(\dfrac{1}{k},\dfrac{1}{2k+1},\dfrac{1}{4k-1}) in H.P.

⇒ (k,\ 2k+1,\ 4k-1) in A.P.

[ 2(2k+1)=k+(4k-1)\implies 4k+2=5k-1\implies k=3 ]

Answer: (k=3)