Problem 55 - Contemporary Precalculus

Question

The sequence in which \(a_{n}=\) per capita amount spent on health services and supplies in year \(n\) with \(n=1\) corresponding to \(1999,\) is approximately arithmetic. (a) If the per capita amount was \(\$ 4154\) in 1999 and \(\$ 5864\) in \(2004,\) find a formula for \(a_{n}\) (b) Use the sequence to estimate the per capita amount in 2005 and 2008

Step-by-Step Solution

Answer

Answer: The estimated per capita amounts spent on health services and supplies in 2005 and 2008 are approximately $6,226 and $7,222, respectively.

Step 1: Determine the common difference

Since the sequence is approximately arithmetic, there is a common difference, denoted by \(d\), between consecutive terms. We have the information for 1999 (year 1) and 2004 (year 6). So, we can find the difference between the per capita amount in 2004 and 1999, and then divide it by the number of years between them. \(d = \frac{a_{6} - a_{1}}{6 - 1} = \frac{5864 - 4154}{5} = 342\) The common difference is approximately \(342\).

Step 2: Write the formula for the sequence

Now that we know the common difference, we can use it to write a formula for the sequence \(a_n\). An arithmetic sequence can be represented by the formula: \(a_n = a_1 + (n - 1)d\) where \(a_1\) is the first term and \(d\) is the common difference. In this problem, \(a_1 = 4154\) and \(d = 342\). So, the formula for the sequence is: \(a_n = 4154 + (n - 1)342\)

Step 3: Estimate the per capita amount for 2005 and 2008

Using the formula we derived in step 2, we can now estimate the per capita amount in 2005 and 2008. For 2005, we have \(n = 7\) (as 1999 corresponds to \(n=1\)): \(a_7 = 4154 + (7 - 1)342 = 4154 + 6 * 342 = 6226\) For 2008, we have \(n = 10\): \(a_{10} = 4154 + (10 - 1)342 = 4154 + 9 * 342 = 7222\) So, the estimated per capita amounts in 2005 and 2008 are \(6226 and \)7222, respectively.

Key Concepts

Common Difference

In the context of arithmetic sequences, the 'common difference' is the constant amount that separates consecutive terms. Imagine you're climbing a ladder and each step brings you equally higher from the ground. This step height is the common difference in arithmetic terms. It's significant because the identical difference helps establish the pattern of the sequence.

When dealing with real-world data like per capita spending, calculating the common difference allows us to evaluate how consistently values increase over time. For example, in our exercise, we determined the common difference based on the per capita amounts spent in 1999 and 2004. We simply subtracted the 1999 value from the 2004 value and divided by the number of years (5) to find this constant gap:

• \(d = \frac{5864 - 4154}{5} = 342\)

This gives us a common difference of approximately 342, meaning each year, spending increased by about 342 dollars per capita.

Arithmetic Sequence Formula

The formula for an arithmetic sequence is fundamental for calculating any term in the sequence. This formula is a versatile tool:

• \(a_n = a_1 + (n-1)d\)

Here, \(a_n\) represents the nth term, \(a_1\) is the initial term, \(n\) is the term number, and \(d\) is the common difference.

Applying this to our problem example, we have an initial per capita spending amount, \(a_1\), of 4154. Using the common difference, 342, we construct the formula:

• \(a_n = 4154 + (n - 1) \times 342\)

This formula provides a quick and efficient way to calculate spending for any given year. For instance, if you'd need to determine what the spending would be for any year past 1999 (\(n=1\)), you can substitute \(n\) with the corresponding year number and get your answer instantly!

Estimation in Sequences

Estimation becomes immensely useful, particularly when exact numbers aren't available or the data for some years are missing. Using the arithmetic sequence formula, we can easily estimate missing data points.

In our example, by knowing the common difference and starting value, we could estimate the per capita spending for 2005 and 2008, years for which we didn't have immediate data. For 2005 (where \(n=7\)):

• \(a_7 = 4154 + (7 - 1) \times 342 = 6226\)

And for 2008 (where \(n=10\)):

• \(a_{10} = 4154 + (10 - 1) \times 342 = 7222\)

These calculations demonstrate how an understanding of sequences allows us to predict trends and make informed estimates. It's a valuable skill, especially when planning for things like budgets or forecasting future expenditures based on past trends.