Problem 4 - Algebra for College Students

Question

Math University had an enrollment of 12,800 students in 1998. Each year the enrollment decreased by 75 students. What was the enrollment in 2005 ? 12,275

Step-by-Step Solution

Answer

The enrollment in 2005 was 12,275 students.

Step 1: Determine the number of years

Identify how many years have passed from 1998 to 2005. We calculate this as 2005 minus 1998, which equals 7 years.

Step 2: Calculate total decrease in enrollment

Since the enrollment decreases by 75 students each year, multiply the annual decrease by the number of years: \[ 75 \times 7 = 525. \]

Step 3: Find the enrollment in 2005

Subtract the total decrease over 7 years from the original enrollment in 1998. The calculation is \[ 12800 - 525 = 12275. \]

Key Concepts

Sequences and Series

When learning mathematics, you'll often encounter the concepts of sequences and series. These terms refer to lists of numbers arranged in a particular order according to a rule or formula.
Understanding sequences and series is crucial because they form the foundation for more complex mathematical concepts such as calculus and analysis.

**Sequences**
A sequence is an ordered list of numbers. Each number in the sequence is called a term. There are different types of sequences, but they all follow a specific pattern or rule that dictates how each term is derived from the previous one.

• Arithmetic Sequence: The difference between consecutive terms is constant. Examples include 2, 4, 6, 8 (where the difference is 2).
• Geometric Sequence: Each term is derived by multiplying the previous term by a fixed number. For example, 3, 9, 27, 81 (where each term is multiplied by 3).

**Series**
In contrast, a series refers to the sum of the terms of a sequence. Imagine writing the terms of a sequence in a linear format, summed together. Understanding these concepts is valuable for solving problems related to patterns, like the one involving enrollment calculations.

Arithmetic Sequence

Arithmetic sequences are one of the most commonly studied types of sequences. As mentioned before, each term in an arithmetic sequence is calculated by adding a constant difference to the previous term.
This difference, known as the common difference, is denoted by the letter "d".

To calculate any specific term in an arithmetic sequence, you can employ the formula:

• General formula: \[ a_n = a_1 + (n-1) imes d \]where:
  • \( a_n \) is the nth term you're solving for.
  • \( a_1 \) is the first term in the sequence.
  • \( d \) is the common difference.
  • \( n \) is the position of the term in the sequence.

Applying this to our enrollment problem, if we consider each year as a term in a sequence, the enrollment following an arithmetic sequence decreases by 75 each year, where 12800 is the first term and 7 is the number of terms beyond the first.

Enrollment Calculations

In the context of the given problem, enrollment calculations involve finding out how the number of students changes over a specific period, using arithmetic sequences.
Enrollment can be thought of as a sequence where each year corresponds to a term and each decrease or increase in enrollment is a result of a fixed pattern.

Let's break this down:

• The original enrollment in 1998 is 12,800 students.
• Every year, this number decreases by 75. Thus, the sequence of enrollment each year is arithmetic because the difference between terms is constant.
• The task is to calculate the enrollment in 2005, which is 7 years from 1998.

To find the enrollment in 2005, we use the arithmetic sequence format from before. We'll subtract 525 (the total decrease after 7 years) from 12,800, resulting in an enrollment of 12,275.

Understanding these calculations can help in planning resources or predicting future trends based on historical data.