Let the six distinct natural numbers be denoted as \( a, b, c, d, e, f \) in ascending order.
Given the conditions: \[ \frac{a + b}{2} = 14 \Rightarrow a + b = 28 \] \[ \frac{e + f}{2} = 28 \Rightarrow e + f = 56 \]
To achieve the maximum overall average for these six numbers, the middle values, \( c \) and \( d \), must be maximized.
This yields the sequence: 1, 27, 25, 26, 27, 29. However, the number 27 is repeated, violating the distinctness requirement. Therefore, a revised selection is necessary while maintaining distinctness.
This results in the sequence: 1, 25, 26, 27, 28, 29. However, the sum \( a + b = 1 + 25 = 26 \), which does not equal 28. The set must still satisfy the original conditions: \[ a + b = 28, \quad e + f = 56 \]
Consider the following assignment:
Consider the following assignment:
The set of six distinct numbers is: 2, 25, 26, 27, 28, 29. This set satisfies \( a + b = 28 \) and \( e + f = 56 \). ✅
The total sum is: \[ 2 + 26 + 25 + 27 + 28 + 29 = 137 \]
The average is: \[ \frac{137}{6} \approx 22.83 \]
Consider the following assignment:
Instead, use the following assignment for \( e, f \):
The optimal valid set is determined to be: \( a = 2, b = 26, c = 25, d = 27, e = 28, f = 29 \).
✅ The maximum possible average is \( \frac{137}{6} = \boxed{22.83} \).
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is