Question

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

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

Answer 1

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

Let $\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are H.M. between given numbers.

Then $12,a,b,c,d,42$ will be in A.P.

common difference, $d=\frac{42-12}{5}=6$

Thus $a=18,b=24,c=30,d=36$

Hence third harmonic mean $=\frac{1}{c}=\frac{1}{30}$