If 1/a, 1/b, 1/c, 1/d are arithmetic progression, prove that ab+bc+cd = 3ad

Question

Please show the ways to find out the problem that if 1/a, 1/b, 1/c, 1/d ... an arithmatic progression, prove that ab+bc+cd = 3ad

Answer 1

1/a. 1/b, 1/c are in arithmetic pregression. Let the common difference be 1/f. Then,

1/a + 1/f = 1/b, 1/b + 1/f = 1/c, 1/c + 1/f = 1/d, also 1/a + 3/f = 1/d

doing cross-multiplication on all the above equations gives us,

a - b = abf, b - c = bcf, c - d = cdf, a - d = 3adf

adding the 1st three equations, just above, together,

(a - b) + (b - c) + (c - d) = abf + bcf + cdf

a - d = f(ab + bc + cd)

Now use the 4th equation, a - d = 3adf

3adf = f(ab + bc + cd)

i.e. ab + bc + cd = 3ad