Analyzing Terms in Geometric Progression (GP)

The question states that three terms, \(2^{\frac{1}{c}}, 2^{\frac{b}{a c}},\) and \(2^{\frac{1}{a}},\) are in Geometric Progression (GP). We need to determine the relationship between \(a, b,\) and \(c\).

Understanding Geometric Progression

A sequence of non-zero numbers is in Geometric Progression (GP) if the ratio of any term to its preceding term is constant. This constant ratio is called the common ratio.

If three terms \(x, y, z\) are in GP, then the ratio of the second term to the first term is equal to the ratio of the third term to the second term. Mathematically, this is expressed as:

\(\frac{y}{x} = \frac{z}{y}\)

This equation can be rearranged to give the characteristic property of three terms in GP:

\(y^2 = xz\)

The square of the middle term is equal to the product of the first and third terms.

Applying the GP Property

We are given the three terms:
First term \(x = 2^{\frac{1}{c}}\)
Second term \(y = 2^{\frac{b}{a c}}\)
Third term \(z = 2^{\frac{1}{a}}\)

Using the GP property \(y^2 = xz\), we substitute the given terms:

\(\left(2^{\frac{b}{a c}}\right)^2 = 2^{\frac{1}{c}} \times 2^{\frac{1}{a}}\)

Simplifying the Equation Using Exponent Rules

We will use the following exponent rules to simplify the equation:

• \((m^p)^q = m^{pq}\)
• \(m^p \times m^q = m^{p+q}\)

Applying the first rule to the left side of the equation:

\(\left(2^{\frac{b}{a c}}\right)^2 = 2^{\frac{b}{a c} \times 2} = 2^{\frac{2b}{a c}}\)

Applying the second rule to the right side of the equation:

\(2^{\frac{1}{c}} \times 2^{\frac{1}{a}} = 2^{\frac{1}{c} + \frac{1}{a}}\)

Now, we can rewrite the equation:

\(2^{\frac{2b}{a c}} = 2^{\frac{1}{c} + \frac{1}{a}}\)

Equating the Exponents

Since the bases are equal (both are 2), the exponents must also be equal:

\(\frac{2b}{a c} = \frac{1}{c} + \frac{1}{a}\)

Let's simplify the right side by finding a common denominator, which is \(ac\):

\(\frac{1}{c} + \frac{1}{a} = \frac{a}{ac} + \frac{c}{ac} = \frac{a + c}{ac}\)

So the equation becomes:

\(\frac{2b}{a c} = \frac{a + c}{a c}\)

Assuming \(a \neq 0\) and \(c \neq 0\) (otherwise the original exponents would be undefined or zero base issues arise depending on the base), we can multiply both sides by \(ac\):

\(2b = a + c\)

Interpreting the Relationship

The equation \(2b = a + c\) is the defining condition for three numbers \(a, b, c\) to be in Arithmetic Progression (AP). In an AP, the middle term is the arithmetic mean of the first and third terms, or equivalently, the difference between consecutive terms is constant (\(b - a = c - b\), which simplifies to \(2b = a + c\)).

Therefore, if \(2^{\frac{1}{c}}, 2^{\frac{b}{a c}}, 2^{\frac{1}{a}}\) are in GP, it implies that \(a, b, c\) are in AP.

Conclusion

Based on the analysis of the geometric progression, the relationship between \(a, b,\) and \(c\) is that they form an Arithmetic Progression.

The correct option is the one stating that \(a, b, c\) are in AP.

Revision Table: Progression Relationships

Additional Information: Types of Progressions

Understanding different types of sequences and series like AP, GP, and HP is fundamental in mathematics.

• Arithmetic Progression (AP): In an AP, each term after the first is obtained by adding a fixed number, called the common difference, to the preceding term. Example: 2, 5, 8, 11, ... (common difference = 3).
• Geometric Progression (GP): In a GP, each term after the first is obtained by multiplying the preceding term by a fixed non-zero number, called the common ratio. Example: 3, 6, 12, 24, ... (common ratio = 2).
• Harmonic Progression (HP): A sequence is in HP if the reciprocals of its terms are in AP. Example: 1, 1/2, 1/3, 1/4, ... (reciprocals 1, 2, 3, 4 are in AP).

These progressions are important in various areas of mathematics and physics, including finance, growth models, and wave phenomena.