Proving the Relationship Between Terms in Arithmetic and Geometric Progressions

In this article, we will explore the relationship between terms in arithmetic and geometric progressions. Specifically, we will prove that if 1/ab, 1/2b, 1/bc are in arithmetic progression (AP), then 1/ab, 1/bc, 1/2b are in geometric progression (GP).

Understanding the Given Terms

Given the terms 1/ab, 1/2b, 1/bc are in arithmetic progression, we need to prove that they form a geometric progression.

Step 1: Establishing the Arithmetic Progression Condition

First, recall that for three terms to be in arithmetic progression, the common difference between consecutive terms must be equal. Therefore, we have:

1/2b - 1/ab 1/bc - 1/2b

Step 2: Simplifying the Expression

Let's simplify the above equation step-by-step:

1/2b - 1/ab 1/bc - 1/2b

Multiplying through by 2abbc to clear the denominators:

2b - 2a/b 2c - 2b/bc

Simplify further:

2b - 2a/b 2c - 2b/bc

Multiplying through by ab to clear the denominators:

2ab - a^2 b^2 - a^2 / bc

Rearrange the terms:

2b^2 - 2ab a^2 - b^2

Step 3: Deriving the Geometric Progression Condition

To show that 1/ab, 1/bc, 1/2b are in geometric progression, we need to demonstrate that the common ratio between consecutive terms is the same.

Let's take the common ratio q b^2 / ac.

From the earlier steps, we know:

a^2 - b^2 2b^2 - 2ab

Simplifying further:

b^2 ac

This means that the terms 1/ab, 1/bc, 1/2b have a common ratio:

1/bc / 1/ab a / b

1/2b / 1/bc c / b

Hence, a/b b/c, which implies a, b, c are in geometric progression.

Final Verifications

Let's verify the common ratio of the terms:

1/bc / 1/ab a / b

1/2b / 1/bc c / b

Since b^2 ac, the terms 1/ab, 1/bc, 1/2b indeed form a geometric progression.

Conclusion

By following these steps, we have proven that if the terms 1/ab, 1/2b, 1/bc are in arithmetic progression, then they are also in geometric progression when the condition b^2 ac holds true.