Question

$9$ arithmetic means and $9$ harmonic means are inserted between $2$ and $3$ alternatively. If $A_6$ and $H_6$ is the sixth AM and HM respectively, then the value of $A_6+\frac{2}{H_6}$ is

Answer 1

Let $H_1,H_2,...,H_9$ be the $9$ harmonic means between $2$ and $3$.
$\Rightarrow 2,H_1,H_2,...,H_9,3\text{are in H.P}\Rightarrow \frac{1}{2},\frac{1}{H_1},\frac{1}{H_2},...,\frac{1}{H_9},\frac{1}{3}\text{are in A.P with}\text{common difference,}D=(\frac{1}{3}-\frac{1}{2})\left(\frac{1}{9+1}\right)=-\frac{1}{60}\therefore \frac{1}{H_i}=\frac{1}{2}+iD;(i=1,2,..,9)\Rightarrow \frac{1}{H_i}=\frac{1}{2}-\frac{i}{60}$

Let $A_1,A_2,...,A_9$ be the $9$ arithmetic means between $2$ and $3$.
$\Rightarrow 2,A_1,A_2,...,A_9,3\text{are in A.P with}\text{common difference,}d=\frac{3-2}{9+1}=\frac{1}{10}\therefore A_i=2+id;(i=1,2,..,9)\Rightarrow A_i=2+\frac{i}{10}{A}_i+\frac{2}{H_i}=2+\frac{i}{10}+1-\frac{i}{30}=3+\frac{i}{15}\therefore A_6+\frac{2}{H_6}=3+\frac{6}{15}=3.4$