Problem 42
Question
Find the kth partial sum of the arithmetic sequence \(\left\\{a_{n}\right\\}\) with common difference d. $$k=8, a_{1}=\frac{2}{3}, d=-\frac{4}{3}$$
Step-by-Step Solution
Verified Answer
Answer: The 8th partial sum of the arithmetic sequence is -32.
1Step 1: Find the formula for the nth term of the arithmetic sequence
Given the arithmetic sequence \(\left\\{a_{n}\right\\}\) with common difference \(d\) and first term \(a_1\), the formula for the nth term can be expressed as:
$$a_n = a_1 + (n-1)d$$
Here, we are given that \(a_1 = \frac{2}{3}\) and \(d = -\frac{4}{3}\).
2Step 2: Find the 8th term of the sequence
Using the arithmetic sequence formula (\(a_n = a_1 + (n-1)d\)), we find the 8th term (\(a_8\)) as follows:
$$a_8 = \frac{2}{3} + (8-1)\left(-\frac{4}{3}\right)$$
$$a_8 = \frac{2}{3} + 7\left(-\frac{4}{3}\right)$$
$$a_8 = \frac{2}{3} - \frac{28}{3}$$
$$a_8 = -\frac{26}{3}$$
3Step 3: Find the 8th partial sum
Now, we will find the 8th partial sum \(S_8\) using the formula we mentioned earlier:
$$S_n = n\frac{a_1 + a_n}{2}$$
Plug in the values we've computed for \(a_1\), \(a_8\), and \(k=8\) to get \(S_8\):
$$S_8 = 8\frac{\frac{2}{3} - \frac{26}{3}}{2}$$
$$S_8 = 8\frac{-\frac{24}{3}}{2}$$
$$S_8 = 8\left(-\frac{8}{2}\right)$$
$$S_8 = 8(-4)$$
$$S_8 = -32$$
The 8th partial sum of the given arithmetic sequence is -32.
Key Concepts
Partial Sum of an Arithmetic SequenceCommon DifferenceNth Term of an Arithmetic Sequence
Partial Sum of an Arithmetic Sequence
The concept of the partial sum of an arithmetic sequence is integral to understanding how arithmetic sequences work and is particularly important when adding up a certain number of terms. An arithmetic sequence, as you may know, is a sequence of numbers in which each term after the first is obtained by adding a constant value, known as the common difference, to the previous term.
By definition, the partial sum, denoted as \( S_k \) when referring to the sum of the first \( k \) terms, can be found using the formula:
\[S_k = \frac{k}{2}(2a_1 + (k - 1)d) \]Or, alternatively, as the sum of the first and the \( k \)th term multiplied by \( k \) divided by two:
\[S_k = \frac{k}{2}(a_1 + a_k) \]In this formula, \( a_1 \) represents the first term of the sequence, \( d \) is the common difference, and \( a_k \) is the \( k \)th term. This result stems from the fact that an arithmetic sequence has a linear nature which implies that the average of the first and \( k \)th terms, multiplied by the number of terms \( k \) effectively gives us the sum of all \( k \) terms.
As seen in our original problem, you'd replace \( a_1 \) with \( \frac{2}{3} \) and \( d \) with \( -\frac{4}{3} \) to calculate the partial sum for the first 8 terms. Since arithmetic sequences are linear patterns, this calculation provides a swift way to aggregate multiple terms.
By definition, the partial sum, denoted as \( S_k \) when referring to the sum of the first \( k \) terms, can be found using the formula:
\[S_k = \frac{k}{2}(2a_1 + (k - 1)d) \]Or, alternatively, as the sum of the first and the \( k \)th term multiplied by \( k \) divided by two:
\[S_k = \frac{k}{2}(a_1 + a_k) \]In this formula, \( a_1 \) represents the first term of the sequence, \( d \) is the common difference, and \( a_k \) is the \( k \)th term. This result stems from the fact that an arithmetic sequence has a linear nature which implies that the average of the first and \( k \)th terms, multiplied by the number of terms \( k \) effectively gives us the sum of all \( k \) terms.
