Understanding the Problem: Sum of Consecutive Even Integers

The goal is to find the largest integer in a sequence of fifty consecutive even integers, given that their sum is 3250. This problem can be approached using the principles of arithmetic series and basic algebra. Let's break down the steps and solve it using different methods.

Step-by-Step Solution Using Algebra

To find the largest of the fifty consecutive even integers, let's denote the first integer in the sequence as x. The sequence would then be:

x, x 2, x 4, ..., x 98

The sum of these integers can be formulated as:

S x (x 2) (x 4) ... (x 98)

This is an arithmetic series where:

First term (a) x Last term (l) x 98 Number of terms (n) 50

The sum of an arithmetic series can be calculated using the formula:

S (n/2) * (a l)

Substituting the values, we get:

S (50/2) * (x (x 98)) 25 * (2x 98) 5 2450

We are given that the sum S is 3250:

5 2450 3250

Solving for x, we get:

5 800

x 16

Therefore, the first integer in the sequence is 16. The largest integer in the sequence is:

x 98 16 98 114

Using the J Programming Language

To verify the solution using a different approach, we can use the J programming language to generate all possible sets of 50 consecutive even integers from 1 to 1000 and find the one that sums to 3250. The code can be written as:

./n~3250
114

The answer is 114, which confirms our earlier calculation.

Using the Concept of Triangular Numbers

Another method involves recognizing that the sum of the first 50 even numbers is 2450. We can use the sum of the first 49 triangular numbers:

1 2 3 ... 49 (49 * 50) / 2 1225

Multiplying by 2 to convert these triangular numbers into even numbers, we get:

2 * 1225 2450

The sum we need is 3250, so we calculate:

3250 - 2450 800

x 800 / 50 16

The largest integer in the sequence is:

x 98 16 98 114

Thus, the largest integer in the sequence is 114.