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Chapter 12: Problem 31

Determine the common ratio, the fifth term, and the \(n\) th term of the geometric sequence. $$144,-12,1,-\frac{1}{12}, \dots$$

Short Answer

Expert verified
The common ratio is \(-\frac{1}{12}\), the fifth term is \(\frac{1}{144}\), and the nth term is \(a_n = 144 \times \left(-\frac{1}{12}\right)^{n-1}\).

Step by step solution

01

Identify the Common Ratio

A geometric sequence progresses by a constant factor called the common ratio. To find the common ratio, divide any term by the previous term. Let's calculate the common ratio using the first two terms: \(-12 / 144\). This gives us the common ratio \(r = -\frac{1}{12}\).
02

Calculate the Fifth Term

We use the formula for the nth term of a geometric sequence: \(a_n = a_1 \times r^{n-1}\). To find the fifth term, substitute \(a_1 = 144\), \(r = -\frac{1}{12}\), and \(n = 5\): \(a_5 = 144 \times \left(-\frac{1}{12}\right)^{4}\). Simplifying, we have \(a_5 = 144 \times \left(\frac{1}{20736}\right) = \frac{1}{144}\).
03

Derive the General Formula for the Nth Term

The nth term of a geometric sequence is calculated by the formula \(a_n = a_1 \times r^{n-1}\). For this sequence, substitute \(a_1 = 144\) and \(r = -\frac{1}{12}\). Therefore, the nth term is \(a_n = 144 \times \left(-\frac{1}{12}\right)^{n-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Ratio
The common ratio in a geometric sequence is a crucial element that defines the sequence's progression. It is the constant factor by which each term is multiplied to get the next term. Finding the common ratio is simple: you divide one term by the previous one.
For example, in the sequence given in the exercise, where the first term is 144 and the second term is -12, the common ratio is calculated as follows: \[ r = \frac{-12}{144} = -\frac{1}{12} \]This value of \(-\frac{1}{12}\) remains the same throughout the sequence. It means each term is obtained by multiplying the previous term by \(-\frac{1}{12}\). A negative common ratio results in terms that alternate in sign, as seen in this sequence.
Nth Term Formula
The nth term formula of a geometric sequence helps in finding any term in the sequence. It's a potent tool because it allows you to calculate any term without listing all of the previous ones. The formula is:\[ a_n = a_1 \times r^{n-1} \] Where:
  • \(a_n\) is the term you're solving for.
  • \(a_1\) is the first term of the sequence.
  • \(r\) is the common ratio.
  • \(n\) is the term number.
This formula simplifies the process by providing a direct route to any term in the sequence, rather than needing to multiply repeatedly. For instance, when looking for the fifth term in the exercise, using the formula gives us:\[ a_5 = 144 \times \left(-\frac{1}{12}\right)^{4} = \frac{1}{144} \] This ensures accuracy and saves time by skipping multiple steps.
Geometric Progression
Geometric progression is all about a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Characteristics of a geometric progression include:
  • Consistent multiplication pattern.
  • Terms either steadily increase, decrease, or alternate in sign.
  • Easy to predict future terms using the nth term formula.
In the example from the exercise, the sequence \(144, -12, 1, -\frac{1}{12}, \ldots\) is a geometric progression because each term is obtained by multiplying the previous term by \(-\frac{1}{12}\). Understanding geometric progressions is key in various mathematical and real-world applications, such as calculating interest, analyzing population growth, and much more. Recognizing a sequence as geometric allows you to apply specialized formulas and insights specific to this type of sequence.

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Most popular questions from this chapter

Find the sum. $$\sum_{j=0}^{5} 7\left(\frac{3}{2}\right)^{j}$$

Find the partial sum \(S_{n}\) of the geometric sequence that satisfies the given conditions. $$a_{3}=28, \quad a_{6}=224, \quad n=6$$

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.]

Fibonacci's Rabbits Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair that becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the \(n\) th month? Show that the answer is \(F_{n},\) where \(F_{n}\) is the \(n\) th term of the Fibonacci sequence.

Use the Binomial Theorem to expand the expression. $$(1-x)^{5}$$
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