Problem 67 - College Algebra

Question

Give two examples of arithmetic sequences whose \(10^{\mathrm{th}}\) terms are 206

Step-by-Step Solution

Answer

Two possible sequences: 1) \(a_1 = 50, d = 17.33\) and 2) \(a_1 = 20, d = 20.67\).

Step 1: Understanding Arithmetic Sequences

An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is a constant. This constant is called the common difference, denoted as \(d\). The general formula for the \(n^{ ext{th}}\) term \(a_n\) of an arithmetic sequence is given by: \[a_n = a_1 + (n-1)d\]where \(a_1\) is the first term and \(n\) is the term number.

Step 2: Finding the Expression for the 10th Term

Since we are interested in the \(10^{ ext{th}}\) term of the sequence and it is given as 206, we can use the formula: \[a_{10} = a_1 + (10-1)d = 206\]This simplifies to: \[a_1 + 9d = 206\]This expression will help us generate different sequences.

Step 3: First Example of Arithmetic Sequence

Choose a convenient first term and solve for the common difference. Let's start with \(a_1 = 50\):\[50 + 9d = 206\]Subtracting 50 from both sides: \[9d = 156\]Dividing both sides by 9 gives: \[d = \frac{156}{9} = 17.33\]Thus, one example of an arithmetic sequence is where \(a_1 = 50\) and \(d = 17.33\).

Step 4: Second Example of Arithmetic Sequence

Let's choose another first term, \(a_1 = 20\):\[20 + 9d = 206\]Subtracting 20 from both sides: \[9d = 186\]Dividing both sides by 9 gives: \[d = \frac{186}{9} = 20.67\]Thus, another example of an arithmetic sequence is where \(a_1 = 20\) and \(d = 20.67\).

Key Concepts

10th term

The concept of the "10th term" in an arithmetic sequence is crucial for understanding how sequences progress over time. In an arithmetic sequence, each term grows by adding a fixed amount, known as the common difference. To find the 10th term, we use the general formula for any term in the sequence:\[ a_n = a_1 + (n-1) d \] where

• \(a_n\) is the term we want to find, which in this case is 206.
• \(a_1\) is the first term of the sequence.
• \(d\) is the common difference between consecutive terms.
• \(n\) is the term number, here it is 10.

Substituting the known values for the 10th term, we get \(a_{10} = a_1 + 9d = 206\). The equation \(a_1 + 9d = 206\) forms the basis for finding sequences where the 10th term is known.

common difference

The common difference in an arithmetic sequence is what gives the sequence its consistency. It is the amount we add (or subtract if negative) to each term to get to the next one. In this context, it lets us calculate different terms of the sequence using the formula \( a_n = a_1 + (n-1) d \).

• To find the common difference when the 10th term is 206, we rearrange the equation to get \(d\) alone: \(d = \frac{a_{10} - a_1}{9}\).
• This is because there are 9 intervals between the first term and the 10th term, allowing us to calculate \(d\).

By choosing different starting points (\(a_1\)), we can determine different possible values of \(d\) that still satisfy the condition of a 10th term being 206.

sequence generation

Sequence generation involves creating specific sequences that meet certain criteria, such as having a specific 10th term. Using arithmetic sequences, we can generate sequences by choosing a first term \(a_1\) and calculating the common difference \(d\) that satisfies the equation \(a_1 + 9d = 206\).

• For example, choosing \(a_1 = 50\), we find \(d = 17.33\).
• By choosing \(a_1 = 20\), another possible sequence arises with \(d = 20.67\).

Sequence generation allows flexibility in creation, as adjusting the first term leads to different common differences while the necessary 10th term remains constant at 206. Each new pair \((a_1, d)\) generates an exciting sequence pathway anchored by the known 10th term.