Question

If nine arithmetic means and nine harmonic means are inserted between $2$ and $3$ alternatively, then prove that $A+6/H=5$ (where $A$ is any of the A.M's and $H$ is the corresponding H.M.

Answer 1

Given,

A.M.:

$T_{11}=3=2+10d\Rightarrow d=\frac{1}{10}$

$\therefore A=2+\frac{i-1}{10}$ i th term

H.M.:

$T_{11}=\frac{1}{3}=\frac{1}{2}+10d\Rightarrow d=-\frac{1}{60}$

$\therefore \frac{1}{H}=\frac{1}{2}-\frac{i-1}{60}$ i th term

Given,

$A+\frac{6}{H}$

$=2+\frac{i-1}{10}+6\times (\frac{1}{2}-\frac{i-1}{60})$

$=2+\frac{i-1}{10}+3-\frac{i-1}{10}$

$=5$

Hence proved.