Of the three numbers, the first is twice the second and the second is twice the third. The average of the reciprocal of the numbers is $$\frac{7}{{72}}$$. The numbers are:

Solution (By Examveda Team)

$$\eqalign{ & {\text{Let}}\,{\text{three}}\,{\text{numbers}}\,{\text{be}}\,x,\,y,\,z. \cr & {\text{Given}}, \cr & x = 2y \cr & \Rightarrow x = 4z \cr & \Rightarrow y = 2z \cr & \Rightarrow z = z \cr & {\text{The}}\,{\text{average}}\,{\text{of}}\,{\text{reciprocal}}\,{\text{numbers}}\,{\text{is}}\,\frac{7}{{72}} \cr & \frac{{ { {\frac{1}{x}} + {\frac{1}{y}} + {\frac{1}{z}} } }}{3} = \frac{7}{{72}} \cr & \Rightarrow \frac{{ {yz + xz + xy} }}{{3xyz}} = \frac{7}{{72}} \cr & \Rightarrow \frac{{2z \times z + 4z \times z + 4z \times 2z}}{{3\left( {4z \times 2z \times z} \right)}} = \frac{7}{{72}} \cr & \Rightarrow \frac{{2{z^2} + 4{z^2} + 8{z^2}}}{{3 \times 8{z^3}}} = \frac{7}{{72}} \cr & \Rightarrow \frac{{14{z^2}}}{{24{z^3}}} = \frac{7}{{72}} \cr & \Rightarrow 504 = 84z \cr & z = 6 \cr & {\text{So}},\,x = 4z = 4 \times 6 = 24, \cr & \Rightarrow y = 2z = 2 \times 6 = 12 \cr & {\text{Thus}}\,{\text{the}}\,{\text{numbers}}\,{\text{are}}\,24,\,12,\,6 \cr} $$