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Chapter 9: Problem 23

Find the specified term of the geometric sequence that has the two given terms. $$a_{10} ; \quad a_{4}=4, \quad a_{1}=12$$

Short Answer

Expert verified
The tenth term \(a_{10}\) is \(\frac{4}{9}\).

Step by step solution

01

Understanding the Problem

We have a geometric sequence where the first term \(a_1 = 12\) and the fourth term \(a_4 = 4\). We need to find the tenth term \(a_{10}\).
02

Recall Geometric Sequence Formula

The general formula for the \(n\)-th term of a geometric sequence is \(a_n = a_1 imes r^{n-1}\), where \(r\) is the common ratio.
03

Set Up Equation for Known Terms

Using the formula, the fourth term is given by \(a_4 = a_1 imes r^{3}\). Substituting the known values, we have \(4 = 12 imes r^3\).
04

Solve for Common Ratio

Divide both sides by 12 to find \(r^3\): \[4 = 12 imes r^3 \r^3 = \frac{4}{12} = \frac{1}{3}\]Now solve for \(r\) by taking the cube root: \[r = \sqrt[3]{\frac{1}{3}}\].
05

Calculate the Tenth Term

Use the geometric sequence formula \(a_{10} = a_1 imes r^{9}\):\[a_{10} = 12 imes \left(\sqrt[3]{\frac{1}{3}}\right)^9\]Simplify the expression:\[a_{10} = 12 imes \left(\frac{1}{3}\right)^3\]\[a_{10} = 12 imes \frac{1}{27} = \frac{12}{27} = \frac{4}{9}\].
06

Final Answer

The tenth term of the sequence is \(\frac{4}{9}\).

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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 vital component. It is the constant factor between consecutive terms of the sequence.

To find the common ratio in a sequence, divide any term by the previous term. This ratio remains consistent throughout the sequence.
  • Given: First term (\(a_1 = 12\)) and fourth term (\(a_4 = 4\)).
  • Find the ratio: Since \(a_4 = a_1 \times r^3\), substitute the values to find \(r^3\).
  • Equation: \(4 = 12 \times r^3\), solving gives \(r^3 = \frac{1}{3}\).
Here, solving for \(r\) involves understanding that to move from the first to the fourth term, the sequence is multiplied by the same number three times.
Remember, the common ratio \(r\) can be a fraction or a whole number, but it stays constant.
Sequence Term Formula
The sequence term formula is the blueprint for finding any term in a geometric sequence. It tells us that each term \(a_n\) can be determined if we know the first term and the common ratio.

The formula is: \(a_n = a_1 \times r^{n-1}\). This means multiplying the first term by the common ratio raised to the power of one less than the term's position.
  • To find \(a_{10}\), plug into the formula: \(a_{10} = 12 \times r^{9}\).
  • We already determined \(r = \sqrt[3]{\frac{1}{3}}\), which simplifies the finding of \(a_{10}\).
  • Simplify: \(a_{10} = 12 \times \left(\frac{1}{3}\right)^3\).
Understanding this formula is crucial, as it connects any term to its sequence roots, utilizing both the initial term and the common ratio.
Cube Root Calculation
Calculating a cube root might seem tricky, but it is a manageable process.

When given \(r^3 = \frac{1}{3}\), finding \(r\) requires taking the cube root of both sides.
  • This step is essential: \(r = \sqrt[3]{\frac{1}{3}}\).
  • The cube root undoes cubing a number, distilling \(r^3\) back to \(r\).
A cube root answers the question, "What number, multiplied by itself twice, gives this product?" It ensures precision in sequences requiring fractional power computations.
Practice, and this calculation will become instant and intuitive.

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Most popular questions from this chapter

If \(f\) is a linear function, show that the sequence with \(n\) th term \(a_{n}=f(n)\) is an arithmetic sequence.

Express the sum in terms of summation notation. (Answers are not unique.) $$4+11+18+\cdots+466$$

Find the first four terms of the sequence of partial sums for the given sequence. $$\left\\{(-1)^{n}(1 / 2)^{n}\right\\}$$

(a) Find the number of negative integers greater than \(-500\) that are divisible by 33 (b) Find their sum.

For the given \(n\) th term \(a_{n}=f(n)\) of a sequence, use the graph of \(y=f(x)\) on the interval \([1,100]\) to verify that as \(n\) increases without bound, \(a_{n}\) approaches some real number \(c\) $$a_{n}=\left(1+\frac{1}{n}+\frac{1}{2 n^{2}}\right)^{n}$$
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