Understanding Geometric Progression (GP) Properties

A sequence of non-zero numbers is said to be in Geometric Progression (GP) if the ratio of any term to its preceding term is constant throughout the sequence. This constant ratio is called the common ratio, denoted by \(r\).

If three numbers a, b, and c are in GP, then the ratio of consecutive terms is equal:

\(\frac{b}{a} = \frac{c}{b}\)

This condition can be rearranged to get a useful property:

\(b^2 = ac\)

Given that a, b, and c are positive terms in a GP, we will examine each of the given statements to determine if the resulting sequences are also in GP.

Analyzing Statement 1: a\(\textsuperscript{2}\), b\(\textsuperscript{2}\), c\(\textsuperscript{2}\) in GP

For a\(\textsuperscript{2}\), b\(\textsuperscript{2}\), c\(\textsuperscript{2}\) to be in GP, the square of the middle term must equal the product of the other two terms. That is, we need to check if \((b^2)^2 = (a^2)(c^2)\).

We know from the given information that a, b, c are in GP, so \(b^2 = ac\).

Let's substitute this into the equation we need to check:

\((b^2)^2 = (ac)^2\)

\((b^2)^2 = a^2 c^2\)

This shows that \((b^2)^2 = (a^2)(c^2)\). Therefore, a\(\textsuperscript{2}\), b\(\textsuperscript{2}\), c\(\textsuperscript{2}\) are in GP.

Statement 1 is correct.

Analyzing Statement 2: \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) in GP

For \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) to be in GP, the square of the middle term must equal the product of the other two terms. That is, we need to check if \((\frac{1}{b})^2 = (\frac{1}{a})(\frac{1}{c})\).

Let's simplify both sides of the equation:

Left side: \((\frac{1}{b})^2 = \frac{1}{b^2}\)

Right side: \((\frac{1}{a})(\frac{1}{c}) = \frac{1}{ac}\)

We know that a, b, c are in GP, so \(b^2 = ac\).

Since \(b^2 = ac\), it follows that \(\frac{1}{b^2} = \frac{1}{ac}\).

Thus, \((\frac{1}{b})^2 = (\frac{1}{a})(\frac{1}{c})\). Therefore, \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in GP.

Statement 2 is correct.

Analyzing Statement 3: \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) in GP

Given that a, b, c > 0, their square roots \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) are real and positive numbers.

For \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) to be in GP, the square of the middle term must equal the product of the other two terms. That is, we need to check if \((\sqrt{b})^2 = (\sqrt{a})(\sqrt{c})\).

Let's simplify both sides of the equation:

Left side: \((\sqrt{b})^2 = b\)

Right side: \((\sqrt{a})(\sqrt{c}) = \sqrt{ac}\)

We know that a, b, c are in GP, so \(b^2 = ac\).

Since a, b, c are positive, we can take the positive square root of both sides of \(b^2 = ac\):

\(\sqrt{b^2} = \sqrt{ac}\)

\(b = \sqrt{ac}\)

Thus, \((\sqrt{b})^2 = (\sqrt{a})(\sqrt{c})\). Therefore, \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) are in GP.

Statement 3 is correct.

Summary of Geometric Progression Properties

Based on our analysis, all three statements are correct when a, b, and c are positive terms in a Geometric Progression.

• Statement 1: a\(\textsuperscript{2}\), b\(\textsuperscript{2}\), c\(\textsuperscript{2}\) are in GP.
• Statement 2: \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in GP.
• Statement 3: \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) are in GP (for positive a, b, c).

Therefore, the correct option is the one that includes statements 1, 2, and 3.

Revision Table: GP Properties

This table summarizes the key property used to check if three terms are in GP and the findings from the analysis.

Additional Information: Extending GP Properties

The properties observed in this question are examples of how certain operations on a geometric progression maintain its geometric progression property.

• Raising to a power: If a, b, c are in GP, then \(a^n, b^n, c^n\) are also in GP for any real number \(n\), provided the terms are well-defined (e.g., \(a^n\) is real). This is because if \(b/a = c/b = r\), then \(b = ar\) and \(c = br = ar^2\). Then \((b^n)/(a^n) = (ar)^n / a^n = a^n r^n / a^n = r^n\) and \((c^n)/(b^n) = (ar^2)^n / (ar)^n = a^n r^{2n} / a^n r^n = r^n\). The common ratio becomes \(r^n\). Statements 1 and 3 are specific cases of this (\(n=2\) and \(n=1/2\)).
• Taking reciprocals: If a, b, c are in GP with common ratio \(r\), then \(1/a, 1/b, 1/c\) are also in GP with common ratio \(1/r\). This was shown in the analysis of Statement 2.
• These properties demonstrate that GPs are closed under exponentiation (and thus reciprocals and roots).