Question

If there are four harmonic means between $\frac{1}{12},\frac{1}{42}$, then the third harmonic mean is

• A. $18$
• B. $24$
• C. $\frac{1}{30}$
• D. $36$

Answer 1

The correct option is C $\frac{1}{30}$

Let $H_1,H_2,H_3,H_4$ be four harmonic mean between $\frac{1}{12}$ and $\frac{1}{42}$

Then $\frac{1}{12},H_1,H_2,H_3,H_4,\frac{1}{42}$ is a H.P

$\Rightarrow 12,\frac{1}{H_1},\frac{1}{H_2},\frac{1}{H_3},\frac{1}{H_4},42$ is an A.P with $a=12$ and $a_6=42$

Now, $a_6=42$

$\Rightarrow a+5d=42$

$\Rightarrow 12+5d=42$

$\Rightarrow 5d=30$

$\Rightarrow d=6$

So, $a_4=\frac{1}{H_3}$

$\Rightarrow a+3d=\frac{1}{H_3}$

$\Rightarrow 12+18=\frac{1}{H_3}$

$\Rightarrow H_3=\frac{1}{30}$

Hence $\frac{1}{30}$ is the required third harmonic mean