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

Let the positive numbers $a, b, c, d$ be in $A.P.$,then $abc, abd, acd, bcd$ are

If $(b+c), (c+a), (a+b)$ are in $H.P.$,then $a^2, b^2, c^2$ are in.......

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

If $\log (x + z) + \log (x + z - 2y) = 2\log (x - z),$ then $x, y, z$ are in

If $a_1, a_2, a_3, \dots, a_n$ form a harmonic progression,find the value of $a_1a_2 + a_2a_3 + \dots + a_{n-1}a_n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module