If 9 arithmetic means (A.M.s) and 9 harmonic | vedclass

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(C) Let $A_j$ and $H_j$ denote the $j^{th}$ $A.M.$ and $H.M.$ respectively,where $j = 1, 2, \dots, 9$,inserted between $2$ and $3$.For the $A.M.s$,the

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$5$

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(C) Let $A_j$ and $H_j$ denote the $j^{th}$ $A.M.$ and $H.M.$ respectively,where $j = 1, 2, \dots, 9$,inserted between $2$ and $3$.For the $A.M.s$,the sequence $2, A_1, A_2, \dots, A_9, 3$ is in $A.P.$ with $11$ terms.Common difference $d = \frac{3 - 2}{9 + 1} = \frac{1}{10}$.Thus,$A_j = 2 + j 	imes d = 2 + \frac{j}{10}$.For the $H.M.s$,the sequence $2, H_1, H_2, \dots, H_9, 3$ is in $H.P.$Therefore,$\frac{1}{2}, \frac{1}{H_1}, \frac{1}{H_2}, \dots, \frac{1}{H_9}, \frac{1}{3}$ is in $A.P.$Common difference $D = \frac{\frac{1}{3} - \frac{1}{2}}{9 + 1} = \frac{-\frac{1}{6}}{10} = -\frac{1}{60}$.Thus,$\frac{1}{H_j} = \frac{1}{2} + j 	imes D = \frac{1}{2} - \frac{j}{60}$.Now,calculate $A_j + \frac{6}{H_j}$:$A_j + 6 	imes \frac{1}{H_j} = (2 + \frac{j}{10}) + 6 	imes (\frac{1}{2} - \frac{j}{60})$$= 2 + \frac{j}{10} + 3 - \frac{6j}{60}$$= 5 + \frac{j}{10} - \frac{j}{10} = 5$.

If $9$ arithmetic means $(A.M.s)$ and $9$ harmonic means $(H.M.s)$ are inserted between $2$ and $3$,and if the harmonic mean $H$ corresponds to the arithmetic mean $A$ (i.e.,the $j^{th}$ $A.M.$ and $j^{th}$ $H.M.$),then $A + \frac{6}{H} = $

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