Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, then difference of first and third numbers is

1. Define the numbers based on the relationships:To avoid fractions, it is best to start by defining the second number as $x$.Second number: $x$First number: $2x$ (since it is twice the second)Third number: $4x$ (since the first number, $2x$, is half of the third, the third must be $2 \times 2x = 4x$)2. Set up the Average Equation:The average of these three numbers is given as 56.$$\text{Average} = \frac{\text{Sum of numbers}}{\text{Total count}} = 56$$$$\frac{2x + x + 4x}{3} = 56$$3. Solve for $x$:Combine the terms in the numerator:$$\frac{7x}{3} = 56$$Multiply both sides by 3:$$7x = 168$$Divide by 7:$$x = 24$$4. Find the First and Third numbers:First number ($2x$): $2 \times 24 = 48$Third number ($4x$): $4 \times 24 = 96$5. Calculate the difference:The question asks for the difference between the first and third numbers:$$96 - 48 = \mathbf{48}$$