Problem 68 - College Algebra

Question

Use the formula for the general term (the nth term) of a geometric sequence to solve 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. You are offered a job that pays \(\$ 30,000\) for the first year with an annual increase of \(5 \%\) per year beginning in the second year. That is, beginning in year \(2,\) your salary will be 1.05 times what it was in the previous year. What can you expect to earn in your sixth year on the job?

Step-by-Step Solution

Answer

The salary for the sixth year can be calculated as \$30,000 \cdot 1.05^5.

Step 1: Part 1: Identify the Geometric Sequence

The first sequence is a geometric sequence, where each term is doubled (or multiplied by 2) from the previous one. The formula to find the nth term of a geometric sequence is \(a_n = a_1 \cdot r^{(n-1)}\) where \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number. In this case, \(a_1 = $1\) and \(r = 2\). The formula becomes \(a_n = 1 \cdot 2^{n-1}\).

Step 2: Part 2: Find the 30th term of the Geometric Sequence

We can use the formula derived in 'Step 1' to find the 30th term. Here, \(n = 30\), so substituting \(n = 30\) into the formula, we get \(a_{30} = 1 \cdot 2^{30-1} = 1 \cdot 2^{29}\). We aren't asking for a particular term in the question, but this information is useful for illustrating how the formula works.

Step 3: Part 3: Identify the Salary Increase Pattern

The salary starts at \$30,000 in year 1, and increases by \(5\%\) each year starting from year 2. This is also handled as a geometric sequence because the salary amount changes by a constant ratio each year, just like the initial daily savings pattern. Here, the first term \(a_1\) is \$30,000, and the common ratio \(r\) is \(1 + 5\% = 1.05\).

Step 4: Part 4: Find the Salary for the Sixth Year

We use a similar formula as in step 1 to find the salary for the sixth year. In this case, the term number \(n = 6\), the first term \(a_1 = \$30,000\), and the common ratio \(r = 1.05\). Plugging these into the formula \(a_n = a_1 \cdot r^{(n - 1)}\), we find that \(a_{6} = \$30,000 \cdot 1.05^{(6 - 1)} = \$30,000 \cdot 1.05^5\). Calculate the result to find the salary for the sixth year.

Key Concepts

nth term formula

The nth term formula in a geometric sequence is essential for predicting terms in a pattern where each term after the first is multiplied by a constant. This formula is expressed as \(a_n = a_1 \cdot r^{(n-1)}\), where:

• \(a_n\) represents the nth term we want to find.
• \(a_1\) is the first term in the sequence.
• \(r\) is the common ratio, the number by which we multiply each term to get the next one.
• \(n\) is the term number.

Consider the savings example: by saving \(\\(1\) on the first day and doubling this amount each subsequent day, our sequence becomes \(1, 2, 4, 8, ...\). Here, \(a_1 = \\)1\), \(r = 2\), and the nth term can be calculated using the formula as \(a_n = 1 \cdot 2^{(n-1)}\). This allows us to determine how much we save on any given day. For instance, for day 30, we use \(a_{30} = 1 \cdot 2^{29}\) to find the total amount saved by then.

common ratio

The common ratio is a pivotal aspect of any geometric sequence, as it determines how each term in the sequence progresses. In a geometric sequence, the common ratio \(r\) is calculated by dividing any term in the sequence by the preceding term. In the savings example, the sequence \(1, 2, 4, 8, ...\) clearly shows a common ratio of \(2\), because each term is twice the previous one.
Similarly, in the salary increase example, the yearly adjustment forms a geometric progression where the initial salary of \(\$30,000\) increases by a factor of \(1.05\) each year. Consequently, the common ratio in this case is \(1.05\). This ratio signifies that every year's salary is \(5\%\) higher than the last, or put differently, is multiplied by \(1.05\). Finding the common ratio is crucial to understanding the pattern of the sequence and calculating further terms using the nth term formula.

geometric progression

A geometric progression is a sequence of numbers where each term after the first is obtained by multiplying the previous term by a constant non-zero number known as the common ratio. Geometric progressions are a neat way of representing exponential growth or decay.
In our examples, the two distinct geometric progressions showcase different real-life scenarios. The daily saving example follows a geometric progression of \(1, 2, 4, 8, ...\) where the growth factor is \(2\). This demonstrates how savings can quickly add up when doubled each day.
On the contrary, the salary progression illustrates a smaller, yet steady geometric increase. Starting with \(\$30,000\) and increasing by a ratio of \(1.05\) annually, it models a common scenario of salary increments in jobs and is represented as a geometric progression.

• This progression offers insights into predicting future earnings.
• Both scenarios highlight the simplicity and power of geometric progressions in modeling cyclically increasing or decreasing events.

Understanding the nature of geometric progressions helps make predictions about series' future terms, whether in finance, mathematics, or other applications.