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Difference between revisions of "1959 AHSME Problems/Problem 33"

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Given HP = <math>3</math> <math>,</math> <math>4</math> <math>,</math> <math>6</math> \\
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== Problem ==
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So, <math>\tfrac {1} {3}</math>,<math>\tfrac {1} {4}</math>,<math>\tfrac {1} {6}</math> are in <math>AP</math>. \\
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Then, common difference <math> (d) </math> <math>=</math> <math>\tfrac {1} {4}</math> <math>-</math> <math>\tfrac {1} {3}</math> <math>=</math> <math>\tfrac {1} {6}</math> <math>-</math> <math>\tfrac {1} {4}</math> <math>=</math> <math>-</math> <math>\tfrac {1} {12}</math> \\
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A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression.
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Finding the fourth term of this <math>AP</math> <math>=</math> <math>\tfrac{1}{12}</math> by <math>AP</math> is trivial. \\
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Let <math>S_n</math> represent the sum of the first <math>n</math> terms of the harmonic progression; for example <math>S_3</math> represents the sum of
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So, fourth term of Harmonic Progression <math>=</math> <math>12</math> <math>.</math> \\
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the first three terms. If the first three terms of a harmonic progression are <math>3,4,6</math>, then:
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Now, <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> <math>(B)</math>
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<math>\textbf{(A)}\ S_4=20 \qquad\textbf{(B)}\ S_4=25\qquad\textbf{(C)}\ S_5=49\qquad\textbf{(D)}\ S_6=49\qquad\textbf{(E)}\ S_2=\frac{1}2 S_4  </math>
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== Solution ==
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From the problem, we know that <math>\tfrac {1} {3}</math>,<math>\tfrac {1} {4}</math>,<math>\tfrac {1} {6}</math>, ... are in arithmetic progression.
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Then, the common difference <math>d</math> of this arithmetic progression is <math>\tfrac {1} {4}</math> <math>-</math> <math>\tfrac {1} {3}</math> <math>=</math> <math>\tfrac {1} {6}</math> <math>-</math> <math>\tfrac {1} {4}</math> <math>=</math> <math>-</math> <math>\tfrac {1} {12}</math>
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Thus, the fourth term of the arithmetic progression is <math>\tfrac{1}{12}</math>.
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So, the fourth term of the harmonic progression is <math>12</math>.  
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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>.
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== See also ==
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{{AHSME 50p box|year=1959|num-b=32|num-a=34}}
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{{MAA Notice}}

Latest revision as of 14:42, 21 July 2024

Problem

A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression. Let $S_n$ represent the sum of the first $n$ terms of the harmonic progression; for example $S_3$ represents the sum of the first three terms. If the first three terms of a harmonic progression are $3,4,6$, then: $\textbf{(A)}\ S_4=20 \qquad\textbf{(B)}\ S_4=25\qquad\textbf{(C)}\ S_5=49\qquad\textbf{(D)}\ S_6=49\qquad\textbf{(E)}\ S_2=\frac{1}2 S_4$

Solution

From the problem, we know that $\tfrac {1} {3}$,$\tfrac {1} {4}$,$\tfrac {1} {6}$, ... are in arithmetic progression.

Then, the common difference $d$ of this arithmetic progression is $\tfrac {1} {4}$ $-$ $\tfrac {1} {3}$ $=$ $\tfrac {1} {6}$ $-$ $\tfrac {1} {4}$ $=$ $-$ $\tfrac {1} {12}$

Thus, the fourth term of the arithmetic progression is $\tfrac{1}{12}$.

So, the fourth term of the harmonic progression is $12$.

Now, we can see that $S_4$ $=$ $3$ $+$ $4$ $+$ $6$ $+$ $12$ $=$ $25$, which corresponds to answer $\fbox{\textbf{(B)}}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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