Problem: Among fourteen consecutive integers, the sum of the last seven integers is 161. The sum of the first 7 integers is?

Solution:

Let's denote the first integer by x. Then, the next 13 consecutive integers will be x + 1, x + 2, ..., x + 13.

We know that the sum of the last seven integers is 161. Therefore,

(x + 7) + (x + 8) + ... + (x + 13) = 161

Now, we can simplify this expression by using the formula for the sum of an arithmetic sequence:

[(first term + last term) * number of terms] / 2

Using this formula, we can rewrite the left-hand side of the equation as:

[(x + 7) + (x + 13)] * 7 / 2

Simplifying this expression, we get:

(2x + 20) * 7 / 2 = 161

Simplifying further, we get:

2x + 20 = 46

2x = 26

x = 13

Therefore, the first integer is 13.

Sum of the First 7 Integers:

Now that we know the first integer, we can easily find the sum of the first seven integers by adding them up:

13 + 14 + 15 + 16 + 17 + 18 + 19 = 112

Therefore, the sum of the first seven integers is 112.

Conclusion:

Among fourteen consecutive integers, the sum of the last seven integers is 161. The sum of the first 7 integers is 112.

112