Problem 67 - Algebra and Trigonometry

Question

Use the formula for the general term (the nth term) of a geometric sequence to solve Exercises \(65-68\) In Exercises \(65-66,\) suppose you save \(\$ 1\) the first day of a month, \(\$ 2\) the second day, \(\$ 4\) the third day, and so on. That is, each day you save twice as much as you did the day before. A professional baseball player signs a contract with a beginning salary of \(\$ 3,000,000\) for the first year and an annual increase of \(4 \%\) per year beginning in the second year. That is, beginning in year \(2,\) the athlete's salary will be 1.04 times what it was in the previous year. What is the athlete's salary for year 7 of the contract? Round to the nearest dollar.

Step-by-Step Solution

Answer

For the first problem, you will have saved $512 by the 10th day. For the second problem, the player's salary in the 7th year of the contract will be approximately $3,663,442.

Step 1: Identify the geometric sequence for savings

The first exercise provides a geometric progression where each day's saving is double of the previous day's saving. We begin by saving $1 on the first day, $2 on the second day, $4 on the third day, and so on. The general term of a geometric sequence is given by \(a_n = a_1 \cdot r^{(n-1)}\), where \(a_n\) is the nth term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

Step 2: Apply the formula for the geometric sequence for savings

Let's use the formula to find out how much will be saved on a specific day. For instance, on the 10th day, \(a_{10} = 1 \cdot 2^{(10-1)} = 1 \cdot 2^9 = 512$}

Step 3: Identify the geometric sequence for the salary

The second exercise is similar but deals with an annual increase of 4% in the salary of a baseball player. The player's salary was $3,000,000 for the first year. So, starting from the second year, the salary is expected to be 1.04 times the previous years. So, this is also a geometric sequence, with \(a_1 = 3000000$ and \(r = 1.04$.

Step 4: Apply the formula for the geometric sequence for salary

Again, we use the formula to get the 7th year's salary, keeping in mind that the sequence starts from the second year: \(a_{7-1} = 3000000 \cdot 1.04^{(7-2)} = 3000000 \cdot 1.04^5\). Rounded to the nearest dollar, this equals $3663442.

Key Concepts

Geometric Progression

A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. To visualize this, imagine a set of dominos where each is twice the size of the last; this set is growing in a geometric fashion.

In the case of the savings exercise, the amount of money saved each day forms a geometric progression. On the first day, you save \(1, on the second day \)2, on the third day \(4, and so on. Notice how each term grows by a factor of 2 - that's the common ratio at play. Consequently, the progression for the savings looks like this: \)1, \(2, \)4, ...

Common Ratio

In the context of a geometric sequence, the common ratio is the multiplier that defines the rate at which the sequence grows or shrinks. To find the common ratio, divide any term by its preceding term. It's worth noting that the common ratio can be any real number, positive or negative, but it cannot be zero, since division by zero is undefined.

For the savings example, dividing the second day's savings (\(2) by the first day's savings (\)1) gives us a common ratio of 2. In contrast, the salary progression of the baseball player, which increases by an annual rate of 4%, has a common ratio of 1.04. This means each year the salary is multiplied by 1.04 to get the next year's salary.

Nth Term of a Geometric Sequence

The formula for the nth term of a geometric sequence is a powerful tool as it allows us to calculate the value of any term without having to know all the preceding terms. The nth term is given by the formula: \(a_n = a_1 \cdot r^{(n-1)}\), where \(a_n\) is the nth term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

In application, to find the amount saved on the 10th day, we use the formula with \(a_1 = 1\) dollar, \(r = 2\), and \(n = 10\). This yields \(a_{10} = 1 \cdot 2^{9} = 512\) dollars. Similarly, for the athlete's seventh-year salary, we calculate it using \(a_1 = 3,000,000\) dollars, \(r = 1.04\), but here, since the progression starts in year 2, we use \(n = 6\) (for the seventh year’s salary). Thus, we arrive at \(a_{7-1} = 3,000,000 \cdot 1.04^{5} = 3,663,442\) dollars, when rounded to the nearest dollar.