Insert 6 harmonic means between 3 and `(6)/(23)`

Explanation

Let `H_(1),H_(2),H_(3),H_(4),H_(5),H_(6)` be 6 HM's batween `(6)/(23)`.
Then, `3,H_(1),H_(2),H_(3),H_(4),H_(5),H_(6),(6)/(23)` are in HP.
` implies (1)/(3),(1)/(H_(1)),(1)/(H_(2)),(1)/(H_(3)),(1)/(H_(4)),(1)/(H_(5)),(1)/(H_(6)),(23)/(6)` are in AP.
Let common difference of this AP be D.
`therefore D=((23)/(6)-(1)/(3))/(7)=((23-2))/(7xx6)=(21)/(7xx6)=(1)/(2)`
`therefore D=(1)/H_(1)=(1)/(3)+D=(1)/(3)+(1)/(2)=(5)/(6)`
`implies H_(1)=(6)/(5)=1(1)/(5)`
`(1)/(H_(2))=(1)/(3)+2D=(1)/(3)+1=(4)/(3) implies H_(2)=(3)/(4)`
` (1)/(H_(3))=(1)/(3)+3D=(1)/(3)+(3)/(2)=(11)/(6) implies H_(3)=(6)/(11)`
` (1)/(H_(4))=(1)/(3)+4D=(1)/(3)+2=(7)/(3)implies H_(4)=(3)/(7)`
` (1)/(H_(5))=(1)/(3)+5D=(1)/(3)+(5)/(2)=(17)/(6)implies H_(5)=(6)/(17)`
and ` (1)/(H_(6))=(1)/(3)+6D=(1)/(3)+3=(10)/(3)implies H_(6)=(3)/(10)`
` therefore " HM's are " 1(1)/(3),(3)/(4),(6)/(11),(3)/(7),(6)/(17),(3)/(10)`.