Problem 87 - Precalculus Mathematics for Calculus

Question

If \(a_{1}, a_{2}\) \(a_{3}, \ldots\) is a geometric sequence with common ratio \(r,\) show that the sequence $$\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, \dots$$ is also a geometric sequence, and find the common ratio.

Step-by-Step Solution

Answer

The new sequence is geometric with common ratio \(\frac{1}{r}\).

Step 1: Understand the Geometric Sequence

In a geometric sequence, each term is obtained by multiplying the previous term by a constant, known as the common ratio \( r \). Let's denote the terms of the sequence as \( a_1, a_2, a_3, \ldots \). Thus, \( a_2 = a_1 \cdot r \), \( a_3 = a_2 \cdot r \), and so on, meaning the general term is given by \( a_n = a_1 \cdot r^{n-1} \).

Step 2: Apply the Definition to the New Sequence

Given the sequence \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \), let's substitute \( a_n \) with \( a_1 \cdot r^{n-1} \). The term \( \frac{1}{a_n} \) becomes \( \frac{1}{a_1 \cdot r^{n-1}} \). This can be rewritten as \( \frac{1}{a_1} \cdot r^{-(n-1)} \).

Step 3: Observe the Relation of Terms

Observe the terms of the new sequence: \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \). The first two terms are \( \frac{1}{a_1} \) and \( \frac{1}{a_2} = \frac{1}{a_1 \cdot r} \). The second divided by the first is \( \frac{1}{a_1 \cdot r} \div \frac{1}{a_1} = \frac{1}{r} \).

Step 4: Determine the Common Ratio for the New Sequence

In the new sequence \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \), the common ratio is \( \frac{1}{r} \). This is because \( \frac{1}{a_2} \div \frac{1}{a_1} = \frac{r^{-(2-1)}}{r^{-(1-1)}} = r^{-1} = \frac{1}{r} \).

Key Concepts

Common Ratio

A geometric sequence is characterized by each term being a constant multiple of the previous term. This constant is known as the common ratio. Understanding the common ratio is crucial, as it defines how the sequence progresses. For a sequence like \(a_1, a_2, a_3, \ldots\), the common ratio \(r\) means:

• \(a_2 = a_1 \cdot r\)
• \(a_3 = a_2 \cdot r\)
• \(a_n = a_1 \cdot r^{n-1}\)

The role of \(r\) is to determine the exact multiplication needed to progress from one term to the next. In our original sequence, every term depends directly on this common ratio, making it fundamental to the structure of the sequence.

Sequence Transformation

When we transform a sequence, such as converting each term into its reciprocal, the essential properties, like the common ratio, can also transform. Taking a geometric sequence \(a_1, a_2, a_3, \ldots\) and forming \(\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots\), the sequence will still maintain a geometric nature. However, the common ratio changes.

• The original sequence has terms \(a_n = a_1 \cdot r^{n-1}\).
• For the transformed sequence, we have \(\frac{1}{a_n} = \frac{1}{a_1 \cdot r^{n-1}} = \frac{1}{a_1} \cdot r^{-(n-1)}\).

This transformation shows how the multiplicative nature of the sequence and the consistency of the common ratio translate to the reciprocal sequence.

Reciprocal Sequence

Turning a sequence into its reciprocal means that each term is converted to its inverse. This creates what we call a reciprocal sequence. When considering a geometric sequence, the reciprocal sequence remains geometric. But how does this happen?Understand that if you have a sequence with terms \(a_1, a_2, a_3, \ldots\), the reciprocals are \(\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots\). The original common ratio \(r\) inverts. Thus, in the reciprocal sequence:

• \(\frac{1}{a_2} = \frac{1}{a_1 \cdot r}\)
• Common ratio for reciprocals becomes \(\frac{1}{r}\), derived from \(\frac{1}{a_2} \div \frac{1}{a_1} = \frac{1}{r}\)

The property of being a geometric sequence is retained, but the new common ratio \(\frac{1}{r}\) reveals how reciprocal sequences evolve their own pattern based on inversion of the original ratio.