The sixth H.M. between 3 and frac{6}{13} is | vedclass

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

Question

(A) The formula for the $n^{th}$ Harmonic Mean $(H.M.)$ between two numbers $a$ and $b$ is given by $H_n = \frac{(n+1)ab}{na+b}$.Here,$a = 3$,$b = \fr

Vedclass · Accepted Answer

$\frac{63}{120}$

Vedclass · Answer

(A) The formula for the $n^{th}$ Harmonic Mean $(H.M.)$ between two numbers $a$ and $b$ is given by $H_n = \frac{(n+1)ab}{na+b}$.Here,$a = 3$,$b = \frac{6}{13}$,and $n = 6$.Substituting these values into the formula:$H_6 = \frac{(6+1) 	imes 3 	imes \frac{6}{13}}{6 	imes 3 + \frac{6}{13}}$$H_6 = \frac{7 	imes 3 	imes \frac{6}{13}}{18 + \frac{6}{13}}$$H_6 = \frac{\frac{126}{13}}{\frac{234+6}{13}}$$H_6 = \frac{126}{240}$Simplifying the fraction by dividing both numerator and denominator by $2$,we get:$H_6 = \frac{63}{120}$.

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

Similar Questions

If the $m^{th}$ terms of the series $63 + 65 + 67 + 69 + \dots$ and $3 + 10 + 17 + 24 + \dots$ are equal,then $m = $

The sum of the series $1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots$ up to $n$ terms is:

If $m$ is a root of the given equation $(1 - ab)x^2 - (a^2 + b^2)x - (1 + ab) = 0$ and $m$ harmonic means are inserted between $a$ and $b$,then the difference between the last and the first of the means equals

The $n^{th}$ term of the series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

If $A = 1 + r^z + r^{2z} + r^{3z} + .......\infty$,then the value of $r$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module