Problem 46 - Algebra 2

Question

The sum of an infinite geometric series is twice its first term. a. Error Analysis A student says the common ratio of the series is \(\frac{3}{2} \cdot\) What is the student's error? b. Find the common ratio of the series.

Step-by-Step Solution

Answer

a) The student's error is claiming the common ratio is \(\frac{3}{2}\), which would make the sum of the infinite geometric series undefined. b) The correct common ratio of the series is \(\frac{1}{2}\).

Step 1: Check the student's error

Substitute the student's suggested common ratio \(\frac{3}{2}\) into the equation: \(2a = \frac{a}{(1 - \frac{3}{2})}\). Solving this, we come up with a term undefined, meaning the student's claim of \(\frac{3}{2}\) being the common ratio is incorrect, as the sum of an infinite geometric series can't be undefined.

Step 2: Calculate the correct common ratio

To obtain the correct common ratio, solve the equation \(2a = \frac{a}{1 - r}\) for r, where 2a is the sum and a is the first term. This gives us \(2 = \frac{1}{1 - r}\). By solving for r, we get \(r = \frac{1}{2}\). This is the correct common ratio.

Key Concepts

Understanding the Common Ratio

In an infinite geometric series, the common ratio is the factor by which you multiply each term to get the next term. This ratio is crucial because it determines whether the series converges to a sum or not. For a series to have a finite sum, the absolute value of the common ratio must be less than 1. This means:

• If the common ratio, \( r \), is between -1 and 1, the series converges.
• If \( |r| \geq 1 \), the series can be infinite or undefined.

In the given problem, the student's error was claiming the common ratio was \( \frac{3}{2} \). Since this value is greater than 1, any series with this ratio would diverge. A correctly converging series should have a ratio less than 1, like \( \frac{1}{2} \) found in this exercise.

Calculating the Sum of the Series

To calculate the sum of an infinite geometric series, use the formula:\[S = \frac{a}{1 - r}\]where \( S \) is the sum, \( a \) is the first term, and \( r \) is the common ratio. The exercise tells us that the sum is twice the first term, which means:\[2a = \frac{a}{1 - r}\]This information helps us determine the correct common ratio. Solving for \( r \), we manipulate the equation:\[2 = \frac{1}{1 - r}\]This simplifies to \( r = \frac{1}{2} \). Hence, when the series' sum is twice its first term, \( r \) must be \( \frac{1}{2} \) to make both sides of the equation equal.

Error Analysis: Spotting Mistakes

Error analysis involves identifying where and why mistakes were made. In this example, the student proposed a common ratio of \( \frac{3}{2} \). By substituting this into the formula, we find:\[2a = \frac{a}{1 - \frac{3}{2}}\]Here, the denominator becomes \( 1 - \frac{3}{2} = -\frac{1}{2} \), resulting in an undefined term. This error shows a misunderstanding of how \( r \) affects convergence. By checking whether \( |r| < 1 \), we ensure that mistakes like these are avoided. Always verify that the proposed ratio allows the series to converge. This practice helps in catching errors early and confirming that calculations are on the right path.