Problem 47 - Precalculus: Mathematics for Calculus

Question

Which Term? The first term of a geometric sequence is 1536 and the common ratio is \(\frac{1}{2} .\) Which term of the sequence is \(6 ?\)

Step-by-Step Solution

Answer

The 9th term of the sequence is 6.

Step 1: Understand the Problem

Identify the key elements of the problem: the first term of the geometric sequence, the common ratio, and the value of the term we need to find. Here, the first term is 1536, the common ratio is \( \frac{1}{2} \), and we need to find which term equals 6.

Step 2: Formula for the nth Term

Recall the formula for the nth term of a geometric sequence: \[ a_n = a_1 \, r^{n-1}\]where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.

Step 3: Substitute Known Values

Substitute the values provided into the formula: \[ 6 = 1536 \left(\frac{1}{2}\right)^{n-1}\] This equation will allow us to solve for \( n \).

Step 4: Isolate the Exponent

Start by dividing both sides by 1536 to isolate the exponential expression:\[ \left(\frac{1}{2}\right)^{n-1} = \frac{6}{1536}\]Simplify the fraction on the right side to get:\[ \left(\frac{1}{2}\right)^{n-1} = \frac{1}{256}\]

Step 5: Solve for n

Recognize that \( \frac{1}{256} = \left(\frac{1}{2}\right)^8 \). Thus, we have:\[ \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^8\]Since the bases are the same, equate the exponents:\[ n-1 = 8\]Solve for \( n \):\[ n = 9\]

Step 6: Conclusion

The term of the sequence that is equal to 6 is the 9th term.

Key Concepts

Common Ratio

In a geometric sequence, the common ratio is the factor by which we multiply each term to get the next term. Understanding the common ratio is crucial because it determines the structure and progression of the sequence. In our exercise, the common ratio is given as \( \frac{1}{2} \). This means that to find the next term in the sequence, we multiply the preceding term by \( \frac{1}{2} \).

• The common ratio can be any non-zero number, positive or negative.
• For a ratio less than 1, the sequence terms decrease.
• If the ratio is greater than 1, the terms grow larger.
• If the ratio equals 1, each term is identical to the first term.

In practice, knowing the common ratio is useful for predicting the long-term behavior of the sequence.

nth Term Formula

The nth term formula of a geometric sequence enables you to find any term in the sequence without listing all previous terms. The formula is given as: \[ a_n = a_1 \cdot r^{n-1} \] where:

• \( a_n \) is the term you want to find.
• \( a_1 \) is the first term of the sequence.
• \( r \) is the common ratio.
• \( n \) is the position of the term in the sequence.

To solve a problem like ours, substitute the known values into this formula. It is a handy tool because it simplifies the process of finding specific terms within geometric progressions.

Geometric Progression

A geometric progression, or geometric sequence, 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. In our example, starting with 1536 and applying a common ratio of \( \frac{1}{2} \) illustrates a sequence shrinking with each term.

• The terms of the sequence decrease exponentially.
• Geometric progressions can model real-world phenomena like radioactive decay.
• Recognizing and understanding the pattern of a geometric sequence is helpful in both algebra and applied mathematics.

Whether is increasing or decreasing, the pattern detection in geometric progression aids in problem-solving.

Exponential Equation Solving

Solving exponential equations involves finding the unknown exponent that makes the equation true. In our problem, we deal with an equation where both sides are written as powers of \( \frac{1}{2} \). This simplifies to an equation where we only need to match exponents.

• First, express the equation in a form where bases are the same.
• Use properties of exponents to simplify, like dividing or multiplying terms.
• Equate the exponents once the bases are matched.
• Solve the resulting simple equation for the unknown variable.

With practice, solving equations where the variable is in the exponent becomes an intuitive process, bolstered by a solid understanding of the properties involved.