We are given the problem statement:
a1 < a2 < a3 < a4 < a5 < a6 < a7
It is known that a4 = 27.
The average of all seven numbers is (a1 + a2 + a3 + a4 + a5 + a6 + a7)/7 = 30.
We are required to find the minimum value of a7.
This implies that we need to distribute the maximum possible values to the other numbers.
First, let's consider a1, a2, and a3.
Since the median is 27,
it follows that a1, a2, a3 < 27.
The maximum possible distribution for these three numbers is
24, 25, 26.
Now we have the sequence:
24, 25, 26, 27, a5, a6, a7.
The sum of the first four numbers is 24 + 25 + 26 + 27 = 102.
Therefore, a5 + a6 + a7 = 108.
Here, we utilize the average of a5, a6, and a7
to deduce the maximum possible values for a5 and a6.
108 / 3 = 36.
This indicates that
a5, a6, and a7 could be 36, 36, 36.
However, these three numbers must be distinct from each other.
Thus, we slightly adjust the possibilities for a5
to obtain
35, 36, 37.
Hence, the minimum value of a7 is determined to be 37.