Problem 25
Question
In Exercises \(23-34,\) write a formula for the general term (the nth term) of each arithmetic sequence. Do not use a recursion formula. Then use the formula for \(a_{n}\) to find \(a_{20}\), the 20 th term of the sequence. $$7,3,-1,-5, \dots$$
Step-by-Step Solution
Verified Answer
The formula for the nth term of the sequence is \(a_{n}= 7 - 4(n - 1)\). The 20th term of the sequence is -69.
1Step 1: Identify the first term and the common difference
From the sequence, it can be seen that the first term, \(a_{1}\), is 7. The common difference, d, can be found by subtracting the first term from the second term, which gives us -4.
2Step 2: Write a formula for the nth term
Substituting \(a_{1}\) = 7 and d = -4 into the formula for an arithmetic sequence (\(a_{n}= a_{1}+ (n-1)d\)) gives the formula for the nth term: \(a_{n}= 7 + (n-1) * (-4)\). Hence, the nth term of the sequence can be written as \(a_{n} = 7 - 4(n - 1)\).
3Step 3: Find the 20th term of the sequence
Substitute n = 20 into the formula \(a_{n} = 7 - 4(n - 1)\) to find the 20th term. This gives \(a_{20} = 7 - 4*(20 - 1) = 7 - 76 = -69\).
Key Concepts
nth term formulacommon differencearithmetic series
nth term formula
Understanding the 'nth term formula' is crucial for anyone studying arithmetic sequences. It's the key to finding any term in the sequence, without the need to write out all the terms. The formula is a direct approach to zero in on a specified term, usually noted as \(a_n\).
To construct this formula, we need two pieces of information: the first term of the sequence (\(a_1\)) and the common difference (\(d\)). The 'nth term' is calculated using the expression \(a_n = a_1 + (n - 1)d\), where \(n\) is the term number we want to find. This compact formula is powerful as it gives us the flexibility to leap directly to any term in the sequence, effortlessly traversing the ordered list of numbers.
When solving for a specific term like \(a_{20}\), for instance, we simply substitute the values into the formula. If the first term is 7 and the common difference is -4, the formula would be \(a_n = 7 + (n - 1)(-4)\). Plug in \(n = 20\), and we swiftly arrive at \(a_{20} = -69\), bypassing the necessity of summing up individual terms.
To construct this formula, we need two pieces of information: the first term of the sequence (\(a_1\)) and the common difference (\(d\)). The 'nth term' is calculated using the expression \(a_n = a_1 + (n - 1)d\), where \(n\) is the term number we want to find. This compact formula is powerful as it gives us the flexibility to leap directly to any term in the sequence, effortlessly traversing the ordered list of numbers.
When solving for a specific term like \(a_{20}\), for instance, we simply substitute the values into the formula. If the first term is 7 and the common difference is -4, the formula would be \(a_n = 7 + (n - 1)(-4)\). Plug in \(n = 20\), and we swiftly arrive at \(a_{20} = -69\), bypassing the necessity of summing up individual terms.
common difference
The 'common difference' is the consistent interval between consecutive terms of an arithmetic sequence, typically represented by \(d\). It's what makes an arithmetic sequence linear, as every step from one term to the next is a fixed stride dictated by this value.
Finding the common difference involves a simple subtraction between any two successive terms. If our sequence begins with 7, 3, and then -1, the common difference is found by subtracting the first term from the second (\(3 - 7 = -4\)), solidifying the pattern of the sequence.
Once \(d\) is identified, it becomes the backbone for determining subsequent terms and unearths the dynamic of growth or decline within the sequence. A positive common difference signifies an increasing sequence, whereas a negative one, such as our example with \(d = -4\), depicts a decreasing sequence. This elementary concept is foundational in exploring the arithmetic sequence more deeply.
Finding the common difference involves a simple subtraction between any two successive terms. If our sequence begins with 7, 3, and then -1, the common difference is found by subtracting the first term from the second (\(3 - 7 = -4\)), solidifying the pattern of the sequence.
Once \(d\) is identified, it becomes the backbone for determining subsequent terms and unearths the dynamic of growth or decline within the sequence. A positive common difference signifies an increasing sequence, whereas a negative one, such as our example with \(d = -4\), depicts a decreasing sequence. This elementary concept is foundational in exploring the arithmetic sequence more deeply.
arithmetic series
Moving beyond individual terms, the 'arithmetic series' is the sum of terms within an arithmetic sequence. It's the result when numbers in a sequence are joined together through addition. Unlike the arithmetic sequence that lists numbers, the series focuses on the cumulative value.
The formula to determine the sum of the first \(n\) terms of an arithmetic series (\(S_n\)) is \(S_n = \frac{n}{2}(2a_1 + (n - 1)d)\), where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) stands for the number of terms to be added.
This calculation is incredibly useful for practical applications, such as financial planning or analyzing patterns. Think of it as doing the legwork to answer 'How much in total?', a common real-world question. The concept finds its strength in its ability to foresee the sum total without laboriously adding each term - a time-saver and insight provider, especially as sequences stretch to great lengths.
The formula to determine the sum of the first \(n\) terms of an arithmetic series (\(S_n\)) is \(S_n = \frac{n}{2}(2a_1 + (n - 1)d)\), where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) stands for the number of terms to be added.
This calculation is incredibly useful for practical applications, such as financial planning or analyzing patterns. Think of it as doing the legwork to answer 'How much in total?', a common real-world question. The concept finds its strength in its ability to foresee the sum total without laboriously adding each term - a time-saver and insight provider, especially as sequences stretch to great lengths.
Other exercises in this chapter
Problem 24
Evaluate each factorial expression. $$\frac{18 !}{16 !}$$
View solution Problem 25
Evaluate each expression. $$ \frac{_{7} C_{3}}{_{5} C_{4}}-\frac{98 !}{96 !} $$
View solution Problem 25
In Exercises 9-30, use the Binomial Theorem to expand each binomial and express the result in simplified form. $$(x-1)^{3}$$
View solution Problem 25
In Exercises 25-34, use mathematical induction to prove that each statement is true for every positive integer \(n.\) 2 is a factor of \(n^{2}-n\).
View solution