Problem 32 - Algebra 2 and Trigonometry

Question

In a geometric sequence, \(a_{3}=1\) and \(a_{7}=9 .\) Find \(a_{1}\)

Step-by-Step Solution

Answer

The first term, \( a_1 \), is \( \frac{1}{3} \).

Step 1: Understand the Formula for a Geometric Sequence

In a geometric sequence, each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The formula for the nth term is \( a_n = a_1 imes r^{n-1} \), where \( a_1 \) is the first term and \( r \) is the common ratio.

Step 2: Set Up Equations Using Given Terms

We know that \( a_3 = 1 \) and \( a_7 = 9 \). Using the formula, these can be written as two equations: \( a_3 = a_1 imes r^2 = 1 \) and \( a_7 = a_1 imes r^6 = 9 \).

Step 3: Solve for Common Ratio \( r \)

Divide the second equation by the first to eliminate \( a_1 \): \[ \frac{a_1 imes r^6}{a_1 imes r^2} = \frac{9}{1} \]. This simplifies to \( r^4 = 9 \). Taking the fourth root gives \( r = 9^{1/4} \).

Step 4: Calculate the First Term \( a_1 \)

Now use \( a_3 = a_1 imes r^2 = 1 \). Substitute \( r = 9^{1/4} \): \( a_1 imes (9^{1/4})^2 = 1 \). This simplifies to \( a_1 imes 9^{1/2} = 1 \). Since \( 9^{1/2} = 3 \), rearrange to find \( a_1 = \frac{1}{3} \).

Key Concepts

Common Ratio

In a geometric sequence, the concept of the common ratio is central. It's the fixed, non-zero number by which each term is multiplied to get the next term in the sequence. If you know one term and the common ratio, you can find any subsequent term easily using multiplication.

• The common ratio, often denoted as "\( r \)," is crucial in determining how a sequence behaves.
• In geometric sequences, the ratio between any two consecutive terms remains the same.
• When determining the common ratio, it's essential to have at least two known terms from the sequence.

Identifying the correct common ratio allows us to predict, explore, and understand the sequence's growth or decay behavior, which is pivotal in solving other parts of the sequence.

Nth Term Formula

The nth term formula for a geometric sequence is the tool that allows us to access any term in the sequence without listing all preceding terms.
The formula is expressed as: \[ a_n = a_1 \times r^{n-1} \]

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

To utilize this formula, you need both the first term and the common ratio. This formula provides a streamlined approach to sequence analysis and proves especially useful in solving for unknowns in the sequence.

Solving Equations

Solving equations is a methodical process, often used to find unknown variables. In the context of geometric sequences, we frequently set up equations to identify the common ratio or an initial term.
Consider the geometric sequence where \( a_3 = 1 \) and \( a_7 = 9 \). We can write:

• \( a_3 = a_1 \times r^2 = 1 \)
• \( a_7 = a_1 \times r^6 = 9 \)

To solve for \( r \), divide the second equation by the first, effectively eliminating \( a_1 \) and allowing us to calculate \( r \) as:\[ \frac{a_1 \times r^6}{a_1 \times r^2} = \frac{9}{1} \]Simplifying: \( r^4 = 9 \). Solving for \( r \) involves finding the fourth root of 9.Equations like these are fundamental in breaking down geometric problems into manageable steps.

Exponents

Exponents are a powerful mathematical tool, used to simplify multiplication of the same number by itself.
In geometric sequences, exponents feature prominently due to the common ratio being raised to a power, which represents term positions.

• In our example, \( r^n \) signifies the common ratio raised to the calculated power for a given term.
• Solving \( r^4 = 9 \) requires understanding that the fourth root (or exponentiation power) of 9 will determine \( r \).

When working with exponents, it's critical to follow arithmetic laws and remember that exponentiation significantly alters numbers' magnitude, explaining the rapid growth or decline of sequences. Exponents make it possible to handle large sequences efficiently and help in calculating precise term values.