Problem 55 - College Algebra with Corequisite Support

Question

At which term does the sequence $$ \\{10,12,14.4,17.28, \quad \ldots\\} $$ exceed \(100 ?\)

Step-by-Step Solution

Answer

The sequence exceeds 100 at the 24th term.

Step 1: Identify the type of sequence

The given sequence is geometric because each term increases by a constant ratio. The first term is \( a_1 = 10 \). We need to find the common ratio \( r \).

Step 2: Determine the common ratio

Divide the second term by the first term to find the common ratio: \( r = \frac{12}{10} = 1.2 \). Hence, the sequence is geometric with \( a_1 = 10 \) and \( r = 1.2 \).

Step 3: Formulate the general term

The general term for a geometric sequence is given by the formula \( a_n = a_1 \, r^{n-1} \). So, for this sequence, the general term is \( a_n = 10 \, (1.2)^{n-1} \).

Step 4: Set up the inequality

We need to find the smallest \( n \) such that \( 10 \, (1.2)^{n-1} > 100 \).

Step 5: Solve the inequality for \( n \)

First, divide by 10: \( (1.2)^{n-1} > 10 \). Take the logarithm on both sides: \( \log((1.2)^{n-1}) > \log(10) \). Use logarithm power rule: \((n-1) \log(1.2) > \log(10) \). Solve for \( n-1 \): \( n-1 > \frac{\log(10)}{\log(1.2)} \). Calculate \( \frac{\log(10)}{\log(1.2)} \).

Step 6: Calculate and interpret the result

Calculate \( n-1 > \frac{1}{0.07918} \approx 22.4687 \). This gives \( n > 23.4687 \), so \( n = 24 \) by rounding up since \( n \) must be an integer.

Key Concepts

Common Ratio

A geometric sequence is defined by its constant ratio between successive terms, known as the **common ratio**. This makes it easy to determine each subsequent term, starting from the first term. In our sequence, we have:

• First term, \( a_1 = 10 \)
• Second term, \( a_2 = 12 \)

To find the common ratio \( r \), we divide the second term by the first term: \[ r = \frac{12}{10} = 1.2 \]This means that each term in the sequence is 1.2 times the term before it. Understanding the common ratio helps us determine the nature and progression of the sequence.

General Term Formula

The general term formula allows us to find any term in a geometric sequence. This formula is essential for understanding the sequence's behavior and solving related problems.For a geometric sequence, the general term \( a_n \) is given by:\[ a_n = a_1 \times r^{n-1} \]Where:

• \( a_1 \) is the first term
• \( r \) is the common ratio
• \( n \) is the term position in the sequence

Using our example, the general term formula becomes:\[ a_n = 10 \times (1.2)^{n-1} \]This formula allows us to calculate any term, \( a_n \), directly, without having to compute all previous terms. It’s a shortcut that’s both powerful and efficient.

Logarithm

Logarithms are very useful when dealing with exponential expressions, such as those found in geometric sequences. They help us solve for unknowns in such expressions.In our problem, we needed to find when the sequence exceeds 100. We reached the expression:\[ (1.2)^{n-1} > 10 \]To solve this inequality, we use logarithms:

• Apply the logarithm to both sides: \( \log((1.2)^{n-1}) > \log(10) \)
• Use the power rule for logarithms: \( (n-1) \cdot \log(1.2) > \log(10) \)

This step simplifies the equation, allowing us to isolate \( n-1 \). Logs make it easier to compare exponential growth with specific values, a key aspect when solving sequence-related questions.

Inequality

Inequalities are crucial for problems that ask for limits or thresholds, like determining when a sequence surpasses a given value. Here, we want to know the smallest term \( n \) such that:\[ 10 \times (1.2)^{n-1} > 100 \]After simplifying, the inequality becomes:\[ (1.2)^{n-1} > 10 \]We calculate:

• Find logs to get \( (n-1) \cdot \log(1.2) > \log(10) \)
• Solve for \( n-1 \): \( n-1 > \frac{\log(10)}{\log(1.2)} \)

This tells us that \( n-1 \) must be greater than approximately 22.4687. Thus, \( n \) must be 24, since \( n \) has to be a whole number. Converting the problem into an inequality and using logarithms lets us handle exponential increases systematically and accurately.