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Chapter 13: Problem 1

Find the second term in a geometric sequence in which the first term is \(9,\) the third term is \(4,\) and the common ratio is positive.

Short Answer

Expert verified
The second term in the sequence is 6.

Step by step solution

01

Understand the Geometric Sequence Formula

In a geometric sequence, each term is found by multiplying the previous term by a constant, known as the common ratio \(r\). The formula for the \(n\)-th term is: \(a_n = a_1 \times r^{(n-1)}\), where \(a_1\) is the first term.
02

Write Information for Given Terms

You are given first term \(a_1 = 9\), and the third term \(a_3 = 4\). Using the formula \(a_3 = a_1 \times r^2\), substitute the known values: \(4 = 9 \times r^2\).
03

Solve for Common Ratio \(r\)

Rearrange the equation \(4 = 9 \times r^2\) to solve for \(r\). Divide both sides by 9: \(r^2 = \frac{4}{9}\). Take the positive square root (since \(r\) is positive), so \(r = \frac{2}{3}\).
04

Find the Second Term

Substitute the known values into the formula for the second term: \(a_2 = a_1 \times r\). With \(a_1 = 9\) and \(r = \frac{2}{3}\), the second term \(a_2 = 9 \times \frac{2}{3} = 6\).
05

Review the Solution

Double-check each step to ensure calculations are correct. With \(a_1 = 9\), the calculated \(r = \frac{2}{3}\) is correct, and substituting to find \(a_2\), the solution for the second term is confirmed to be \(6\).

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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 the factor by which each term is multiplied to produce the next term. Understanding this ratio is key to exploring the behavior of the sequence. The formula for getting the common ratio, given the first term and any other term, is derived by dividing the latter term by the former one raised to the power of the term's positional difference. For example, if the first term is known and the third term is given:

  • If the first term is 9 and third term is 4, the equation for common ratio is given by rearranging formula for nth term as:
  • \[ a_3 = a_1 \times r^2 \rightarrow 4 = 9 \times r^2 \]
  • Solve for \( r^2 \) by dividing both sides by 9:
  • \[ r^2 = \frac{4}{9} \]
  • Then, take the square root to find \( r \):
  • \[ r = \frac{2}{3} \]
Notice here that you take only the positive root because the problem states that the common ratio is positive.

Each calculated term is a clear step using the fundamental axiom of geometric progression.
Nth Term Formula
The nth term formula is a blueprint for navigating through a geometric sequence. It allows you to compute any term without listing all preceding ones, significantly saving time and effort. The formula is straightforward: \[ a_n = a_1 \times r^{(n-1)} \]where:

  • \(a_n\) is the nth term you're trying to compute.
  • \(a_1\) is the first term in the sequence.
  • \(r\) is the common ratio.
  • \(n\) is the position of the term you need to find.
Understanding this formula is essential for anyone working with geometric sequences because it provides the method to find any term efficiently.

For instance, to find the second term using the given sequence:
  • \[ a_2 = a_1 \times r \]
  • Here, \( a_1 = 9 \) and \( r = \frac{2}{3} \), so:
  • \[ a_2 = 9 \times \frac{2}{3} = 6 \]
Once you decode this formula, working through sequences becomes systematic and clear.
Sequence Calculations
Calculating terms in a geometric sequence always involves a clear series of steps based on solid mathematical principles. With the math in hand, let鈥檚 explore how to proceed:

  • Start by identifying the given terms and what you need to find.
  • Next, use the nth term formula to find unknowns such as the common ratio or any specific term.
These formulae and values create a roadmap to solving any geometric sequence problem.

Step by step, you unravel the sequence:
  • Determine the first term given in the problem.
  • Use the additional term (like the third term given here) to find the common ratio.
  • Check your results against the original sequence requirements, adjusting calculations for accuracy.
Every calculation builds on the previous one, setting a foundation for each subsequent step. Remember, practice your calculations to develop confidence in using these sequences consistently.

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Most popular questions from this chapter

Carry out the indicated operations. Express your results in rectangular form for those cases in which the trigonometric functions are readily evaluated without tables or a calculator. $$\left(-\frac{1}{2}+\frac{1}{2} \sqrt{3} i\right)^{6}$$

Carry out the indicated expansions. $$(\sqrt{2}+\sqrt{3})^{5}$$

Rewrite the sums using sigma notation. $$x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}$$

Rewrite the sums using sigma notation. $$2-2^{2}+2^{3}-2^{4}+2^{5}$$

An important model that is used in population biology and ecology is the Ricker model. The Canadian biologist William E. Ricker introduced this model in his paper Stock and Recruitment (Journal of the Fisheries Research Board of Canada, \(11(1954) 559-623\) ). For information on Ricker himself, see the web page The general form of the Ricker model that we will use here is defined by a recursive sequence of the form \(P_{0}=\) initial population at time \(t=0\) \(P_{t}=r P_{t-1} e^{-k P_{t-1}} \quad\) for \(t \geq 1,\) and where \(r\) and \(k\) are positive constants (a) 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$$P_{t}=5 P_{t-1} e^{-P_{t-1} / 1000} \quad(t \geq 1)$$ 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 \(t=0,1, \ldots, 5 .\) Describe in complete sentences how the size of the population changes over this period. Does the population seem to be approaching an equilibrium level? (b) Using a graphing utility, compute the sizes of the population in part (a) through the end of the year \(t=20\) and draw the corresponding scatter plot. Note that the population seems to be approaching an equilibrium level of \(1609(\text { or } 1610)\) (c) Determine the equilibrium population algebraically by solving the following equation for \(P_{t-1} .\) For the final answer, use a calculator and round to the nearest integer. $$P_{t-1}=5 P_{t-1} e^{-P_{t-1} / 1000}$$
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