Problem 12 - Precalculus: Mathematics for Calculus

Question

Find the \(n\) th term of the geometric sequence with given first term \(a\) and common ratio \(r .\) What is the fourth term? $$a=\sqrt{3}, \quad r=\sqrt{3}$$

Step-by-Step Solution

Answer

The fourth term is 9.

Step 1: Understand the Geometric Sequence Formula

In a geometric sequence, the nth term is given by the formula: \( a_n = a \times r^{(n-1)} \), where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.

Step 2: Identify Given Values

We are given \( a = \sqrt{3} \) and \( r = \sqrt{3} \). We need to find the 4th term, so \( n = 4 \).

Step 3: Substitute Into the Formula

Substitute the given values into the geometric sequence formula: \( a_4 = \sqrt{3} \times (\sqrt{3})^{4-1} \).

Step 4: Simplify the Exponent

Calculate the power of the common ratio: \( (\sqrt{3})^3 = (\sqrt{3}) \times (\sqrt{3}) \times (\sqrt{3}) = 3 \times \sqrt{3} = 3\sqrt{3} \).

Step 5: Calculate the Fourth Term

Now, calculate the 4th term: \( a_4 = \sqrt{3} \times 3\sqrt{3} = 3 \times (\sqrt{3}) \times (\sqrt{3}) = 3 \times 3 = 9 \).

Key Concepts

nth term formula

The nth term formula is pivotal when dealing with geometric sequences. This formula allows us to calculate any term in the sequence given the first term and the common ratio. The formula is expressed as \( a_n = a \times r^{(n-1)} \), where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the term number you wish to find.To break it down:

• \( a \): This is the starting point of your sequence. In our exercise, it is \( \sqrt{3} \).
• The exponent \( n-1 \): It represents the number of times the common ratio is multiplied by itself. If you are looking for the 4th term, you use \( 4-1 \), which equals 3.
• \( a_n \): This represents the term you are trying to calculate or find.

Understanding this formula is crucial because it integrates both the initial term of the sequence and the repetitive multiplication defined by the common ratio. Each term in a geometric sequence builds upon the previous one, exemplifying exponential growth or decay.

common ratio

The "common ratio" is an essential element in any geometric sequence. It determines how the sequence progresses through multiplication from one term to the next. In a given sequence, each term is obtained by multiplying the previous term by this common value.### Role of Common Ratio

• The common ratio \( r \) subtly controls the evolution of the sequence. In our context, \( r = \sqrt{3} \).
• This ratio must remain constant throughout the sequence, ensuring its geometric nature.
• It could be a positive number, negative number, or even a fraction, each introducing unique characteristics to the sequence.

The common ratio dictates whether a sequence increases, decreases, or oscillates. If it's greater than 1, the sequence generally grows. If it's between 0 and 1, the sequence shrinks towards zero. Understanding this concept is key to predicting how sequences evolve.

simplifying exponents

Simplifying exponents is a critical step in evaluating terms in a geometric sequence. When we substitute values into the nth term formula, we often encounter expressions with exponents that need simplification.For example, in our exercise:- We have \( (\sqrt{3})^3 \).- To simplify, consider \( \sqrt{3} \times \sqrt{3} \times \sqrt{3} \).- Multiply the first two: \( \sqrt{3} \times \sqrt{3} = 3 \).- Then multiply the result by the remaining \( \sqrt{3} \), resulting in \( 3\sqrt{3} \).Simplifying exponents can seem challenging, but using these basic principles of multiplication and powers will make it much easier. It transforms complex expressions into simpler calculations, making it feasible to handle larger sequences effortlessly. This step often involves power rules and occasionally requires recognizing square roots and cubes in the numerical operations.