Number of G.P's having 5,9 and 11 as its three terms is equal to

Options

A. exactly two
B. almost two
C. at least one
D. none of these

Explanation

To find the number of geometric progressions (G.P.s) that can have the terms 5, 9, and 11, we can follow these steps:

### Step 1: Define the terms in the G.P.
Let \( A \) be the first term of the G.P. and \( R \) be the common ratio. The terms can be represented as:
- First term: \( A \)
- Second term: \( AR \)
- Third term: \( AR^2 \)

### Step 2: Assign the given values to the terms
We can assign the values 5, 9, and 11 to the terms of the G.P. in different orders. Let's assume:
- \( AR^{l-1} = 5 \)
- \( AR^{m-1} = 9 \)
- \( AR^{n-1} = 11 \)

Where \( l, m, n \) are the positions of these terms in the G.P.

### Step 3: Set up the equations
From the equations above, we can express \( A \) in terms of \( R \) and the respective terms:
1. From \( AR^{l-1} = 5 \), we have \( A = \frac{5}{R^{l-1}} \)
2. From \( AR^{m-1} = 9 \), we have \( A = \frac{9}{R^{m-1}} \)
3. From \( AR^{n-1} = 11 \), we have \( A = \frac{11}{R^{n-1}} \)

### Step 4: Equate the expressions for \( A \)
Since all three expressions for \( A \) must be equal, we can set them equal to each other:
\[ \frac{5}{R^{l-1}} = \frac{9}{R^{m-1}} = \frac{11}{R^{n-1}} \]

### Step 5: Formulate ratios
From the first two equations:
\[ \frac{5}{9} = R^{l-m} \]
From the second and third equations:
\[ \frac{9}{11} = R^{m-n} \]
From the first and third equations:
\[ \frac{5}{11} = R^{l-n} \]

### Step 6: Solve for \( R \)
Now we have three equations:
1. \( R^{l-m} = \frac{5}{9} \)
2. \( R^{m-n} = \frac{9}{11} \)
3. \( R^{l-n} = \frac{5}{11} \)

### Step 7: Analyze the equations
To find the number of valid G.P.s, we need to check if there are integer solutions for \( l, m, n \) such that:
- \( l - m = k_1 \)
- \( m - n = k_2 \)
- \( l - n = k_3 \)

Where \( k_1, k_2, k_3 \) are integers derived from the ratios.

### Step 8: Check for contradictions
If we assume \( l, m, n \) are distinct integers, we will find contradictions since the ratios lead to equal powers of \( R \), which cannot hold for distinct integers.

### Conclusion
Since we cannot have distinct integers \( l, m, n \) that satisfy the conditions derived from the ratios, we conclude that there are **no valid G.P.s** that can have 5, 9, and 11 as terms.

### Final Answer
The number of G.P.s having 5, 9, and 11 as its three terms is **0**.

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