As seen in our original problem, you'd replace \( a_1 \) with \( \frac{2}{3} \) and \( d \) with \( -\frac{4}{3} \) to calculate the partial sum for the first 8 terms. Since arithmetic sequences are linear patterns, this calculation provides a swift way to aggregate multiple terms.
Common Difference
The 'common difference' is a term that frequently crops up when dealing with arithmetic sequences. It's the consistent interval or gap between any two consecutive terms in the sequence. For instance, in the sequence 2, 5, 8, 11... the common difference is 3, as each term increases by 3 from the one before it.
In the formula for the \( n \)th term, \( a_n = a_1 + (n-1)d \), the common difference is represented by \( d \). It's the engine that drives the sequence, moving it forward step by precise step. In our textbook problem, the common difference \( d \) is \( -\frac{4}{3} \) which indicates that each term is subtracted by \( \frac{4}{3} \) from its preceding term, revealing that the sequence is decreasing.
Understanding the common difference is key to predicting the behavior of the arithmetic sequence, whether it's increasing, decreasing, or constant (when the common difference is zero). It also plays a crucial role in computing both the \( n \)th term and the partial sum of the sequence.
In the formula for the \( n \)th term, \( a_n = a_1 + (n-1)d \), the common difference is represented by \( d \). It's the engine that drives the sequence, moving it forward step by precise step. In our textbook problem, the common difference \( d \) is \( -\frac{4}{3} \) which indicates that each term is subtracted by \( \frac{4}{3} \) from its preceding term, revealing that the sequence is decreasing.
Understanding the common difference is key to predicting the behavior of the arithmetic sequence, whether it's increasing, decreasing, or constant (when the common difference is zero). It also plays a crucial role in computing both the \( n \)th term and the partial sum of the sequence.
Nth Term of an Arithmetic Sequence
Identifying the \( n \)th term is a routine task when dealing with arithmetic sequences. It refers to finding any term's value at a particular position denoted by \( n \) within a sequence. For any arithmetic sequence, the \( n \)th term can be determined using a specific formula:
\[a_n = a_1 + (n-1)d \]Here, we take the first term, \( a_1 \), and add the common difference \( d \) multiplied by \( n-1 \) (because we've already included the first term). In plain language, you're adding the common difference to the first term as many times as needed to reach the \( n \)th term.
For example, in our exercise, to find the 8th term (when \( n = 8 \) ) we took the first term \( \frac{2}{3} \) and added 7 times the common difference \( -\frac{4}{3} \) to it, resulting in \( -\frac{26}{3} \).
This concept is critical because knowing the \( n \)th term allows you to understand the structure of the sequence and calculate other properties, such as partial sums or whether a specific number is part of the sequence.
\[a_n = a_1 + (n-1)d \]Here, we take the first term, \( a_1 \), and add the common difference \( d \) multiplied by \( n-1 \) (because we've already included the first term). In plain language, you're adding the common difference to the first term as many times as needed to reach the \( n \)th term.
For example, in our exercise, to find the 8th term (when \( n = 8 \) ) we took the first term \( \frac{2}{3} \) and added 7 times the common difference \( -\frac{4}{3} \) to it, resulting in \( -\frac{26}{3} \).
This concept is critical because knowing the \( n \)th term allows you to understand the structure of the sequence and calculate other properties, such as partial sums or whether a specific number is part of the sequence.
Other exercises in this chapter
Problem 42
Expand and simplify the rule of the function \(g .\) $$f(x)=2 x^{3}-x+4 \text { and } g(x)=f(x+3)$$
View solution Problem 42
Find the sum. $$\sum_{k=1}^{5}(-k)$$
View solution Problem 43
In Exercises \(43-48,\) find the sum. $$\sum_{n=1}^{7} 2^{n}$$
View solution Problem 43
Find the indicated term of the expansion of the given expression. third, \((x+y)^{5}\)
View solution