1959 AMC 12 Problem 33

Question

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:

S_4=20

S_4=25

S_5=49

S_6=49

S_2=\dfrac12S_4

Write the reciprocals

\frac{1}{3},\frac{1}{4},\frac{1}{6}

and identify their common difference

Continue the arithmetic progression of reciprocals one more term

The reciprocals begin

\frac13,\quad\frac14,\quad\frac16,

with common difference

-\frac{1}{12}.

The next reciprocal is

\frac{1}{12},

so the fourth term is

12.

Therefore

S_4=3+4+6+12=25.

Thus, the correct answer is B.

Problem 33 in other years