(\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)
11TH•Mathematics•Sequences And Series•Question 5
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If the numbers $\dfrac{1}{k},\ \dfrac{1}{2k+1}$ and $\dfrac{1}{4k-1}$ are in harmonic sequence, find $k$.
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