Problem 26 - Algebra 2

Question

Determine whether each series is arithmetic or geometric. Then evaluate the finite series for the specified number of terms. \(2+4+6+8+\ldots ; n=20\)

Step-by-Step Solution

Answer

The given series is an arithmetic series. The sum of the first 20 terms is 420.

Step 1: Identify the type of series

By observing the given series, \(2+4+6+8+\ldots\), it can be noticed that each consecutive term is obtained by adding a constant difference of 2. Therefore, the given series is an arithmetic series.

Step 2: Write down the formula for the sum of an arithmetic series

The sum \(S\) of the first \(n\) terms of an arithmetic series can be found using the formula \(S = n/2 [2a + (n − 1)d]\), where \(a\) is the first term and \(d\) is the common difference. For the given arithmetic series, \(a = 2\) and \(d = 2\).

Step 3: Substitute the values into the formula and solve

Substitute \(n = 20\), \(a = 2\), and \(d = 2\) into the formula to find the sum of the first 20 terms of the series. Doing the calculations, \(S = 20/2 [2*2 + (20 - 1)*2] = 10 * [4 + 38] = 10 * 42 = 420\).

Key Concepts

Series Evaluation

In mathematics, a series represents the sum of the terms of a sequence. Determining the type of series is crucial to finding the correct method for evaluating it. To evaluate a series is essentially to calculate its sum. There are two primary types of series: arithmetic and geometric. Recognizing the difference is the first step in series evaluation. - **Arithmetic Series:** In this series, the difference between consecutive terms remains constant. For example, in the series 2, 4, 6, 8..., each term increases by 2. - **Geometric Series:** In contrast, this series multiplies each term by a constant to get to the next term. Evaluating a given series means knowing whether it's arithmetic or geometric as this directs the calculation method. In our example, since the series has a constant addition of 2 between terms, we can confirm it as arithmetic. Using specific formulas tailored to each type ensures a correct evaluation for finding the sum.

Common Difference

An essential feature in arithmetic series is the common difference. This is the constant amount by which each term in the series increases from the previous term. Understanding and finding this difference is crucial for both identifying a series as arithmetic and calculating its sum.- **Calculating the Common Difference (d):** Subtract any term from the term succeeding it. For example, in the series 2, 4, 6, 8..., the common difference is calculated as: \( d = 4 - 2 = 2 \). Each term advances by this fixed value, reinforcing the pattern of the series. The common difference not only helps in identifying the series type but is also fundamental in applying formulas, such as finding the sum of the series.Consistently identifying and using this difference enhances your problem-solving efficiency and accuracy in evaluating arithmetic series.

Sum of Arithmetic Series

Finding the sum of an arithmetic series involves using a straightforward formula. The sum is calculated over a specified number of terms, denoted as \( n \).Here’s how you do it:1. **Formula:** The sum \( S \) of the first \( n \) terms in an arithmetic series is given by the formula: \[ S = \frac{n}{2} \times [2a + (n − 1) \times d] \]. 2. **Parameters:** - \( a \): The first term of the series. - \( d \): The common difference.To find the sum of our series 2, 4, 6, 8... for 20 terms:- First Term \( a = 2 \).- Common Difference \( d = 2 \).- Number of terms \( n = 20 \).Substitute these values into the formula:- \[ S = \frac{20}{2} \times [2 \times 2 + (20 − 1) \times 2] \]- Simplifies to: \( S = 10 \times [4 + 38] = 10 \times 42 = 420 \).This formula efficiently gives you the total sum, saving time and potential errors in manual summation. Understanding how each component of the formula plays into finding the overall sum ties together the process of series evaluation, common difference identification, and successful calculation of the sum.