The harmonic mean of 4, 8, 16 is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The harmonic mean $(H.M.)$ of $n$ numbers $x_1, x_2, ..., x_n$ is given by $\frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$.For the numbers $4, 8, 16$,we

Vedclass · Accepted Answer

$6.85$

Vedclass · Answer

(C) The harmonic mean $(H.M.)$ of $n$ numbers $x_1, x_2, ..., x_n$ is given by $\frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$.For the numbers $4, 8, 16$,we have $n = 3$.$H.M. = \frac{3}{\frac{1}{4} + \frac{1}{8} + \frac{1}{16}}$$= \frac{3}{\frac{4+2+1}{16}}$$= \frac{3}{\frac{7}{16}}$$= \frac{3 	imes 16}{7} = \frac{48}{7} \approx 6.857$.Rounding to two decimal places,we get $6.86$. Given the options,the closest value is $6.85$.

The harmonic mean of $4, 8, 16$ is

Similar Questions

If $a, b, c$ are in $H.P.$,then the value of $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$ is

If $(y - x)$,$2(y - a)$,and $(y - z)$ are in $H.P.$,then $(x - a)$,$(y - a)$,and $(z - a)$ are in

What is the harmonic mean of the reciprocals of the first $n$ natural numbers?

The sixth $H.M.$ between $3$ and $\frac{6}{13}$ is

If $\frac{1}{4}, a, b, \frac{1}{19}$ form a $H.P.$,then the values of $a$ and $b$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module