Problem 40
Question
The common ratio in a geometric sequence is \(\frac{3}{2},\) and the fifth term is \(1 .\) Find the first three terms.
Step-by-Step Solution
Verified Answer
The first three terms are \(\frac{16}{81}\), \(\frac{8}{27}\), and \(\frac{4}{9}\).
1Step 1: Understand the Problem
In a geometric sequence, each term is found by multiplying the previous term by a constant called the "common ratio." We are given the common ratio \( r = \frac{3}{2} \) and the fifth term \( a_5 = 1 \). We are tasked with finding the first three terms of this sequence.
2Step 2: Use the Formula for the nth Term of a Geometric Sequence
The formula for the nth term of a geometric sequence is \( a_n = a_1 \, r^{(n-1)} \), where \( a_n \) is the nth term, \( a_1 \) is the first term, and \( r \) is the common ratio. For the fifth term where \( n=5 \), the formula is \( a_5 = a_1 \, r^{4} \).
3Step 3: Solve for the First Term
We know \( a_5 = 1 \) and \( r = \frac{3}{2} \). Using the formula from Step 2, \( 1 = a_1 \, \left(\frac{3}{2}\right)^{4} \). The expression \( \left(\frac{3}{2}\right)^{4} \) simplifies to \( \frac{81}{16} \). Thus, \( 1 = a_1 \, \frac{81}{16} \). Solve for \( a_1 \) by multiplying both sides by \( \frac{16}{81} \): \[ a_1 = 1 \, \times \, \frac{16}{81} = \frac{16}{81} \].
4Step 4: Calculate the Second and Third Terms
Now that we have \( a_1 = \frac{16}{81} \), we can find the subsequent terms using the common ratio \( r = \frac{3}{2} \). For the second term, use \( a_2 = a_1 \, r = \frac{16}{81} \, \times \, \frac{3}{2} = \frac{48}{162} = \frac{8}{27} \). For the third term, use \( a_3 = a_2 \, r = \frac{8}{27} \, \times \, \frac{3}{2} = \frac{24}{54} = \frac{4}{9} \).
5Step 5: Verify the Calculations
Check each step for any calculation errors. Reconfirm that the sequence continues to yield the correct fifth term by calculating \( a_4 = a_3 \, r = \frac{4}{9} \, \times \, \frac{3}{2} = \frac{12}{18} = \frac{2}{3} \) and \( a_5 = a_4 \, r = \frac{2}{3} \, \times \, \frac{3}{2} = 1 \). Since calculations match the given fifth term, our solutions are consistent.
Key Concepts
Common RatioNth Term FormulaSequence Calculation
Common Ratio
In a geometric sequence, the common ratio is a crucial element. This is the factor by which you multiply each term in the sequence to get the next term. For example, if the common ratio is \( \frac{3}{2} \), like in our problem, each term is multiplied by \( \frac{3}{2} \) to find the subsequent term.
Understanding the common ratio helps simplify sequence calculations and predict how a sequence will progress.
Understanding the common ratio helps simplify sequence calculations and predict how a sequence will progress.
- If the common ratio is greater than 1, each term is larger than the one before.
- If smaller than 1, terms decrease in size.
Nth Term Formula
The nth term formula in a geometric sequence allows us to find any term when we know the first term \( a_1 \) and the common ratio \( r \). The formula is \( a_n = a_1 \, r^{(n-1)} \).
This formula means that if you need the nth term, you start with the first term and multiply it by the common ratio raised to the power of one less than the term number.
This formula means that if you need the nth term, you start with the first term and multiply it by the common ratio raised to the power of one less than the term number.
- For example, to find the fifth term \( a_5 \) when \( r = \frac{3}{2} \), you compute \( a_5 = a_1 \, \left(\frac{3}{2}\right)^4 \).
- Working through powers of ratios is key in locating particular points within the sequence.
Sequence Calculation
Sequence calculation involves both using the nth term formula and understanding the relation between terms via the common ratio. To determine any specific term, you start by solving for the first term, if needed.
For instance, given \( a_5 = 1 \) and the common ratio \( r = \frac{3}{2} \), we found the first term \( a_1 \) using the equation \( 1 = a_1 \, \left(\frac{3}{2}\right)^4 \). Solving gives \( a_1 = \frac{16}{81} \).
Once you have \( a_1 \), finding other terms like \( a_2 \) and \( a_3 \) becomes straightforward:
For instance, given \( a_5 = 1 \) and the common ratio \( r = \frac{3}{2} \), we found the first term \( a_1 \) using the equation \( 1 = a_1 \, \left(\frac{3}{2}\right)^4 \). Solving gives \( a_1 = \frac{16}{81} \).
Once you have \( a_1 \), finding other terms like \( a_2 \) and \( a_3 \) becomes straightforward:
- Calculate \( a_2 = a_1 \, r \)
- Proceed to \( a_3 = a_2 \, r \)
Other exercises in this chapter
Problem 39
Find the first four partial sums and the \(n\) th partial sum of the sequence \(a_{n^{*}}\) $$a_{n}=\sqrt{n}-\sqrt{n+1}$$
View solution Problem 40
Find the term containing \(y^{3}\) in the expansion of \((\sqrt{2}+y)^{12}\).
View solution Problem 40
The 20 th term of an arithmetic sequence is \(101,\) and the common difference is \(3 .\) Find a formula for the \(n\) th term.
View solution Problem 40
Find the first four partial sums and the \(n\) th partial sum of the sequence \(a_{n^{*}}\) \(a_{n}=\log \left(\frac{n}{n+1}\right) \quad[\text {Hint}:\) Use a
View solution