Question

in three numbers . first is double of the second and second number is double of the third number.the averag of theirreciprocal is 7/72 find the numbers​

Answer 1

Step-by-step explanation:

Let three numbers be $x, y,z$

Given $x=2y⇒x=4z;y=2z;z=z$

The average of reciprocal numbers is \(\dfrac{7}{72}\)

\(\dfrac{1/x+1/y+1/z}{3}=\dfrac{7}{72}\)

\(\dfrac{xy+yz+zy}{3xyz}=\dfrac{7}{72}\)

\(\dfrac{2z^2+4z^2+8z^2}{3×4z×2z×z}=\dfrac{7}{72}\)

\(\dfrac{7}{12z}=\dfrac{7}{72}\)

\(z=\dfrac{504}{84}\)

\(z=6\)

$⇒x=4×6$ = 24