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Chapter 9: Problem 36

Write an expression for the \(n\) th term of the geometric sequence. Then find the indicated term. $$a_{1}=1, r=\sqrt{3}, n=8$$

Short Answer

Expert verified
The 8th term of the geometric sequence is \((\sqrt{3})^{7}\).

Step by step solution

01

Understand the Properties of a Geometric Sequence

In a geometric sequence, the terms follow a pattern where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the 'common ratio'. In this task, the first term \(a_1 = 1\) and the common ratio \(r = \sqrt{3}\). You are asked to find the 8th term, \(a_8\).
02

Apply the Formula for the nth Term

The formula for the nth term of a geometric sequence is \(a_n = a_1 \cdot r^{(n-1)}\). You need to fill in the given values into the formula to calculate the 8th term: Substituting \(a_1 = 1\), \(r = \sqrt{3}\), and \(n = 8\) into the formula, we get \(a_8 = 1 \cdot (\sqrt{3})^{(8-1)}\)
03

Simplify the Expression

Simplify the expression to get the 8th term of the sequence. The exponent calculation in the expression (\(\sqrt{3}^{7}\)) means the square root of 3 is raised to the power of 7. Carry out the calculation to get \(a_8 = (\sqrt{3})^{7}\).

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Key Concepts

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

nth term formula
In understanding geometric sequences, one of the most important aspects is the formula for finding the nth term. This formula provides a way to compute any term in the sequence if you know the first term and the common ratio. It's expressed as:
  • \(a_n = a_1 \cdot r^{(n-1)}\)
Here, \(a_n\) is the nth term of the sequence, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number you want to find.
To apply this, follow these steps:
  • First, identify \(a_1\) and \(r\) from the sequence.
  • Then, substitute these values along with \(n\) into the formula.
  • Finally, calculate the result by evaluating the expression.
This formula is a powerful tool in sequence analysis, simplifying what could otherwise be complex calculations.
common ratio
The common ratio in a geometric sequence is a crucial element. It's the constant factor between consecutive terms. For any sequence, once you know the first term and the common ratio, you can build the entire sequence. In our example, the common ratio is given as \(\sqrt{3}\).
Here鈥檚 how the common ratio functions in a sequence:
  • To find the next term, multiply the current term by the common ratio.
  • To find a previous term, divide a known term by the common ratio.
  • The common ratio remains constant throughout the sequence.
Choosing a non-zero and consistent common ratio keeps the sequence valid and predictable, allowing the sequence to grow or shrink in exponential steps.
exponents
Exponents play a critical role in the calculations involved in geometric sequences. They determine how many times the common ratio is multiplied by itself. In the nth term formula of a geometric sequence \(a_n = a_1 \cdot r^{(n-1)}\), exponents are what make the scale of term increase or decrease rapidly.
Understanding how to handle exponents is essential:
  • When an exponent is positive, it means repeated multiplication.
  • An exponent of 1 leaves the base number unchanged.
  • When dealing with square roots and fractional exponents, it implies a degree of root being applied.
In the given problem, calculating \((\sqrt{3})^{7}\) involves recognizing that \(\sqrt{3}\) can be rewritten as \(3^{0.5}\), making the calculation \(3^{0.5 \times 7} = 3^{3.5}\). Managing such expressions accurately ensures the correct computation of terms within the sequence.

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Most popular questions from this chapter

The table shows the average prices \(f(t)\) (in cents per kilowatt hour) of residential electricity in the United States from 2003 through 2010 . (Source: U.S. Energy Information Administration ).$$\begin{array}{|c|c|}\hline \text { Year } & \text { Abcrage Poids }(10) \\\\\hline 2003 & 8.72 \\\2004 & 8.95 \\\2005 & 9.45 \\\2006 & 10.40 \\\2007 & 10.65 \\\2008 & 11.26 \\\2009 & 11.51 \\\2010 & 11.58 \\\\\hline\end{array}$$.(a) Use the regression feature of a graphing utility to find a cubic model for the data. Let \(t\) represent the year, with \(t=3\) corresponding to 2003 (b) Use the graphing utility to plot the data and the model in the same viewing window. (c) You want to adjust the model so that \(t=3\) corresponds to 2008 rather than \(2003 .\) To do this, you shift the graph of \(f\) five units to the left to obtain \(g(t)=f(t+5) .\) Use binomial coefficients to write \(g(t)\) in standard form. (d) Use the graphing utility to graph \(g\) in the same viewing window as \(f\) (e) Use both models to predict the average price in \(2011 .\) Do you obtain the same answer? (f) Do your answers to part (e) seem reasonable? Explain. (g) What factors do you think may have contributed to the change in the average price?

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