Problem 14
Question
\text { The 8th term of } \frac{243}{32}, \frac{81}{16}, \frac{27}{8}, \frac{9}{4}, \ldots, \frac{4}{9}
Step-by-Step Solution
Verified Answer
The 8th term is \( \frac{4}{9} \).
1Step 1: Identify the Pattern
First, we need to identify if this is a geometric sequence. In a geometric sequence, each term is a constant multiple of the previous term. Here, check the ratio between successive terms: \( \frac{81/16}{243/32} = \frac{81 \times 32}{16 \times 243} = \frac{2}{3} \). Repeat this with the next interval: \( \frac{27/8}{81/16} = \frac{27 \times 16}{8 \times 81} = \frac{2}{3} \). This pattern confirms that the sequence is geometric with a common ratio of \( \frac{2}{3} \).
2Step 2: Find the First Term
The first term \( a_1 \) of the sequence is given as \( \frac{243}{32} \). This is our starting value for finding the subsequent terms of the sequence.
3Step 3: Use the Formula for the n-th Term
The formula to find the n-th term \( a_n \) of a geometric sequence is \( a_n = a_1 \times r^{n-1} \), where \( r \) is the common ratio. For this sequence, \( r = \frac{2}{3} \) and \( a_1 = \frac{243}{32} \). We want to find the 8th term (\( n = 8 \)).
4Step 4: Substitute Values into the Formula
Substitute the known values into the formula: \( a_8 = \frac{243}{32} \times \left( \frac{2}{3} \right)^{7} \).
5Step 5: Calculate the Common Ratio Power
Calculate \( \left( \frac{2}{3} \right)^{7} \): \( \left( \frac{2}{3} \right)^7 = \frac{2^7}{3^7} = \frac{128}{2187} \).
6Step 6: Find the Product of Terms
Multiply the first term by the 7th power of the common ratio: \( a_8 = \frac{243}{32} \times \frac{128}{2187} = \frac{243 \times 128}{32 \times 2187} \).
7Step 7: Simplify the Expression
Simplify \( \frac{243 \times 128}{32 \times 2187} \):- Cancel common factors between 243 and 2187 since both can be divided by 243:\( 2187 \div 243 = 9 \), so the expression becomes \( \frac{128}{32 \times 9} = \frac{128}{288} \).- Simplify \( \frac{128}{288} \) by dividing by 32:\( \frac{128 \div 32}{288 \div 32} = \frac{4}{9} \).
8Step 8: Conclusion
The 8th term of the given geometric sequence is \( \frac{4}{9} \).
Key Concepts
n-th term formulacommon ratiosequence simplification
n-th term formula
In a geometric sequence, finding any term in the sequence requires an understanding of the n-th term formula. This formula allows you to calculate the value of any term if the first term and the common ratio are known. The formula is represented as:
- \( a_n = a_1 \times r^{n-1} \)
- \( a_n \) is the n-th term, the one you wish to find.
- \( a_1 \) is the first term of the sequence.
- \( r \) is the common ratio, the factor by which we multiply each term to get the next one.
- \( n \) is the position of the term you want within the sequence.
common ratio
The common ratio in a geometric sequence is the constant factor you multiply by each term to get the next term. Identifying this ratio is crucial because it tells us how the sequence is growing or shrinking. Finding the common ratio involves dividing one term by the preceding term:
- For example, \( r = \frac{81/16}{243/32} = \frac{2}{3} \).
sequence simplification
Simplification in a geometric sequence refers to reducing complex expressions to their simplest form. This process is often necessary after performing multiplications as needed by the n-th term formula. After calculating using the formula, you’ll often find yourself with a fraction or expression that can be reduced:
- Start by looking for common factors in the numerator and denominator.
- Cancel these factors to simplify.
- In the provided example, \( \frac{243 \times 128}{32 \times 2187} \) simplifies by canceling 243 and then dividing both the numerator and denominator by 32, which results in \( \frac{4}{9} \).
Other exercises in this chapter
Problem 13
Solve \(A=P+P r t\) for \(P\), given that \(A=\$ 326, r=7 \%\), and \(t=9\) years.
View solution Problem 14
Find the sum of the odd whole numbers between 11 and 193 , inclusive. 9384
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Eric saved a nickel the first day of a month, a dime the second day, and 20 cents the third day and then continued to double his daily savings each day for 14 d
View solution Problem 14
The 7 th term of \(2,6,18,54, \ldots\) 1458
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