The correct option is C 3ad
Since a, b, c, d are in H.P. therefore 1a,1b,1c,1d are in AP.
∴1b−1a=1c−1b=1d−1c=λ(say)∴a−b=λab,b−c=λbc and c−d=λcd
Adding the above three, we get,
a−d=λ(ab+bc+cd)⋯(i)
Now 1d=1a+3λ⇒a−d=3λad
Substituting in (i) , we get ab + bc + cd = 3ad.