Problem 51 - Contemporary Precalculus

Question

In Exercises \(49-54,\) you are asked to find a geometric sequence. In each case, round the common ratio r to four decimal places. The value of all group life insurance (in billions of dollars) in year \(n\) can be approximated by a geometric sequence \(\left\\{c_{n}\right\\},\) where \(n=1\) corresponds to \(1991 .\) (a) If there was \(\$ 3.9631\) billion in effect in 1991 and \(\$ 4.1672\) billion in \(1992,\) find a formula for \(c_{n}\) (b) How much group life insurance is in effect in \(2000 ?\) In \(2004 ?\) In \(2008 ?\)

Step-by-Step Solution

Answer

Answer: In 2000, approximately $6.1286$ billion dollars of group life insurance was in effect. In 2004, approximately $8.1185$ billion dollars of group life insurance was in effect. In 2008, approximately $10.7749$ billion dollars of group life insurance was in effect.

Step 1: Find the common ratio

To find the common ratio, we'll divide the value of the second year by the value of the first year, and round the result to 4 decimal places: \(r = \frac{c_2}{c_1}\) Plug in values: \(r = \frac{4.1672}{3.9631} = 1.0515\)

Step 2: Determine the formula for c_n

Now we know the common ratio, and we can find the general formula for the geometric sequence \(c_n\): \(c_n = c_1 * r^{n-1}\) We'll substitute given values: \(c_n = 3.9631 * (1.0515)^{n-1}\)

Step 3: Calculate group life insurance in effect for year 2000

We are asked to find the group life insurance in effect for year 2000, which is \(n=10\) since \(n=1\) corresponds to 1991. We will substitute the value of n in the formula: \(c_{10} = 3.9631 * (1.0515)^{10-1} = 3.9631 * (1.0515)^9 = 6.1286\) In 2000, approximately \(6.1286\) billion dollars of group life insurance was in effect.

Step 4: Calculate group life insurance in effect for year 2004

Now we will substitute the value of n for 2004. Since \(n=1\) corresponds to 1991, we have \(n=14\): \(c_{14} = 3.9631 * (1.0515)^{14-1} = 3.9631 * (1.0515)^{13} = 8.1185\) In 2004, approximately \(8.1185\) billion dollars of group life insurance was in effect.

Step 5: Calculate group life insurance in effect for year 2008

Finally, we will substitute the value of n for 2008. Since \(n=1\) corresponds to 1991, we have \(n=18\): \(c_{18} = 3.9631 * (1.0515)^{18-1} = 3.9631 * (1.0515)^{17} = 10.7749\) In 2008, approximately \(10.7749\) billion dollars of group life insurance was in effect.

Key Concepts

Common Ratio

In geometric sequences, the "common ratio" is a crucial element. It is the factor that connects consecutive terms. You calculate the common ratio (\( r \)) by dividing a term in the sequence by the previous term.

It's essential to round your result to four decimal places for accuracy in complex calculations.- In the exercise, we have the values for 1991 and 1992, \( 3.9631 \) and \( 4.1672 \) billion dollars respectively.- Thus, the formula to find the common ratio becomes:\[ r = \frac{c_2}{c_1} \]- Substituting the provided values yields:\[ r = \frac{4.1672}{3.9631} = 1.0515 \]Understanding the common ratio helps in unfolding the geometric sequence's pattern, whether it's in financial data, population growth, or physics. Remember, the common ratio stays constant throughout the entire sequence, making prediction and analysis much easier.

Exponential Growth

Exponential growth in geometric sequences indicates how rapidly a quantity increases over time. Each term grows by a factor of the common ratio (\( r \)).

In the context of life insurance value:- Using the formula \( c_n = c_1 \cdot r^{n-1} \),we describe the exponential increase.- Here, \( c_1 = 3.9631 \) billion dollars and \( r = 1.0515 \).This equation signifies:- Initial value \( c_1 \) when \( n=1 \).- Each subsequent year: multiply the current value by \( r = 1.0515 \).This continuous multiplication demonstrates exponential growth: a rapid escalation depicted as an upward curve when plotted on a graph. Such growth patterns are apparent in many real-world applications, highlighting the power and utility of mathematical models.

Mathematical Modeling

Mathematical modeling transforms real-world situations into mathematical terms, enabling predictions and insights. Using geometric sequences to approximate financial trends, like life insurance values, is a classic example.

In our exercise, we aim to forecast how the value changes over time using the formula \( c_n = c_1 \cdot r^{n-1} \).- This model lets us project future insurance figures based on past data.- For example, calculating the insurance value for 2000, 2004, and 2008.- Each year corresponds to an \( n \) value: - 2000: \( n=10 \) - 2004: \( n=14 \) - 2008: \( n=18 \)Mathematical modeling like this provides a framework for decision-makers to plan, assess risk, and allocate resources effectively, underscoring its significant role in strategy and analysis across industries.