36given, a,b,c,d are in H.P
so, 1/a, 1/b, 1/c, 1/d are in A.P
so, 1/b - 1/a = 1/c - 1/b = 1/d - 1/c = k (say), where k is the common difference
=> (a - b)/ab = (b - c)/bc = (c - d)/cd = k
=> k = (a - b) + (b - c) + (c - d)/(ab + bc + cd)
=> ab + bc + cd = (a - d) / k -------- (i)
Again, 1/a + 3k = 1/d
=> 3k = 1/d - 1/a = (a - d)/ad
=> k = (a - d)/3ad
putting the value of k in (i) we get....
ab + bc + cd = 3ad proved....
Firstly i was not able to figure out the soln.. but at last i did it finally...
well that's not a big deal