The number 1, 5 and 25 can be three terms (not necessarily consecutive) of

Options

A. only one AP
B. more than one but `hat(u)nite` numbers of `APs`
C. `"in"hat(u)"nite"` number of `APs`
D. `hat(u)"nite"` number of `GPs`

Explanation

To solve the problem, we need to determine whether the numbers 1, 5, and 25 can form an arithmetic progression (AP) or a geometric progression (GP).

### Step 1: Check for Arithmetic Progression (AP)

In an arithmetic progression, the difference between consecutive terms is constant. The general formula for the n-th term of an AP is given by:

\[ T_n = A + (n-1)D \]

where \(A\) is the first term, \(D\) is the common difference, and \(n\) is the term number.

Let’s denote:
- The first term as \(T_P = 1\)
- The second term as \(T_Q = 5\)
- The third term as \(T_R = 25\)

Using the formula for the n-th term, we can write:

1. For the P-th term:
\[ A + (P-1)D = 1 \quad \text{(1)} \]

2. For the Q-th term:
\[ A + (Q-1)D = 5 \quad \text{(2)} \]

3. For the R-th term:
\[ A + (R-1)D = 25 \quad \text{(3)} \]

### Step 2: Set Up Equations

Now, we can subtract these equations to find relationships between \(P\), \(Q\), and \(R\):

From (1) and (2):
\[ (A + (Q-1)D) - (A + (P-1)D) = 5 - 1 \]
This simplifies to:
\[ (Q - P)D = 4 \quad \text{(4)} \]

From (2) and (3):
\[ (A + (R-1)D) - (A + (Q-1)D) = 25 - 5 \]
This simplifies to:
\[ (R - Q)D = 20 \quad \text{(5)} \]

### Step 3: Analyze the Results

From equations (4) and (5), we have:
- \(D(Q - P) = 4\)
- \(D(R - Q) = 20\)

If \(D \neq 0\), we can divide both equations by \(D\):
\[ Q - P = \frac{4}{D} \quad \text{(6)} \]
\[ R - Q = \frac{20}{D} \quad \text{(7)} \]

### Step 4: Check for Integer Solutions

For \(P\), \(Q\), and \(R\) to be integers, both \(\frac{4}{D}\) and \(\frac{20}{D}\) must also be integers, which implies \(D\) must be a divisor of both 4 and 20. The common divisors are \(1, 2, 4\).

### Step 5: Conclusion for AP

Since we can find integer values for \(P\), \(Q\), and \(R\) that satisfy these equations, we conclude that the numbers 1, 5, and 25 can form an infinite number of arithmetic progressions.

### Step 6: Check for Geometric Progression (GP)

In a geometric progression, the ratio of consecutive terms is constant. The general formula for the n-th term of a GP is given by:

\[ T_n = A \cdot r^{(n-1)} \]

where \(A\) is the first term and \(r\) is the common ratio.

Using the same terms:
1. For the P-th term:
\[ A \cdot r^{(P-1)} = 1 \quad \text{(8)} \]

2. For the Q-th term:
\[ A \cdot r^{(Q-1)} = 5 \quad \text{(9)} \]

3. For the R-th term:
\[ A \cdot r^{(R-1)} = 25 \quad \text{(10)} \]

### Step 7: Set Up Ratios

From (8) and (9):
\[ \frac{A \cdot r^{(Q-1)}}{A \cdot r^{(P-1)}} = \frac{5}{1} \]
This simplifies to:
\[ r^{(Q-P)} = 5 \quad \text{(11)} \]

From (9) and (10):
\[ \frac{A \cdot r^{(R-1)}}{A \cdot r^{(Q-1)}} = \frac{25}{5} \]
This simplifies to:
\[ r^{(R-Q)} = 5 \quad \text{(12)} \]

### Step 8: Analyze the Results for GP

From equations (11) and (12):
- \(r^{(Q-P)} = 5\)
- \(r^{(R-Q)} = 5\)

This implies that both \(Q-P\) and \(R-Q\) must be such that their ratios yield a consistent value for \(r\).

### Step 9: Conclusion for GP

Since we can find rational values for \(r\) that satisfy these equations, we conclude that the numbers 1, 5, and 25 can also form an infinite number of geometric progressions.

### Final Answer

The numbers 1, 5, and 25 can form:
- Infinite Arithmetic Progressions (AP)
- Infinite Geometric Progressions (GP)