Question

If nine harmonic means are inserted between $2$ and $3,$ then the value of $A+\frac{6}{H}$ $($where $A$ is any of the A.M.'s and $H$ the corresponding H.M.'s$)$ is:

• A. $8$
• B. $5$
• C. $10$
• D. None of these

Answer 1

The correct option is B $5$

Let $H_1,H_2,...,H_9$ be the harmonic means between $2$ and $3$, then, $H_1,H-2,...,H_9,3$ are in $H.P.$

$\therefore \frac{1}{2},\frac{1}{H_1},\frac{1}{H_2},...,\frac{1}{H_9},\frac{1}{3}$ are in $A.P.$ with common difference

$D=\frac{2-3}{(9+1)\times 2\times 3}=-\frac{1}{60}[\because D=\frac{a-b}{(n+1)ab}]$

$\therefore \frac{1}{H_i}=\frac{1}{2}+iD;i=1,2,3,...,9\Rightarrow \frac{1}{H_i}=\frac{1}{2}-\frac{i}{60}\Rightarrow \frac{6}{H_i}=3-\frac{i}{10}$ ...(1)

Let $A_1,A_2,...,A_9$ be the nine $A.M{.}^{\prime }s$ between $2$ and $3$.

Then, $2,A_1,A_2,...,A_9,3$ are in $A.P.$ with common difference

$d=\frac{3-2}{9+1}=\frac{1}{10}[\because d=\frac{b-a}{n+1}]$

$\therefore A_i=2+id;i=1,2,...,9\Rightarrow A_i=2+\frac{i}{10}$ ...(2)

From (1) and (2), we have

$A_i+\frac{6}{H_i}=2+\frac{i}{10}+3-\frac{i}{10}=5$ for $i=1,2,3,...,9$