Problem 42 - Precalculus: Mathematical for Calculus

Question

Find the partial sum \(S_{n}\) of the geometric sequence that satisfies the given conditions. $$a_{2}=0.12, \quad a_{5}=0.00096, \quad n=4$$

Step-by-Step Solution

Answer

The partial sum \( S_4 = 0.7488 \).

Step 1: Determine the Common Ratio

To find the common ratio \( r \) of the sequence, we use the formula for the \( n \)-th term of a geometric sequence: \( a_{n} = a_{1}r^{n-1} \). We have two terms given: \( a_2 = 0.12 \) and \( a_5 = 0.00096 \). Using the formula, we get two equations: \( a_2 = a_1 r^1 \) and \( a_5 = a_1 r^4 \). Divide these equations: \( \frac{a_5}{a_2} = r^3 \). Substitute the values: \( \frac{0.00096}{0.12} = r^3 \), which gives \( r^3 = 0.008 \). Solving for \( r \) by taking the cube root, we find \( r = 0.2 \).

Step 2: Find the First Term

We need to find the first term \( a_1 \) using the equation for the second term, \( a_{2} = a_{1} imes r = 0.12 \). We know \( r = 0.2 \), so substituting in gives \( a_1 \times 0.2 = 0.12 \). Solving for \( a_1 \), we get \( a_1 = \frac{0.12}{0.2} = 0.6 \).

Step 3: Use the Partial Sum Formula

The formula for the partial sum of the first \( n \) terms of a geometric sequence is \( S_n = a_1 \frac{1 - r^n}{1 - r} \). Given \( a_1 = 0.6 \), \( r = 0.2 \), and \( n = 4 \), substitute these values into the formula: \( S_4 = 0.6 \frac{1 - (0.2)^4}{1 - 0.2} \).

Step 4: Calculate the Partial Sum

Calculate \( (0.2)^4 = 0.0016 \) and substitute it back: \( S_4 = 0.6 \frac{1 - 0.0016}{0.8} \). Simplifying gives \( S_4 = 0.6 \frac{0.9984}{0.8} = 0.6 \times 1.248 = 0.7488 \).

Key Concepts

Partial Sum

In the context of geometric sequences, a *partial sum* is the sum of a specific number of terms at the start of the sequence. It's useful when you want to determine how these terms accumulate to form a total. For example, in our exercise, we wanted the partial sum of 4 terms from the sequence.

The formula used to compute the partial sum for a geometric sequence is:

• \( S_n = a_1 \frac{1 - r^n}{1 - r} \)

Here, **\( S_n \)** is the sum of the first \( n \) terms, **\( a_1 \)** is the first term, and **\( r \)** is the common ratio.

It's important to note that this formula assumes that the common ratio \( r \) isn't equal to one. Otherwise, the sequence would be arithmetic, not geometric.

Common Ratio

The *common ratio* in a geometric sequence is the constant factor between consecutive terms, denoted by **\( r \)**. It's the multiplier that allows us to move from one term to the next in the sequence.

To find the common ratio, you can use any two known terms of the sequence. If you have the terms \( a_{m} \) and \( a_{n} \), the ratio \( r \) can be calculated as:

• \( r = \left( \frac{a_{n}}{a_{m}} \right)^{\frac{1}{n-m}} \)

In the given exercise, with terms \( a_2 = 0.12 \) and \( a_5 = 0.00096 \), we derived \( r^3 = 0.008 \) and solved it to get \( r = 0.2 \).

Understanding the common ratio is key as it dictates how fast or slow the terms of the sequence grow or decay. A value greater than 1 indicates growth, whereas a value between 0 and 1 indicates decay.

Geometric Series Formula

The *geometric series formula* allows us to add up terms of a geometric sequence. Unlike arithmetic series, where terms add up linearly, geometric series grow or shrink exponentially depending on the common ratio. This is due to each term being a product of the previous one and a fixed ratio.

In practical terms, the geometric series formula expressed as the partial sum formula is:

• \( S_n = a_1 \frac{1 - r^n}{1 - r} \)

It efficiently sums up **\( n \)** terms of the sequence. This formula greatly simplifies the summation when dealing with many terms.

The geometric nature of the sequence is also clear here—the terms don't simply add but rather accumulate based on multiplicative changes, dictated by the common ratio **\( r \)**. This exponential nature offers powerful applications like calculating interest on investments, where values increase or decrease exponentially over time.