1959 AHSME Problems/Problem 33

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

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

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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1959 AHSME Problems/Problem 33

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