Problem 33
Question
In Exercises \(31-38,\) use the given information about the geometric sequence \(\left\\{a_{n}\right\\}\) to find as and a formula for \(a_{n}\). $$a_{1}=1 / 2, a_{2}=5$$
Step-by-Step Solution
Verified Answer
Answer: The general formula for the given sequence (an) is \(a_n = \frac{1}{2} \cdot 10^{(n-1)}\), and the value of 'as' can be found using the formula \(as = \frac{1}{2} \cdot 10^{(s-1)}\).
1Step 1: Find the common ratio
To find the common ratio, divide \(a_2\) by \(a_1\). The common ratio is denoted by 'r'.
Let's find the common ratio (r):
$$r = \frac{a_{2}}{a_{1}} = \frac{5}{\frac{1}{2}} = 10$$
2Step 2: Find the formula for \(a_n\)
The formula for a geometric sequence is given by:$$a_n = a_1 \cdot r^{(n-1)}$$
Now we can plug in our value for \(a_1\) and \(r\) to get the formula:
$$a_n = \left(\frac{1}{2}\right) \cdot 10^{(n-1)}$$
3Step 3: Calculate the value of as
Now that we have the formula for \(a_n\), we can calculate the value of 'as' by using the formula and replacing n with s, which stands for the term's number in the sequence:
$$as = \left(\frac{1}{2}\right) \cdot 10^{(s-1)}$$
So, the given sequence can be represented by the formula:$$a_n = \left(\frac{1}{2}\right) \cdot 10^{(n-1)}$$
and the value of 'as' can be found using the formula:$$as = \left(\frac{1}{2}\right) \cdot 10^{(s-1)}$$
Key Concepts
Common RatioSequence FormulaPrecalculus
Common Ratio
The common ratio in a geometric sequence is a crucial factor that determines the pattern of the sequence. It is the factor by which each term is multiplied to obtain the next term. For example, if each term in the sequence is double the previous term, the common ratio is 2. In our exercise, the sequence begins with its first term, \( a_1 = \frac{1}{2} \), followed by the second term, \( a_2 = 5 \).
To calculate the common ratio \( r \), we divide the second term by the first term:
\[ r = \frac{a_{2}}{a_{1}} = \frac{5}{\frac{1}{2}} = 10 \]
Thus, \( r = 10 \) signifies that to move from one term to the next in this sequence, we multiply by 10. This multiplication factor remains consistent throughout the entire sequence, which is a defining feature of geometric sequences.
To calculate the common ratio \( r \), we divide the second term by the first term:
\[ r = \frac{a_{2}}{a_{1}} = \frac{5}{\frac{1}{2}} = 10 \]
Thus, \( r = 10 \) signifies that to move from one term to the next in this sequence, we multiply by 10. This multiplication factor remains consistent throughout the entire sequence, which is a defining feature of geometric sequences.
Sequence Formula
A geometric sequence can be described using a concise formula that allows us to find any term in the sequence without needing to know all of the preceding terms. The general formula for the \( n \)-th term of a geometric sequence is given by:
\[ a_n = a_1 \cdot r^{(n-1)} \]
Where \( a_n \) is the \( n \)-th term, \( a_1 \) is the first term, and \( r \) is the common ratio. Using the common ratio we found earlier (\( r = 10 \)) and the first term \( a_1 = \frac{1}{2} \), we can write the specific sequence formula for our problem:
\[ a_n = \left(\frac{1}{2}\right) \cdot 10^{(n-1)} \]
This equation empowers students to easily calculate any term in the sequence by simply plugging in the term's position, \( n \). It eliminates the need for repetitive multiplication and showcases the beauty of mathematical expressions in capturing patterns succinctly.
\[ a_n = a_1 \cdot r^{(n-1)} \]
Where \( a_n \) is the \( n \)-th term, \( a_1 \) is the first term, and \( r \) is the common ratio. Using the common ratio we found earlier (\( r = 10 \)) and the first term \( a_1 = \frac{1}{2} \), we can write the specific sequence formula for our problem:
\[ a_n = \left(\frac{1}{2}\right) \cdot 10^{(n-1)} \]
This equation empowers students to easily calculate any term in the sequence by simply plugging in the term's position, \( n \). It eliminates the need for repetitive multiplication and showcases the beauty of mathematical expressions in capturing patterns succinctly.
Precalculus
Precalculus is a mathematical course designed to prepare students for calculus. It covers a variety of topics including functions, algebra, trigonometry, and also sequences and series. Understanding geometric sequences, which form a part of series, is one of the foundational precalculus topics. These sequences are not only a stepping stone for more advanced concepts in calculus but also frequently appear in real-world scenarios like banking, physics, and computer science.
In precalculus, it's essential to grasp the concept of a sequence's behavior, such as how quickly it grows or shrinks, and to use formulas efficiently to find terms in the sequence. Exercises like the one discussed here build the analytical skills needed to tackle more complex functions and change, which are the core of calculus study. With a solid precalculus foundation, students can progress to understand the infinitesimal and the infinite — key themes of calculus.
In precalculus, it's essential to grasp the concept of a sequence's behavior, such as how quickly it grows or shrinks, and to use formulas efficiently to find terms in the sequence. Exercises like the one discussed here build the analytical skills needed to tackle more complex functions and change, which are the core of calculus study. With a solid precalculus foundation, students can progress to understand the infinitesimal and the infinite — key themes of calculus.
Other exercises in this chapter
Problem 32
Find the first five terms of the recursively defined sequence. $$a_{1}=1, a_{2}=3, \text { and } a_{n}=2 a_{n-1}+3 a_{n-2} \text { for } n \geq 3$$
View solution Problem 32
The first term \(a_{1}\) and the common difference d of an arithmetic sequence are given. Find the fifth term and the formula for the nth term. $$a_{1}=-.1, d=-
View solution Problem 33
Expand and (where possible) simplify the expression. $$\left(x^{-3}+x\right)^{4}$$
View solution Problem 33
Use the given information about the arithmetic sequence with common difference d to find a and a formula for \(a_{n}\). $$a_{4}=12, d=2$$
View solution