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Magazine Chapter 11: Problem 450
How many terms are in the finite geometric sequence \(-5, \quad-\frac{5}{3}, \quad-\frac{5}{9}, \ldots,-\frac{5}{59,049} ?\)
Short Answer
Expert verified
There are 11 terms in the sequence.
Step by step solution
01
Identify the geometric sequence components
In a geometric sequence, each term after the first is the product of the previous term and a constant called the common ratio (\(r\)). The given sequence is \(-5, -\frac{5}{3}, -\frac{5}{9}, \ldots,-\frac{5}{59,049}\). The first term \(a\) is \(-5\).
02
Find the common ratio
To find the common ratio \(r\), divide the second term by the first term: \[ r = \frac{-\frac{5}{3}}{-5} = \frac{1}{3}. \]
03
Set up the formula for the n-th term
The formula for the n-th term of a geometric sequence is \[ a_n = a \times r^{(n-1)}. \] Here, \(a = -5\) and \(r = \frac{1}{3}\).
04
Substitute values for the last term to find n
The last term provided is \(-\frac{5}{59,049}\). Set this equal to the formula for the n-th term and solve for \(n\): \[ -\frac{5}{59,049} = -5 \left(\frac{1}{3}\right)^{n-1}.\] Divide both sides by \(-5\): \[ \frac{1}{59,049} = \left(\frac{1}{3}\right)^{n-1}. \]
05
Express 59,049 as a power of 3
Rewrite 59,049 as powers of 3: \[ 59,049 = 3^{10}. \] Thus, the equation becomes \[ \left(\frac{1}{3}\right)^{10} = \left(\frac{1}{3}\right)^{n-1}. \]
06
Determine the number of terms
Since the bases and exponents are the same on both sides of the equation, equate the exponents: \[ 10 = n - 1. \] Solve for \(n\): \[ n = 11. \] Therefore, the number of terms in the sequence is 11.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Common Ratio
In a geometric sequence, the common ratio is a constant that each term is multiplied by to get the next term. It's essential because it defines the pattern of the sequence. For the sequence given in the exercise \(-5, -\frac{5}{3}, -\frac{5}{9}, \ldots,-\frac{5}{59,049}\), the first step is to identify this ratio.
- To find the common ratio, take any term (other than the first) and divide it by the previous term.
Geometric Sequence Formula
The formula for finding any term in a geometric sequence is \[ a_n = a \times r^{(n-1)}, \]where:
The first term is \(a = -5\), and the common ratio \(r\) is \(\frac{1}{3}\). Using the formula, you can compute any term's value in the sequence by substituting \(n\) for the desired term.
- \(a_n\) is the \(n\)-th term of the sequence,
- \(a\) is the first term,
- \(r\) is the common ratio,
- \(n\) is the term number.
The first term is \(a = -5\), and the common ratio \(r\) is \(\frac{1}{3}\). Using the formula, you can compute any term's value in the sequence by substituting \(n\) for the desired term.
Finite Sequence
A finite sequence is a sequence with a limited number of terms. Unlike infinite sequences, they do not go on forever. They have a clear end. In this exercise, we are dealing with a finite geometric sequence given by \(-5, -\frac{5}{3}, -\frac{5}{9}, \ldots,-\frac{5}{59,049}\).
Identifying the number of terms in a finite sequence is important to understand its extent and how many times a pattern or rule has been applied.
Since the sequence ends at \(-\frac{5}{59,049}\), we know it's finite and can determine its total number of terms.
Identifying the number of terms in a finite sequence is important to understand its extent and how many times a pattern or rule has been applied.
Since the sequence ends at \(-\frac{5}{59,049}\), we know it's finite and can determine its total number of terms.
Number of Terms
Calculating the number of terms in a finite geometric sequence is a crucial part of understanding the whole sequence. Using the formula for the \(n\)-th term, \[ a_n = a \times r^{(n-1)}, \]we set \(a_n\) equal to the last term of the sequence to solve for \(n\).
For the sequence, the last term given is \(-\frac{5}{59,049}\). Set up the equation:\[ -\frac{5}{59,049} = -5 \left(\frac{1}{3}\right)^{n-1}. \]Solving this equation involves rewriting \(59,049\) in terms of powers of \(3\) (which is \(3^{10}\)) and equating the exponents since the bases are the same.
Thus, \(n = 11\), showing there are a total of 11 terms in this progression. This demonstrates that our sequence does not extend indefinitely but concludes after 11 terms.
For the sequence, the last term given is \(-\frac{5}{59,049}\). Set up the equation:\[ -\frac{5}{59,049} = -5 \left(\frac{1}{3}\right)^{n-1}. \]Solving this equation involves rewriting \(59,049\) in terms of powers of \(3\) (which is \(3^{10}\)) and equating the exponents since the bases are the same.
Thus, \(n = 11\), showing there are a total of 11 terms in this progression. This demonstrates that our sequence does not extend indefinitely but concludes after 11 terms.
