Problem 38 - Algebra 2

Question

Find \(S_{n}\) for each geometric series described. $$ a_{1}=162, r=\frac{1}{3}, n=6 $$

Step-by-Step Solution

Answer

The sum of the first 6 terms, \( S_6 \), is approximately 243.52.

Step 1: Understand the Formula for Sum of a Geometric Series

The sum of the first \( n \) terms of a geometric series is given by the formula: \[ S_n = a_1 \frac{1-r^n}{1-r} \] where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

Step 2: Substitute the Values into the Formula

Substitute the given values into the formula: \( a_1 = 162 \), \( r = \frac{1}{3} \), and \( n = 6 \). This gives:\[ S_6 = 162 \frac{1-(\frac{1}{3})^6}{1-\frac{1}{3}} \]

Step 3: Calculate \( r^n \)

Calculate the value of \( (\frac{1}{3})^6 \):\[ (\frac{1}{3})^6 = \frac{1}{729} \]

Step 4: Calculate the Numerator

Substitute \( (\frac{1}{3})^6 = \frac{1}{729} \) into the numerator:\[ 1 - (\frac{1}{3})^6 = 1 - \frac{1}{729} = \frac{728}{729} \]

Step 5: Calculate the Denominator

Calculate the denominator:\[ 1 - \frac{1}{3} = \frac{2}{3} \]

Step 6: Find \( S_n \) by Substituting Back

Now substitute the values back into the formula for \( S_6 \):\[ S_6 = 162 \cdot \frac{ \frac{728}{729} }{ \frac{2}{3} } \]

Step 7: Simplify the Expression

Simplify the expression:\[ S_6 = 162 \cdot \frac{728}{729} \cdot \frac{3}{2} \]Calculate:\[ S_6 = 162 \cdot \frac{728}{729} \cdot 1.5 = 162 \cdot \frac{364}{243} \]

Step 8: Perform the Final Calculation

Calculate \( S_6 = 162 \cdot \frac{364}{243} \):\[ S_6 = 162 \cdot 1.498 = 242.52 \] Thus, \( S_6 = 243.52 \) (rounded to two decimal places).

Key Concepts

Sum of a Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The sum of the first \( n \) terms of this sequence is what we call the sum of a geometric series. To find this sum, we use the formula: \\[ S_n = a_1 \frac{1-r^n}{1-r} \] \where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. \
The formula derives from the fact that each term in a geometric series builds upon the previous one based on a consistent pattern defined by the common ratio. \

• The numerator \( 1 - r^n \) acts to subtract a small fraction of the total given the decreasing nature of a convergent geometric series when \(|r| < 1\).
• The denominator \( 1 - r \) essentially reverses the multiplication by \( r \), thus cumulatively summarizing the series when repeated with each term.

\This formula allows us to find the sum efficiently without having to add each term individually.

Common Ratio

The common ratio in a geometric series is the factor by which we multiply each term to get the next term in the sequence. It is usually represented by the letter \( r \). In our given exercise, the common ratio is \( \frac{1}{3} \). \
Understanding the common ratio is crucial because:

• The value of \( r \) determines the nature of the series—if the terms will grow larger, shrink, or remain constant.
• A common ratio greater than 1 tends to make the series grow exponentially.
• A common ratio between 0 and 1 will typically cause the terms to decrease towards zero, making it a convergent series.
• If the common ratio is negative, the terms will alternate in sign.

\By knowing \( r \), we can also evaluate if the series converges or diverges, which impacts whether the series will sum to a finite number.

Number of Terms

The number of terms, postulated with the symbol \( n \), denotes how many elements of the series are included when computing the sum. In our example exercise, \( n = 6 \), indicating that the first six terms of the series should be considered. The total \( n \) affects the extent and accuracy of the geometric series' sum, as it directly influences the formula \( S_n = a_1 \frac{1-r^n}{1-r} \). \
More terms equate to a more precise approximation of the series' behavior, especially in finite series. \

• For a given \( n \), if the common ratio \( r \) is a fraction, an increased \( n \) will make \( r^n \) very small, leading the series sum to stabilize or converge.
• In contrast, if \( r \) is greater than 1, the series sum can grow very large with a bigger \( n \).
• For finite sequences, the number of terms provides an endpoint, ultimately giving a practical boundary to an otherwise infinite mathematical model.

\Thus, understanding \( n \) allows us clarity in how far into the series we are summing, impacting both theoretical digestion and practical computation.