/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Find the indicated term of the g... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the indicated term of the given geometric sequence. $$2 / 3,4 / 9,8 / 27, \ldots ; a_{8}$$

Short Answer

Expert verified
The 8th term \( a_8 \) is \( \frac{256}{6561} \).

Step by step solution

01

Identify the First Term

The first term of the geometric sequence is given as \( a_1 = \frac{2}{3} \).
02

Calculate the Common Ratio

The common ratio \( r \) is the factor you multiply by to go from one term to the next in a geometric sequence. To find \( r \), divide the second term by the first term: \( r = \frac{4/9}{2/3} = \frac{4 \times 3}{9 \times 2} = \frac{2}{3} \).
03

Write the Formula for the nth Term

The formula for the nth term \( a_n \) of a geometric sequence is \( a_n = a_1 \cdot r^{n-1} \).
04

Substitute Known Values into the Formula

To find the 8th term, substitute the known values into the formula: \( a_8 = \left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right)^{7} \).
05

Calculate the Power of the Common Ratio

Compute \( \left(\frac{2}{3}\right)^7 \). This can be calculated as \( \frac{2^7}{3^7} = \frac{128}{2187} \).
06

Multiply to Find the 8th Term

Now multiply the first term by the power of the common ratio: \( a_8 = \frac{2}{3} \cdot \frac{128}{2187} = \frac{256}{6561} \).
07

Simplify the Result

Simplify the fraction if possible. \( \frac{256}{6561} \) does not simplify further, so this is the final result.

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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 is a crucial part of a geometric sequence. It is the consistent factor that you multiply by to move from one term to the next in the sequence. To determine the common ratio, you divide any term by the one preceding it.

For instance, if you have a geometric sequence like the one provided: \(2/3, 4/9, 8/27, \ldots \), you can find the common ratio by dividing the second term \(4/9\) by the first term \(2/3\). Here's how it works:
  • Calculate \( r = \frac{4/9}{2/3} = \frac{4 \times 3}{9 \times 2} = \frac{2}{3} \).
The result \( \frac{2}{3} \) is the common ratio. This means each term in the sequence is \( \frac{2}{3} \) of the previous one, which confirms the geometric nature of the sequence.
nth Term Formula
In a geometric sequence, you often need to find a specific term without listing all preceding terms. This is where the nth term formula is your go-to tool. The formula is:
  • \( a_n = a_1 \cdot r^{n-1} \)
Here, \( a_n \) is the nth term, \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number you're solving for.

Let's apply it to our example to find the 8th term:
  • First term (\( a_1 \)) is \( \frac{2}{3} \).
  • Common ratio (\( r \)) is \( \frac{2}{3} \).
  • We want the 8th term (\( n = 8 \)).
Substitute these into the formula: \( a_8 = \left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right)^{7} \). The formula allows you to find the \( a_8 \) efficiently without computing each preceding term.
Exponentiation
Exponentiation empowers you to handle repeated multiplication of a number by itself, which is especially useful when applying the nth term formula for higher terms in a geometric sequence. When using exponentiation, you are basically raising the common ratio to a power.

In this case, to find \( a_8 \), we need \( \left(\frac{2}{3}\right)^7 \), which involves multiplying \( \frac{2}{3} \) by itself six more times after the initial existence, like this:
  • \( \left(\frac{2}{3}\right)^7 = \frac{2^7}{3^7} \)
Each base (2 for the numerator and 3 for the denominator) is raised to the 7th power:
  • \( 2^7 = 128 \)
  • \( 3^7 = 2187 \)
Therefore, \( \left(\frac{2}{3}\right)^7 = \frac{128}{2187} \). This step ensures you've applied the repeated multiplication necessary for solving the geometric sequence's particular term.
Fraction Simplification
Fraction simplification is the process of reducing a fraction to its simplest form, where the numerator and denominator no longer have any common divisors other than 1. This is done by dividing both the numerator and the denominator by their greatest common divisor (GCD).

In the final stages of finding the 8th term in our geometric sequence, you ended up with the fraction \( \frac{256}{6561} \). Always check if this result can be simplified. However, in this case, both 256 and 6561 have no common divisors other than 1, which means the fraction is already in its simplest form.

Knowing when a fraction can no longer be simplified helps in providing the most precise and understandable answer. Always attempt simplification as the last step, ensuring clarity and correctness in your mathematical solutions.

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Most popular questions from this chapter

If the common difference in an arithmetic sequence is twice the first term, show that $$\frac{S_{n}}{S_{m}}=\frac{n^{2}}{m^{2}}$$

(a) Complete the following table. $$\begin{array}{cccccccccc} \hline k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \left(\begin{array}{l}8 \\ k\end{array}\right) & & & & & & & & & & \\ \hline \end{array}$$ (b) Use the results in part (a) to verify that $$\left(\begin{array}{l} 8 \\ 0 \end{array}\right)+\left(\begin{array}{l} 8 \\ 1 \end{array}\right)+\left(\begin{array}{l} 8 \\ 2 \end{array}\right)+\cdots+\left(\begin{array}{l} 8 \\ 8 \end{array}\right)=2^{8}$$ (c) By taking \(a=b=1\) in the expansion of \((a+b)^{n},\) show that $$\left(\begin{array}{l} n \\ 0 \end{array}\right)+\left(\begin{array}{l} n \\ 1 \end{array}\right)+\left(\begin{array}{l} n \\ 2 \end{array}\right)+\cdots+\left(\begin{array}{l} n \\ n \end{array}\right)=2^{n}$$

(The Ricker model continued) Suppose that the initial size of a population is \(P_{0}=300\) and that the size of the population at the end of year \(t\) is given by $$ P_{t}=10 P_{t-1} e^{-P_{t-1} / 1000} \quad(t \geq 1) $$ (a) Use a graphing utility to compute the population sizes through the end of year \(t=5 .\) (As in Example 5, round the final answers to the nearest integers.) Then use the graphing utility to draw the population scatter plot for \(t=0,1, \ldots, 5 .\) Give a general description (in complete sentences) of how the size of the population changes over this period. (b) Use a graphing utility to compute the population sizes through the end of year \(t=20,\) and draw the scatter plot. To help you see the pattern, use the option on your graphing utility that connects adjacent dots in a scatter plot with line segments. Describe the population trend that emerges over the period \(t=15\) to \(t=20\) (c) For a clearer view of the long-term population behavior, use a graphing utility to compute the population sizes for the period \(t=25\) to \(t=35,\) and draw the scatter plot. As in part (b), use the option on your graphing utility that connects adjacent dots with line segments. Summarize (in complete sentences) what you observe.

Find the coefficient of \(B^{-10}\) in the expansion of \(\left[\left(B^{2} / 2\right)-\left(3 / B^{3}\right)\right]^{10}\).

Carry out the indicated expansions. $$\left(\frac{1}{2}-2 a\right)^{6}$$

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