Problem 30
Question
Find the indicated term in each arithmetic sequence. $$ \text { 70th term of } 2 \sqrt{5}, 4 \sqrt{5}, 6 \sqrt{5}, \ldots $$
Step-by-Step Solution
Verified Answer
The 70th term is \(140\sqrt{5}\).
1Step 1: Identify the first term and common difference
In an arithmetic sequence, we start by identifying the first term (\(a\)) and the common difference (\(d\)). Here, the first term (\(a\)) is \(2\sqrt{5}\) and the common difference (\(d\)) is \(4\sqrt{5} - 2\sqrt{5} = 2\sqrt{5}\).
2Step 2: Use the formula for the nth term of an arithmetic sequence
The formula for the nth term (\(a_n\)) of an arithmetic sequence is given by: \[a_n = a + (n-1) \times d\]. Here, we need to find the 70th term (\(a_{70}\)).
3Step 3: Substitute the known values into the formula
Substitute \(a = 2\sqrt{5}\), \(d = 2\sqrt{5}\), and \(n = 70\) into the formula: \[a_{70} = 2\sqrt{5} + (70-1) \times 2\sqrt{5}\].
4Step 4: Simplify the expression
Simplify the expression: \[a_{70} = 2\sqrt{5} + 69 \times 2\sqrt{5}\ = 2\sqrt{5} + 138\sqrt{5}\ = 140\sqrt{5}\].
Key Concepts
first termcommon differenceformula for nth term
first term
The first term of an arithmetic sequence is very important. It sets the starting point from which all other terms are derived. In our example, the first term is given as \(2\sqrt{5}\). This means the series begins with \(2\sqrt{5}\).
The first term is often denoted by the letter \(a\). So, if you're asked to find the first term, you're simply looking for the initial number in the sequence.
Example: In the sequence \(2\sqrt{5}, 4\sqrt{5}, 6\sqrt{5}, \ldots\), the first term is \(2\sqrt{5}\).
The first term is often denoted by the letter \(a\). So, if you're asked to find the first term, you're simply looking for the initial number in the sequence.
Example: In the sequence \(2\sqrt{5}, 4\sqrt{5}, 6\sqrt{5}, \ldots\), the first term is \(2\sqrt{5}\).
common difference
The common difference in an arithmetic sequence is the amount by which each term increases from the previous term. It is consistent, meaning it remains the same throughout the sequence.
In our example, we calculate the common difference \(d\) by subtracting the first term from the second term:
\[d = 4\sqrt{5} - 2\sqrt{5} = 2\sqrt{5}\]
Knowing the common difference helps us understand how the sequence progresses. If the common difference is positive, the terms will increase. If it's negative, the terms will decrease.
Example: For the sequence \(2\sqrt{5}, 4\sqrt{5}, 6\sqrt{5}, \ldots\), the common difference is \(2\sqrt{5}\). This means each term is \(2\sqrt{5}\) more than the one before it.
In our example, we calculate the common difference \(d\) by subtracting the first term from the second term:
\[d = 4\sqrt{5} - 2\sqrt{5} = 2\sqrt{5}\]
Knowing the common difference helps us understand how the sequence progresses. If the common difference is positive, the terms will increase. If it's negative, the terms will decrease.
Example: For the sequence \(2\sqrt{5}, 4\sqrt{5}, 6\sqrt{5}, \ldots\), the common difference is \(2\sqrt{5}\). This means each term is \(2\sqrt{5}\) more than the one before it.
formula for nth term
To find any term in an arithmetic sequence, we use a specific formula. This formula helps us find the value of the term at any given position in the sequence.
The formula for the nth term \(a_n\) is:
\[a_n = a + (n-1) \times d\]
Here, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the position of the term in the sequence that you want to find.
Example: To find the 70th term in our sequence, substitute \(a = 2\sqrt{5}\), \(d = 2\sqrt{5}\), and \(n = 70\) into the formula:
\[a_{70} = 2\sqrt{5} + (70-1) \times 2\sqrt{5}\] \[a_{70} = 2\sqrt{5} + 69 \times 2\sqrt{5} = 2\sqrt{5} + 138\sqrt{5} = 140\sqrt{5}\]
So, the 70th term of the sequence is \(140\sqrt{5}\). This formula is incredibly useful for quickly finding terms in long sequences without having to list out every single term.
The formula for the nth term \(a_n\) is:
\[a_n = a + (n-1) \times d\]
Here, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the position of the term in the sequence that you want to find.
Example: To find the 70th term in our sequence, substitute \(a = 2\sqrt{5}\), \(d = 2\sqrt{5}\), and \(n = 70\) into the formula:
\[a_{70} = 2\sqrt{5} + (70-1) \times 2\sqrt{5}\] \[a_{70} = 2\sqrt{5} + 69 \times 2\sqrt{5} = 2\sqrt{5} + 138\sqrt{5} = 140\sqrt{5}\]
So, the 70th term of the sequence is \(140\sqrt{5}\). This formula is incredibly useful for quickly finding terms in long sequences without having to list out every single term.
Other exercises in this chapter
Problem 30
Use the Binomial Theorem to find the indicated coefficient or term. The coefficient of \(x^{3}\) in the expansion of \((x-3)^{10}\)
View solution Problem 30
The given pattern continues. Write down the nth term of a sequence \(\left\\{a_{n}\right\\}\) suggested by the pattern. \(\frac{2}{3}, \frac{4}{9}, \frac{8}{27}
View solution Problem 31
Use mathematical induction to prove that $$ \begin{aligned} a+(a+d)+(a+2 d) & \\ +\cdots+[a+(n-1) d] &=n a+d \frac{n(n-1)}{2} \end{aligned} $$
View solution Problem 31
Find the indicated term of each geometric sequence. 8th term of \(0.4,0.04,0.004, \ldots\)
View solution