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Chapter 12: Problem 35

Determine the common ratio, the fifth term, and the \(n\) th term of the geometric sequence. $$1, s^{2 / 7}, s^{4 / 7}, s^{6 / 7}, \dots$$

Short Answer

Expert verified
Common ratio is \(s^{2/7}\); fifth term is \(s^{8/7}\); \(n\)-th term is \(s^{2(n-1)/7}\).

Step by step solution

01

Identify the Common Ratio

In a geometric sequence, each term is obtained by multiplying the previous term by a constant, called the common ratio. To find it, divide the second term by the first term.The first term is 1, and the second term is \(s^{2/7}\). Thus, the common ratio \(r\) is given by:\[r = \frac{s^{2/7}}{1} = s^{2/7}\]
02

Find the Fifth Term

The general formula for the \(n\)-th term of a geometric sequence is \(a_n = a_1 \cdot r^{n-1}\), where \(a_1\) is the first term and \(r\) is the common ratio.Plug in the known values to find the fifth term:\[a_5 = 1 \cdot (s^{2/7})^{5-1} = s^{8/7}\]
03

Derive the Formula for the nth Term

Using the formula for the \(n\)-th term \(a_n = a_1 \cdot r^{n-1}\), where \(a_1 = 1\) and \(r = s^{2/7}\), we can deduce:\[a_n = 1 \cdot (s^{2/7})^{n-1} = s^{2(n-1)/7}\]This gives us the expression for the \(n\)-th term of the sequence.

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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 key element that defines the relationship between its consecutive terms. The common ratio remains constant throughout the sequence. Understanding the common ratio is essential for fully grasping how a geometric sequence progresses. To determine it, we divide the second term by the first term.

Let's consider the sequence provided: the first term is 1 and the second term is \(s^{2/7}\). By dividing these, we find:
  • Common Ratio \(r\) = \(\frac{s^{2/7}}{1} = s^{2/7}\)
This common ratio of \(s^{2/7}\) tells us that each term in the sequence is created by multiplying the previous term by \(s^{2/7}\). This multiplication pattern is what makes the sequence a geometric one.
N-th Term
The nth term of a geometric sequence is critical because it provides a formulaic way to find any specific term in the sequence without writing out all preceding terms. The nth term formula answers the question: "What is the value of the term at position \(n\)?"

For our sequence, the formula is derived from the first term and the common ratio:
  • The first term \(a_1\) is 1.
  • The common ratio \(r\) is \(s^{2/7}\).
  • The formula for the nth term is \(a_n = a_1 \cdot r^{n-1}\).
Substituting the known values yields:
  • \(a_n = 1 \cdot (s^{2/7})^{n-1} = s^{2(n-1)/7}\)
This formula provides a shortcut to any term in the sequence by plugging in the value of \(n\), allowing us to bypass the step-by-step calculations.
Geometric Sequence Formula
The geometric sequence formula is an invaluable tool for analyzing sequences. It serves as a universal pattern that allows us to compute any term without sequential multiplication. This formula is:
  • \(a_n = a_1 \cdot r^{n-1}\)
In this equation:
  • \(a_n\) represents the nth term.
  • \(a_1\) is the first term of the sequence.
  • \(r\) is the common ratio.
  • \(n\) is the term number in the sequence.
Using this formula, one can promptly determine any term's value by simply knowing the position \(n\), first term, and common ratio, illustrating the power of mathematical generalization.
Sequence Terms
Understanding the terms in a geometric sequence involves recognizing that each term is generated by a particular relationship: the multiplication of the previous term by the common ratio. This multiplication is repeated for each term, building up the sequence.

For the given sequence \(1, s^{2/7}, s^{4/7}, s^{6/7}, \ldots\), we start with an initial term:
  • The first term \(a_1 = 1\)
  • Subsequent terms: each is \(s^{2/7}\) times the previous one, such as \(s^{2/7}, s^{4/7}, s^{6/7}\), and so forth.
Understanding sequence terms helps in visualizing the pattern and form mathematical predictions or calculations for future terms using the common ratio and any starting point. By recognizing this structured growth, listeners and learners can apply the concept of geometric progress to various mathematical and real-world contexts.

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Most popular questions from this chapter

Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. $$1+\frac{3}{2}+\left(\frac{3}{2}\right)^{2}+\left(\frac{3}{2}\right)^{3}+\cdots$$

Amortizing a Mortgage When they bought their house, John and Mary took out a \(\$ 90,000\) mortgage at \(9 \%\) interest, repayable monthly over 30 years. Their payment is \(\$ 724.17\) per month (check this, using the formula in the text). The bank gave them an amortization schedule, which is a table showing how much of each payment is interest, how much goes toward the principal, and the remaining principal after each payment. The table below shows the first few entries in the amortization schedule. $$\begin{array}{|c|c|c|c|c|}\hline \begin{array}{c}\text { Payment } \\\\\text { number }\end{array} & \begin{array}{c}\text { Total } \\\\\text { payment }\end{array} & \begin{array}{c}\text { Interest } \\\\\text { payment }\end{array} & \begin{array}{c}\text { Principal } \\\\\text { payment }\end{array} & \begin{array}{c}\text { Remaining } \\\\\text { principal }\end{array} \\\\\hline 1 & 724.17 & 675.00 & 49.17 & 89,950.83 \\\2 & 724.17 & 674.63 & 49.54 & 89,901.29 \\\3 & 724.17 & 674.26 & 49.91 & 89,851.38 \\\4 & 724.17 & 673.89 & 50.28 & 89,801.10 \\\\\hline\end{array}$$ After 10 years they have made 120 payments and are wondering how much they still owe, but they have lost the amortization schedule. (a) How much do John and Mary still owe on their mortgage? [Hint: The remaining balance is the present value of the \(240 \text { remaining payments. }]\) (b) How much of their next payment is interest, and how much goes toward the principal? [Hint: since \(9 \% \div 12=0.75 \%,\) they must pay \(0.75 \%\) of the remaining principal in interest each month.]

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Find the first three terms in the expansion of \(\left(x+\frac{1}{x}\right)^{40}\).

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