Given information:
- The average of twelve consecutive even numbers is 113.

To find:
- The sum of the 3rd and 8th numbers among them.

Solution:
Let's assume the first even number in the sequence as 'x'.

Since the numbers are consecutive even numbers, the next number will be (x + 2), the number after that will be (x + 4), and so on.

The sum of the twelve consecutive even numbers can be calculated as follows:
x + (x + 2) + (x + 4) + ... + (x + 20) = 12 * 113

Simplifying the equation:
12x + 2 + 4 + ... + 20 = 12 * 113
12x + (2 + 4 + ... + 20) = 12 * 113
12x + 2(1 + 2 + ... + 10) = 12 * 113
12x + 2(10 * 11 / 2) = 12 * 113
12x + 110 = 12 * 113
12x = 12 * 113 - 110
12x = 1356 - 110
12x = 1246
x = 1246 / 12
x = 103.83

Since x is not an integer, it means that our assumption was incorrect.

Let's assume the first even number in the sequence as 'y', where y is an integer.

The sum of the twelve consecutive even numbers can be calculated as follows:
y + (y + 2) + (y + 4) + ... + (y + 20) = 12 * 113

Simplifying the equation:
12y + 2 + 4 + ... + 20 = 12 * 113
12y + (2 + 4 + ... + 20) = 12 * 113
12y + 2(1 + 2 + ... + 10) = 12 * 113
12y + 2(10 * 11 / 2) = 12 * 113
12y + 110 = 12 * 113
12y = 12 * 113 - 110
12y = 1356 - 110
12y = 1246
y = 1246 / 12
y = 103.83

Again, y is not an integer, which means our assumption was incorrect again.

Let's assume the first even number in the sequence as 'z', where z is an integer.

The sum of the twelve consecutive even numbers can be calculated as follows:
z + (z + 2) + (z + 4) + ... + (z + 20) = 12 * 113

Simplifying the equation:
12z + 2 + 4 + ... + 20 = 12 * 113
12z + (2 + 4 + ... + 20) = 12 * 113
12z + 2(1 + 2 + ... + 10) = 12 * 113
12z + 2(10 * 11 / 2) = 12 * 113
12z + 110 = 12 * 113
12z = 12 * 113 -