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

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== Problem ==
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A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression.
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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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the first three terms. If the first three terms of a harmonic progression are <math>3,4,6</math>, then:
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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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Given HP = <math>3</math> <math>,</math> <math>4</math> <math>,</math> <math>6</math> \\
 
Given HP = <math>3</math> <math>,</math> <math>4</math> <math>,</math> <math>6</math> \\
 
So, <math>\tfrac {1} {3}</math>,<math>\tfrac {1} {4}</math>,<math>\tfrac {1} {6}</math> are in <math>AP</math>. \\
 
So, <math>\tfrac {1} {3}</math>,<math>\tfrac {1} {4}</math>,<math>\tfrac {1} {6}</math> are in <math>AP</math>. \\

Revision as of 14:03, 16 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

Given HP = $3$ $,$ $4$ $,$ $6$ \\ So, $\tfrac {1} {3}$,$\tfrac {1} {4}$,$\tfrac {1} {6}$ are in $AP$. \\ Then, common difference $(d)$ $=$ $\tfrac {1} {4}$ $-$ $\tfrac {1} {3}$ $=$ $\tfrac {1} {6}$ $-$ $\tfrac {1} {4}$ $=$ $-$ $\tfrac {1} {12}$ \\ Finding the fourth term of this $AP$ $=$ $\tfrac{1}{12}$ by $AP$ is trivial. \\ So, fourth term of Harmonic Progression $=$ $12$ $.$ \\ Now, $S_{4}$ $=$ $3$ $+$ $4$ $+$ $6$ $+$ $12$ $=$ $25$ $(B)$

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