Problem 18
Question
Find the nth term of the arithmetic sequence \(\left\\{a_{n}\right\\}\) whose first term \(a_{1}\) and common difference d are given. What is the 51st term? $$ a_{1}=-2 ; \quad d=4 $$
Step-by-Step Solution
Verified Answer
The 51st term is 198.
1Step 1: Understand the Formula
The nth term of an arithmetic sequence can be found using the formula: \[ a_n = a_1 + (n-1) \times d \]where \(a_n\) is the nth term, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number.
2Step 2: Identify Given Values
From the problem, the first term \(a_1\) is given as -2 and the common difference \(d\) is 4. We need to find the 51st term, so \(n = 51\).
3Step 3: Substitute Values into the Formula
Substitute \(a_1 = -2\), \(d = 4\), and \(n = 51\) into the formula: \[ a_n = -2 + (51-1) \times 4 \]
4Step 4: Simplify the Expression
First, compute the term inside the parentheses: \[ 51-1 = 50 \]Now, multiply this result by the common difference \(d\): \[ 50 \times 4 = 200 \]Finally, add this result to the first term: \[ a_{51} = -2 + 200 \]\[ a_{51} = 198 \]
Key Concepts
nth termcommon differencesequence formula
nth term
One of the core concepts in understanding arithmetic sequences is the idea of the 'nth term'. The nth term is a way to find any term in a sequence without listing all the terms. You can think of it as a shortcut.
In our example exercise, we wanted to find the 51st term of the sequence. To do this, we use the formula for the nth term of an arithmetic sequence. The formula helps us jump directly to the term we need, without calculating each intermediate term. Simply put, if you know the first term and the common difference, you can find any term in the sequence by plugging the values into the formula.
In our example exercise, we wanted to find the 51st term of the sequence. To do this, we use the formula for the nth term of an arithmetic sequence. The formula helps us jump directly to the term we need, without calculating each intermediate term. Simply put, if you know the first term and the common difference, you can find any term in the sequence by plugging the values into the formula.
common difference
The 'common difference' is another key concept in arithmetic sequences. It’s the amount that each term in the sequence increases or decreases from the previous term.
For example, in the exercise provided, the common difference (\(d\)) is 4. This means that every term in the sequence is 4 more than the previous one.
Recognizing the common difference allows us to understand the pattern of the sequence. It's what makes the sequence 'arithmetic'.
If the common difference is positive, like in our example, the terms get bigger. If it's negative, the terms get smaller. For zero common difference, all terms are the same.
For example, in the exercise provided, the common difference (\(d\)) is 4. This means that every term in the sequence is 4 more than the previous one.
Recognizing the common difference allows us to understand the pattern of the sequence. It's what makes the sequence 'arithmetic'.
If the common difference is positive, like in our example, the terms get bigger. If it's negative, the terms get smaller. For zero common difference, all terms are the same.
sequence formula
The 'sequence formula' is the general formula used to find the nth term of an arithmetic sequence. In our exercise, the formula is: \[ a_n = a_1 + (n-1) \times d \]
This formula tells you that to find any term (\(a_n\)), you start with the first term (\(a_1\)), then add the common difference (\(d\)) multiplied by one less than the term number you want (\(n-1\)).
Using this formula, we calculated the 51st term as follows:
Finally, we found \(a_{51} = 198 \). This method can be used for any term in an arithmetic sequence by simply adjusting the values in the formula.
This formula tells you that to find any term (\(a_n\)), you start with the first term (\(a_1\)), then add the common difference (\(d\)) multiplied by one less than the term number you want (\(n-1\)).
Using this formula, we calculated the 51st term as follows:
- Substitute given values: \(a_1 = -2\), \(n = 51\), and \(d = 4\)
- Compute inside parentheses: \(51-1 = 50\)
- Multiply by common difference: \(50 \times 4 = 200\)
- Add to the first term: \(-2 + 200 = 198\)
Finally, we found \(a_{51} = 198 \). This method can be used for any term in an arithmetic sequence by simply adjusting the values in the formula.
Other exercises in this chapter
Problem 18
Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{u_{n}\right\\}=\left\\{\frac{2^{n}}{3^{n-1}}\right\\}
View solution Problem 18
List the first five terms of each sequence. \(\left\\{b_{n}\right\\}=\left\\{\frac{2 n+1}{2 n}\right\\}\)
View solution Problem 19
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ n^{2}+n \text { is divisible by } 2 $$
View solution Problem 19
Expand each expression using the Binomial Theorem. $$ (x-2)^{6} $$
View solution