see that <math>S_4</math> <math>=</math> <math>3</math> <math>+</math> <math>4</math> <math>+</math> <math>6</math> <math>+</math> <math>12</math> <math>=</math> <math>25</math>, which corresponds to answer <math>\fbox{\textbf{(B)}}</math>. | Now, we can see that <math>S_4</math> <math>=</math> <math>3</math> <math>+</math> <math>4</math> <math>+</math> <math>6</math> <math>+</math> <math>12</math> <math>=</math> <math>25</math>, which corresponds to answer <math>\fbox{\textbf{(B)}}</math>. |
Latest revision as of 15:42, 21 July 2024
Problem
A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression. Let represent the sum of the first terms of the harmonic progression; for example represents the sum of the first three terms. If the first three terms of a harmonic progression are , then:
Solution
From the problem, we know that ,,, ... are in arithmetic progression.
Then, the common difference of this arithmetic progression is
Thus, the fourth term of the arithmetic progression is .
So, the fourth term of the harmonic progression is .
Now, we can see that , which corresponds to answer .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
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All AHSME Problems and Solutions |
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