Problem 87
Question
Exercises \(85-87\) will help you prepare for the material covered in the next section. Use the formula \(a_{n}-a_{1} 3^{n-1}\) to find the seventh term of the sequence \(11,33,99,297, \dots\)
Step-by-Step Solution
Verified Answer
To find the seventh term of the sequence, calculate \(a_{7} = 11.3^{6}\) using your calculator or by simplifying step by step. This will give you the seventh term of the sequence.
1Step 1: Identify and Understand the Terms
In the provided sequence \(11, 33, 99, 297, \dots\), the first term \(a_{1}\) is 11. The formula to calculate the nth term is \(a_{n}=a_{1}.3^{n-1}\) where \(n\) is the term number. Here, we need to find the seventh term, hence, \(n=7\). So, the known values are: \(a_{1}=11\), \(n=7\).
2Step 2: Substitute values into the formula
Plugging in these values into the given formula, we get: \(a_{n} = a_{1}.3^{n-1}\), which translates to \(a_{7} = 11.3^{7-1} = 11.3^{6}\). This is the equation that we will solve next.
3Step 3: Calculate the value
Solve the equation \(a_{7} = 11.3^{6}\). The result should be the seventh term of the sequence.
Key Concepts
Sequences in algebraExponential expressionsNth term of a sequence
Sequences in algebra
When dealing with sequences in algebra, we are essentially looking at an ordered list of numbers where each number is typically defined by a specific position or term. The beauty of sequences lies in the patterns they follow, which allows us to predict future terms based on the initial terms or the rule of formation.
In the exercise provided, the sequence follows a pattern where each term is three times the previous one, starting with the first term 11. This is an example of a geometric sequence because each term is generated by multiplying the previous term by a constant ratio. Understanding the pattern is the crucial first step in solving sequence-related problems. Once we've identified the pattern, we can use it to find any term in the sequence, a concept also known as the 'nth term of a sequence.'
In algebra, these patterns enable us to model and describe real-world scenarios systematically. For example, geometric sequences can model exponential growth or decay, like population growth or radioactive decay. The key to mastering sequences in algebra is practice, so when you come across sequence exercises, focus on identifying the pattern and applying the correct formula.
In the exercise provided, the sequence follows a pattern where each term is three times the previous one, starting with the first term 11. This is an example of a geometric sequence because each term is generated by multiplying the previous term by a constant ratio. Understanding the pattern is the crucial first step in solving sequence-related problems. Once we've identified the pattern, we can use it to find any term in the sequence, a concept also known as the 'nth term of a sequence.'
In algebra, these patterns enable us to model and describe real-world scenarios systematically. For example, geometric sequences can model exponential growth or decay, like population growth or radioactive decay. The key to mastering sequences in algebra is practice, so when you come across sequence exercises, focus on identifying the pattern and applying the correct formula.
Exponential expressions
Exponential expressions, often denoted by the format where a number is raised to a power, are seen widely across algebra and are especially pertinent when discussing geometric sequences. In our example, we've encountered the expression where the number 3 is raised to the power of the position of the term minus one, denoted as \(3^{n-1}\).
An exponent, in this context, informs us how many times to multiply the base number by itself. So, \(3^{6}\) means we multiply 3 by itself six times. Exponential growth, as represented by these expressions, increases very quickly, which is a fundamental concept essential for understanding not just algebra but also sciences like biology, economics, and many others.
When calculating exponential expressions, we follow the basic principle of powers, where \(a^{m}.a^{n} = a^{m+n}\) and \(\frac{a^{m}}{a^{n}} = a^{m-n}\). It's also important to remember that any number raised to the power of zero is 1. So learning to manipulate these expressions is paramount to working efficiently with geometric sequences and other algebraic problems involving exponentiation.
An exponent, in this context, informs us how many times to multiply the base number by itself. So, \(3^{6}\) means we multiply 3 by itself six times. Exponential growth, as represented by these expressions, increases very quickly, which is a fundamental concept essential for understanding not just algebra but also sciences like biology, economics, and many others.
When calculating exponential expressions, we follow the basic principle of powers, where \(a^{m}.a^{n} = a^{m+n}\) and \(\frac{a^{m}}{a^{n}} = a^{m-n}\). It's also important to remember that any number raised to the power of zero is 1. So learning to manipulate these expressions is paramount to working efficiently with geometric sequences and other algebraic problems involving exponentiation.
Nth term of a sequence
The nth term of a sequence is a formula that tells us the value of an arbitrary term in the sequence without having to list out all preceding terms. It's like a shortcut to any term in the sequence. It's particularly handy in long sequences or when the term we're looking for is far along in the sequence.
In the context of the given exercise, the nth term is derived from the pattern seen in the geometric sequence. Given the first term (\(a_1\)) and the common ratio (which is the constant by which we multiply to get the next term), we can write a general formula for the nth term as \(a_n = a_1 \times r^{(n-1)}\), where \(a_n\) is the nth term we wish to find, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
Using this formula, we have a powerful tool in algebra that allows us to leapfrog over intermediate steps and lands directly on the term we're interested in. For the exercise, plugging in the value of \(n = 7\) gives us the seventh term directly. It's crucial to understand and master the use of the nth term formula, as it comes in handy not just for homework but for understanding sequences and their behaviors in various scientific and mathematical contexts.
In the context of the given exercise, the nth term is derived from the pattern seen in the geometric sequence. Given the first term (\(a_1\)) and the common ratio (which is the constant by which we multiply to get the next term), we can write a general formula for the nth term as \(a_n = a_1 \times r^{(n-1)}\), where \(a_n\) is the nth term we wish to find, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
Using this formula, we have a powerful tool in algebra that allows us to leapfrog over intermediate steps and lands directly on the term we're interested in. For the exercise, plugging in the value of \(n = 7\) gives us the seventh term directly. It's crucial to understand and master the use of the nth term formula, as it comes in handy not just for homework but for understanding sequences and their behaviors in various scientific and mathematical contexts.
Other exercises in this chapter
Problem 86
Determine whether each statement makes sense or does not make sense, and explain your reasoning. I used the permutations formula to determine the number of ways
View solution Problem 86
Exercises 86-88 will help you prepare for the material covered in the next section. $$\text { Evaluate } \frac{n !}{(n-r) !} \text { for } n-20 \text { and } r-
View solution Problem 88
What is a geometric sequence? Give an example with your explanation.
View solution Problem 89
What is the common ratio in a geometric sequence?
View solution