Let fourth therm of an arithmetic progression be 6 and `m^(th)` term be 18. If A.P has intergal terms only then the numbers of such A.P s is `"____________"`

Explanation

To solve the problem, we need to find the number of arithmetic progressions (APs) that satisfy the given conditions. Let's break it down step by step.

### Step 1: Understand the given terms
We know that:
- The 4th term of the AP is 6.
- The m-th term of the AP is 18.

### Step 2: Write the formulas for the terms
The n-th term of an AP can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

Using this formula:
- The 4th term (when \( n = 4 \)) is:
\[ a + 3d = 6 \tag{1} \]
- The m-th term (when \( n = m \)) is:
\[ a + (m-1)d = 18 \tag{2} \]

### Step 3: Express \( a \) in terms of \( d \)
From equation (1):
\[ a = 6 - 3d \tag{3} \]

### Step 4: Substitute \( a \) into the second equation
Substituting equation (3) into equation (2):
\[ (6 - 3d) + (m-1)d = 18 \]
This simplifies to:
\[ 6 - 3d + md - d = 18 \]
\[ 6 - 4d + md = 18 \]
Rearranging gives:
\[ md - 4d = 12 \]
\[ d(m - 4) = 12 \tag{4} \]

### Step 5: Analyze equation (4)
From equation (4), we see that \( d(m - 4) = 12 \). This means \( m - 4 \) must be a divisor of 12.

### Step 6: Find the divisors of 12
The divisors of 12 are:
\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \]

### Step 7: Determine possible values for \( m \)
For each divisor \( k \) of 12, we can find \( m \):
\[ m = k + 4 \]
Calculating for each divisor:
- \( k = 1 \) → \( m = 5 \)
- \( k = -1 \) → \( m = 3 \)
- \( k = 2 \) → \( m = 6 \)
- \( k = -2 \) → \( m = 2 \)
- \( k = 3 \) → \( m = 7 \)
- \( k = -3 \) → \( m = 1 \)
- \( k = 4 \) → \( m = 8 \)
- \( k = -4 \) → \( m = 0 \) (not valid)
- \( k = 6 \) → \( m = 10 \)
- \( k = -6 \) → \( m = -2 \) (not valid)
- \( k = 12 \) → \( m = 16 \)
- \( k = -12 \) → \( m = -8 \) (not valid)

### Step 8: Valid values of \( m \)
The valid values of \( m \) are:
- 5, 3, 6, 2, 7, 1, 8, 10, 16

### Step 9: Count the valid values
The valid values of \( m \) that yield integral terms in the AP are:
- \( m = 1, 2, 3, 5, 6, 7, 8, 10, 16 \)

Thus, there are **9 valid values** of \( m \).

### Conclusion
The total number of arithmetic progressions that satisfy the given conditions is **9**.

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