Question

If $9$ arithmetic means and harmonic means are inserted between $2$ and $3$, then the value of $A+\frac{6}{H}$ is
(where $A$ is any of the A.M.'s and H the corresponding H.M.)

Answer 1

Let $A_i,H_i(i=1,2,\dots 9)$ denote the $9$ A.M's and $9$ H.M's between $2$ and $3$.
Let $d$ denote the common difference of A.P., then
$T_{11}=2+10d\Rightarrow 3=2+10dd=\frac{1}{10}$
Now,
$A_i=2+(i+1-1)d\Rightarrow A=2+\frac{i}{10}$

We know that,
$2,H_1,H_2,\dots H_9,3$ are in H.P.
$\frac{1}{2},\frac{1}{H_1},\frac{1}{H_2},\dots \frac{1}{H_9},\frac{1}{3}$ is in A.P.
Let $d_1$ is the common difference of this A.P., then
$T_{11}=\frac{1}{2}+10d_1\Rightarrow \frac{1}{3}=\frac{1}{2}+10d_1\Rightarrow d_1=-\frac{1}{60}$
Now,
$\frac{1}{H_i}=\frac{1}{2}+(i+1-1)d_1\Rightarrow \frac{1}{H}=\frac{1}{2}-\frac{i}{60}$

Now,
$A+\frac{6}{H}=2+\frac{i}{10}+6(\frac{1}{2}-\frac{i}{60})=2+\frac{i}{10}+3-\frac{i}{10}=5$