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