Question

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

Answer 1

Hint: To solve this question we need to know about harmonic series. Harmonic mean is a type of a numerical average. While solving this question we need to consider the fact that the reciprocal of numbers in harmonic series changes to arithmetic series.

Complete step by step answer:
The question asks us to find the third harmonic mean between $\dfrac{1}{12}$and $\dfrac{1}{42}$. The first step in this process will be to consider four harmonic means which are between these numbers. Let the four harmonic mean be $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}$. So all the harmonic means hence become $\dfrac{1}{12},\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d},\dfrac{1}{42}$ . So the third harmonic mean between the given fraction will be $\dfrac{1}{c}$.
The second step is to write the above mean in its reciprocal as it becomes the numbers in Arithmetic Progression. So the series hence become:
$\Rightarrow 12,a,b,c,d,42$
Since the above group is in arithmetic progression, the common difference will be found. The formula used will be:
$\Rightarrow \dfrac{\text{Last term - First term }}{\text{Total number of common difference between }{{\text{1}}^{\text{st}}}\text{ and last term}}$
On substituting the value in the above formula we get:
$\Rightarrow \dfrac{\text{42-12 }}{5}$
On calculating the fraction we get:
$\Rightarrow \dfrac{30}{5}$
$\Rightarrow 6$
The common difference hence becomes $6$.
The next step is to find the value of $c$, which is the third term. To find it we will use the formula of finding the ${{n}^{th}}$ term, which is ${{T}_{n}}=a+\left( n-1 \right)d$. The series we will use is $12,a,b,c,d,42$. So the term $c$ is in the fourth position. Hence the value becomes:
$\Rightarrow {{T}_{4}}=12+\left( 4-1 \right)6$
$\Rightarrow {{T}_{4}}=12+3\times 6$
$\Rightarrow {{T}_{4}}=12+18$
$\Rightarrow {{T}_{4}}=30$
So the value of $c$ is $30$.
The value of the harmonic mean will become $\dfrac{1}{c}$ , which is $\dfrac{1}{30}$.
$\therefore $ The third harmonic mean between $\dfrac{1}{12},\dfrac{1}{42}$ is $\dfrac{1}{30}$.

Note: As we know that there are three kinds of series, which are harmonic series, arithmetic series and geometric series. Sometimes we are asked to calculate the range of harmonic mean. In that case do remember that $AM > GM > HM$ , where they are arithmetic mean, geometric mean and harmonic mean respectively.