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

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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1959 AHSME Problems/Problem 33

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Solution