Problem 37
Question
Which term of the geometric sequence \(2,6,18, \ldots\) is \(118,098 ?\)
Step-by-Step Solution
Verified Answer
The 11th term of the sequence is 118,098.
1Step 1: Identify the common ratio
A geometric sequence has a constant ratio between consecutive terms. To find the ratio, divide the second term by the first term. Therefore, the common ratio \( r \) is \( r = \frac{6}{2} = 3 \).
2Step 2: Use the formula for the nth term of a geometric sequence
The formula for the nth term in a geometric sequence is given by \( a_n = a_1 \, r^{n-1} \), where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
3Step 3: Set up the equation
We know \( a_1 = 2 \), \( r = 3 \), and \( a_n = 118,098 \). Substitute these values into the formula: \( 118,098 = 2 \, (3^{n-1}) \).
4Step 4: Solve for \( n \)
First, divide both sides by 2 to isolate the power: \[3^{n-1} = \frac{118,098}{2} = 59,049.\]Then, identify that \( 59,049 = 3^{10} \). Thus, we have \( 3^{n-1} = 3^{10} \).
5Step 5: Find the value of \( n \)
Since the bases are the same, set \( n-1 = 10 \). Solving for \( n \) gives:\[n = 10 + 1 = 11.\]
Key Concepts
Common Rationth Term of a Geometric SequenceSolving Equations
Common Ratio
The common ratio in a geometric sequence is a fundamental concept to grasp. It is the factor by which each term is multiplied to obtain the next term in the sequence.
Identifying the common ratio helps in understanding the growth pattern of the sequence. To find the common ratio, simply divide a term by its preceding term.
For example, in the sequence given in the exercise, the terms are 2, 6, and 18. Thus, we find the common ratio by dividing the second term by the first term:
Identifying the common ratio helps in understanding the growth pattern of the sequence. To find the common ratio, simply divide a term by its preceding term.
For example, in the sequence given in the exercise, the terms are 2, 6, and 18. Thus, we find the common ratio by dividing the second term by the first term:
- \(r = \frac{6}{2} = 3\)
- Verifying with the next consecutive terms: \(\frac{18}{6} = 3\)
nth Term of a Geometric Sequence
The nth term of a geometric sequence is an important tool used to find any specific term in the sequence, especially when the terms grow or shrink rapidly. This is expressed using a formula which is crucial for solving such problems:
\[ a_n = a_1 \, r^{n-1} \] Here:
For the example given, the first term \(a_1\) is 2, the common ratio \(r\) is 3, and we need to find the term when \(a_n = 118,098\). This forms the equation to solve.
\[ a_n = a_1 \, r^{n-1} \] Here:
- \(a_n\) is the nth term we want to find
- \(a_1\) is the first term of the sequence
- \(r\) is the common ratio
- \(n\) is the specific term number
For the example given, the first term \(a_1\) is 2, the common ratio \(r\) is 3, and we need to find the term when \(a_n = 118,098\). This forms the equation to solve.
Solving Equations
Solving equations involving geometric sequences often requires manipulating the known formula to find the unknown term number \(n\). This involves transforming the formula based on given values and performing algebraic operations:
1. Divide both sides by the first term to isolate the geometric part: \[3^{n-1} = \frac{118,098}{2} = 59,049\]2. Recognize the result as a power of the common ratio: \[59,049 = 3^{10}\]3. Equate the exponents, as the bases are the same, \(n-1 = 10\).4. Solve for \(n\): \[n = 10 + 1 = 11\].
This method helps us conclude that the 11th term of the sequence is 118,098.
- You start with \(a_n = a_1 \, r^{n-1}\).
- Substitute the known values, like \(118,098 = 2 \, (3^{n-1})\).
1. Divide both sides by the first term to isolate the geometric part: \[3^{n-1} = \frac{118,098}{2} = 59,049\]2. Recognize the result as a power of the common ratio: \[59,049 = 3^{10}\]3. Equate the exponents, as the bases are the same, \(n-1 = 10\).4. Solve for \(n\): \[n = 10 + 1 = 11\].
This method helps us conclude that the 11th term of the sequence is 118,098.
Other exercises in this chapter
Problem 36
Find the first four partial sums and the \(n\)th partial sum of the sequence \(a_{n} .\) \(a_{n}=\frac{1}{n+1}-\frac{1}{n+2}\)
View solution Problem 37
Find the term containing \(b^{8}\) in the expansion of \(\left(a+b^{2}\right)^{12}\)
View solution Problem 37
Which term of the arithmetic sequence \(1,4,7, \ldots\) is 88\(?\)
View solution Problem 37
Find the first four partial sums and the \(n\)th partial sum of the sequence \(a_{n} .\) \(a_{n}=\sqrt{n}-\sqrt{n+1}\)
View solution