If 9 harmonic means be inserted between 2 and 3, then the value of `A+(6)/(H)+5` (where A is any of the AM's and H is the corresponding HM), is

Explanation

Let `H_(1),H_(2),H_(3),"....",H_(n),` are n harmonic means, then `a,H_(1),H_(2),H_(3),"....",H_(n),b` are in HP.
`:.(1)/(a),(1)/(H_(1)),(1)/(H_(2)),(1)/(H_(3)),"....",(1)/(H_(n)),(1)/(b)` are in AP.
If d be the common difference, then `(1)/(c )=(1)/(a)+(n+2-1)d`
`:. d=((a-c))/(ac(n+1))" " "......(i)"`
`implies (1)/(h_(1))=(1)/(a)+d` and `(1)/(h_(n))=(1)/(c)-d`
`:.h_(1)-h_(n)=(a)/(1+ad)-(c)/(1-cd)=(a)/(1+(a-c)/(a(n+1)))-(c)/(1-(a-c)/(a(n+1)))`
`=(ac(n+1))/(an+a)-(ac(n+1))/(an+c)=ac(n+1)((1)/(cn+a)-(1)/(an+c))`
`=ac(n+1)((an+c-cn-a)/(acn^(2)+(a^(2)+c^(2))n+ac))`
`=(ac(a-c)(n^(2)-1))/(acn^(2)+(a^(2)+c^(2))n+ac)" " "....(ii)"`
But given n is a root of
`x^(2)(1-ac)-x(a^(2)+c^(2))-(1+ac)=0`
Then, `n^(2)(1-ac)-n(a^(2)+c^(2))-(1+ac)=0`
`acn^(2)+(a^(2)+c^(2))n+ac=n^(2)-1`
then from Eq. (ii), `h_(1)-h_(1)=(ac+(a-c)(n^(2)-1))/((n^(2)-1))=ac(a-c)`