Problem 9
Question
Find the \(n\)th term, the fifth term, and the eighth term of the geometric sequence. $$4,-6,9,-13.5, \ldots$$
Step-by-Step Solution
Verified Answer
The nth term is \(4 \cdot \left(-\frac{3}{2}\right)^{n-1}\), the fifth term is 20.25, and the eighth term is -68.4375.
1Step 1: Identify the First Term and Common Ratio
The first term of the sequence, denoted as \(a\), is the first number in the sequence, which is \(4\). The common ratio \(r\) is obtained by dividing the second term by the first term: \(-6 \div 4 = -\frac{3}{2}\). Verify this ratio with subsequent terms like \(9 \div (-6) = -\frac{3}{2}\) and \(-13.5 \div 9 = -\frac{3}{2}\).
2Step 2: Write the General Formula for the nth Term
For a geometric sequence, the general term \( a_n \) is given by \( a_n = a \cdot r^{n-1} \). Substitute \(a = 4\) and \(r = -\frac{3}{2}\) into this formula to get \( a_n = 4 \cdot \left(-\frac{3}{2}\right)^{n-1} \).
3Step 3: Calculate the Fifth Term
To find the fifth term \(a_5\), substitute \(n = 5\) into the formula: \( a_5 = 4 \cdot \left(-\frac{3}{2}\right)^{4} \). Calculate \( \left(-\frac{3}{2}\right)^4 = \left(\frac{9}{4}\right)^2 = \frac{81}{16} \). Therefore, \( a_5 = 4 \times \frac{81}{16} = \frac{81}{4} = 20.25 \).
4Step 4: Calculate the Eighth Term
To find the eighth term \(a_8\), substitute \(n = 8\): \( a_8 = 4 \cdot \left(-\frac{3}{2}\right)^{7} \). Calculate \( \left(-\frac{3}{2}\right)^7 = -\left(\frac{2187}{128}\right) \). Thus, \( a_8 = 4 \times \left(-\frac{2187}{128}\right) = -\frac{8748}{128} = -68.4375 \).
Key Concepts
nth term calculationcommon ratiogeneral term formulasequence analysis
nth term calculation
In a geometric sequence, calculating the nth term is a fundamental step. Understanding this helps you pinpoint any term in the sequence without listing all previous terms. The formula for finding the nth term, denoted as \( a_n \), depends on two key components: the first term \( a \) and the common ratio \( r \). The general formula used is:
- \( a_n = a \cdot r^{n-1} \)
common ratio
The common ratio is a crucial element in every geometric sequence. It indicates how each term is related to the previous one. You find it by dividing any term by the one before it.In our example sequence 4, -6, 9, -13.5, ..., the ratio \( r \) was found by calculating \( -6 \div 4 = -\frac{3}{2} \). It is essential to verify this ratio by dividing additional terms, such as \( 9 \div (-6) = -\frac{3}{2} \), ensuring consistency throughout the sequence.Key points in identifying the common ratio:
- Make sure calculations adhere to the order of terms.
- Confirm with multiple consecutive terms to avoid errors.
general term formula
The core tool for understanding geometric sequences is the general term formula. This formula, \( a_n = a \cdot r^{n-1} \), serves as a blueprint for constructing the sequence. It enables the prediction of any term's value without needing previous terms, provided the first term \( a \) and common ratio \( r \) are known. ### Model Example:Consider our sample sequence:
- First Term \( a = 4 \)
- Common Ratio \( r = -\frac{3}{2} \)
sequence analysis
Sequence analysis in geometric sequences involves understanding patterns and behavior beyond individual terms. By analyzing the terms, we can infer general properties of the sequence. When examining the sequence 4, -6, 9, -13.5, ..., start with both the common ratio \( -\frac{3}{2} \) and the pattern it creates:
- Notice the alternating signs of consecutive terms.
- Examine how the magnitude changes based on the ratio to understand growth or decay behavior.
Other exercises in this chapter
Problem 8
Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ (-1)^{1}+(-1)^{2}+(-1)^{3}+\cdots+(-1)^{n}=\frac{(-1)^{n}-1}{2} $$
View solution Problem 8
Exer. 3-10: Find the \(n\)th term, the fifth term, and the tenth term of the arithmetic sequence. $$ x-8, x-3, x+2, x+7, \ldots $$
View solution Problem 9
Exer. 9-10: Find the number of possible color arrangements for the 12 given disks, arranged in a row. 5 black, 3 red, 2 white, 2 green
View solution Problem 9
Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ 1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6} $$
View solution