Problem 43 - College Algebra with Corequisite Support

Question

For the following exercises, find the specified term for the geometric sequence given. Let \(a_{n}=-\left(-\frac{1}{3}\right)^{n-1} \cdot\) Find \(a_{12}\).

Step-by-Step Solution

Answer

The 12th term, \(a_{12}\), is \(\frac{1}{177147}\).

Step 1: Identify the Sequence Formula

The given geometric sequence formula is \(a_n = -\left(-\frac{1}{3}\right)^{n-1}\). Our task is to use this formula to find the 12th term of the sequence, which is \(a_{12}\).

Step 2: Substitute n=12 into the Formula

To find the 12th term, substitute \(n = 12\) into the sequence formula. This gives us \(a_{12} = -\left(-\frac{1}{3}\right)^{12-1}\). Simplify the exponent to proceed with calculations.

Step 3: Calculate the Exponent

The exponent can be simplified to \(n-1 = 12-1 = 11\). Thus, the expression becomes \(-\left(-\frac{1}{3}\right)^{11}\).

Step 4: Solve for the 12th Term

Calculate \((-\frac{1}{3})^{11}\). Since raising a negative number to an odd power yields a negative result, \((-\frac{1}{3})^{11}\) is negative. Specifically, \((-\frac{1}{3})^{11} = -\left(\frac{1}{3}\right)^{11}\).

Step 5: Apply the Negative Sign

Apply the initial negative sign from the sequence formula: \(-\left((-\frac{1}{3})^{11}\right)\). This saves the expression to \(- \cdot -\left(\frac{1}{3}\right)^{11}\) which simplifies to \(\left(\frac{1}{3}\right)^{11}\).

Step 6: Calculate \((\frac{1}{3})^{11}\)

Calculate \((\frac{1}{3})^{11}\). This equals \(\frac{1}{177147}\), as each multiplication divides 1 by 3 raised to the power of 11.

Key Concepts

Understanding the Sequence Formula

When dealing with geometric sequences, the sequence formula is your primary tool. It defines how each term in the sequence is related to the others.
In general, a geometric sequence has the formula

• \( a_n = a_1 imes r^{n-1} \)

Where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number you want to find.

In our exercise, the formula is given as \( a_n = -\left(-\frac{1}{3}\right)^{n-1} \).
This formula already reflects a specific common ratio and an initial value manipulation (involving a negative sign).
To find any term in this sequence, substitute the desired term number for \( n \), then perform the calculations as shown.

The Role of Exponents

Exponents are crucial in geometric sequences because they dictate how the terms change across the sequence.
The exponent \( n-1 \) in the formula \( a_n = -\left(-\frac{1}{3}\right)^{n-1} \) determines the repeated multiplication of the common ratio.

When using exponents, remember these key points:

• Raising a number to an exponent means multiplying that number by itself for the number of times the exponent indicates.
• The exponent reflects the term's position in the sequence beyond the first term, hence \( n-1 \).

Exponents can impact the term's sign, especially when negative numbers are involved, which is explained in the next section.

Working with Negative Numbers

Negative numbers often appear in geometric sequences and can influence both the sign and the magnitude of the terms.
In our specific formula, \( a_n = -\left(-\frac{1}{3}\right)^{n-1} \), the negative in the base \( -\frac{1}{3} \) shows a sign flip characteristic when raised to different powers:

• Even exponents result in a positive product, because of repeated multiplication of negatives.
• Odd exponents keep the product negative since there's one unpaired negative factor left over.

At step 4 in the solution, the exponent is an odd number \( 11 \), which results in a negative base becoming negative initially. However, this is flipped by the original negative sign in front of the whole expression, ultimately making the term positive in the formula \( a_n = -\left(-\frac{1}{3}\right)^{11} \), leading to \( \left(\frac{1}{3}\right)^{11} \).