Problem 32 - Precalculus with Limits

Question

In Exercises 23 - 32, find a formula for for the arithmetic sequence. \( a_5 = 190, a_10 = 115 \)

Step-by-Step Solution

Answer

The formula for the given arithmetic sequence is \[ a_n = a_1 + (n-1)*d \] (replace \(a_1\) and \(d\) with their calculated values to establish the specific formula for this sequence).

Step 1: Calculate the Common Difference

The common difference can be calculated using the formula \[ d = \frac{a_n - a_m}{n - m} \] where \(d\) is the common difference, \(a_m\) and \(a_n\) are the \(m^{th}\) and \(n^{th}\) terms of the sequence respectively. Apply the given terms such that \(n = 10\), \(m = 5\), \(a_n = 115\) and \(a_m = 190\), and solve to find \(d\).

Step 2: Calculate the first term

To calculate the first term, we use the rearranged formula of the \(n^{th}\) term in an arithmetic sequence: \[ a_m = a_1 + (m - 1)*d \] Here, we know \(a_m = 190\), \(m = 5\) and \(d\) (from Step 1). Solve for \(a_1\) to find the first term of the series.

Step 3: Formulate the formula

After obtaining the common difference and first term, you can apply these values to the standard formula for an arithmetic sequence: \[ a_n = a_1 + (n-1)*d \] where \(a_n\) is the \(n^{th}\) term, \(a_1\) is the first term, \(d\) is the common difference and \(n\) is the number of terms. Swap \(a_1\) and \(d\) with their corresponding values to form the formula for the given sequence.

Key Concepts

Common Difference

In the realm of arithmetic sequences, the term 'common difference' is crucial. It denotes the consistent gap between any two subsequent terms in the sequence. To find this value, we use the simple formula:

• \( d = \frac{a_n - a_m}{n - m} \)

This expression invites us to subtract one term from another and divide by the difference in their positions.
In practice, with the given terms \(a_5 = 190\) and \(a_10 = 115\), we can find the difference as follows:

• Plug \(n = 10\), \(m = 5\), \(a_n = 115\), and \(a_m = 190\) into the formula to get:
• \( d = \frac{115 - 190}{10 - 5} \)
• Which simplifies to \(d = -15\).

Thus, the common difference here is \(-15\), indicating each term decreases by 15 as we progress.

First Term Formula

Discovering the first term of an arithmetic sequence provides the foundation for all subsequent terms. This can be achieved by rearranging the general formula for an arithmetic sequence. This is how the first term formula operates:

• Use \( a_m = a_1 + (m - 1)*d \)

We know \(a_5 = 190\) and have calculated \(d = -15\). Now, rearrange to solve for \(a_1\):

• Substitute \(m = 5\), resulting in:
• \( 190 = a_1 + (5-1)(-15) \)
• Simplifying gives \( a_1 = 190 + 60 \)
• Hence, \( a_1 = 250 \).

Our first term \(a_1\) is 250; this initial value starts the sequence.

Arithmetic Sequence Formula

Once you have the common difference and the first term, you can establish a formula for the entire arithmetic sequence. This standard formula is a function of each term's position:

• \(a_n = a_1 + (n-1)*d\)

Using this formula, you can predict any term within the sequence. Here's how it works with our derived values:

• Take \(a_1 = 250\) and \(d = -15\) and insert them into the formula:
• \( a_n = 250 + (n-1)*(-15) \)
• This formula allows us to calculate any \(a_n\) by altering \(n\).

So, with these values, you can determine any term in the sequence, paving the way for a deeper analysis of the pattern within the arithmetic sequence.