Question

Three nonzero real numbers a,b,c are said to be in harmonic progression if 1/a + 1/c = 2/b. Find all three- term harmonic progressions a, b, c of strictly increasing positive integers in which a = 20 and b divides c.

Answer 1

Explanation:

Since 20,b,c are in harmonic progression, we have

1/20 + 1/c = 2/b,

Which reduces to bc + 20b - 40c = 0.This may also be written in the form

(40 - b)(c + 20) = 800.

Thus we must have 20 < b < 40 are equivalently, 0 < 40 - b < 20. Let us consider the factorisation of 800 in which one term is less than 20 : (40 - b)(c + 20) = 800 =1 x 800 = 2 x 400 = 4 x 200 = 5 x 160 = 8 x 100 = 10 x 80 = 16 x 50.

We thus get the pairs 

(b, c) = (39, 780), (38, 380), (36, 180), (35, 140), (32, 80), (30, 60), (24, 30).

Among these 7 pairs, we see that only 5 pairs (39, 780), (38, 3 80), (36, 180), (35, 140), (30, 60) fulfil the condition of divisibility: b divides c. Thus there are 5 triples satisfying the requirement of the problem.