How many geometric progressios is/are possible containing 27.8 and 12 as three of its/their terms ?

Options

A. One
B. Two
C. Four
D. Infinitely many

Answer

To determine how many geometric progressions (GPs) can contain the terms 27, 8, and 12, we can follow these steps:

### Step 1: Define the terms in the GP
Let’s assume that the three terms 27, 8, and 12 are the \(p\)-th, \(q\)-th, and \(r\)-th terms of a GP, respectively. We denote the first term of the GP as \(A\) and the common ratio as \(R\).

### Step 2: Write the equations for the terms
From the properties of a GP, we can write the following equations based on the definitions of the terms:
1. \( A \cdot R^{p-1} = 27 \) (Equation 1)
2. \( A \cdot R^{q-1} = 8 \) (Equation 2)
3. \( A \cdot R^{r-1} = 12 \) (Equation 3)

### Step 3: Divide the equations
To eliminate \(A\), we can divide Equation 1 by Equation 3:
\[ \frac{A \cdot R^{p-1}}{A \cdot R^{r-1}} = \frac{27}{12} \]
This simplifies to:
\[ R^{p-r} = \frac{27}{12} = \frac{9}{4} = \left(\frac{3}{2}\right)^2 \]
Thus, we have:
\[ R^{p-r} = \left(\frac{3}{2}\right)^2 \quad \text{(Equation 4)} \]

Next, we divide Equation 3 by Equation 2:
\[ \frac{A \cdot R^{r-1}}{A \cdot R^{q-1}} = \frac{12}{8} \]
This simplifies to:
\[ R^{r-q} = \frac{12}{8} = \frac{3}{2} \]
Thus, we have:
\[ R^{r-q} = \frac{3}{2} \quad \text{(Equation 5)} \]

### Step 4: Relate the equations
Now we have two equations:
1. \( R^{p-r} = \left(\frac{3}{2}\right)^2 \)
2. \( R^{r-q} = \frac{3}{2} \)

From Equation 5, we can express \(r\) in terms of \(q\):
\[ r = q + \log_R\left(\frac{3}{2}\right) \]

### Step 5: Substitute and simplify
Substituting \(r\) into Equation 4 gives:
\[ R^{p - (q + \log_R\left(\frac{3}{2}\right))} = \left(\frac{3}{2}\right)^2 \]
This can be simplified to:
\[ R^{p - q} \cdot R^{-\log_R\left(\frac{3}{2}\right)} = \left(\frac{3}{2}\right)^2 \]
This implies:
\[ R^{p - q} = \left(\frac{3}{2}\right)^2 \cdot \left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^3 \]

### Step 6: Establish the relationship
From the above, we can derive the relationship:
\[ p - q = 3 \]
Thus, we can express \(p\) in terms of \(q\):
\[ p = q + 3 \]

### Step 7: Conclusion
Since \(p\) and \(q\) can take any natural number values that satisfy this equation, there are infinitely many combinations of \(p\) and \(q\) that can satisfy the conditions of the GP. Therefore, the number of geometric progressions that can contain the terms 27, 8, and 12 is infinite.

### Final Answer
**Infinite geometric progressions are possible containing the terms 27, 8, and 12.**

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