1959 AHSME Problems/Problem 33

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