If x,y,z are three natural numbers in A.P. such that ` x+y+z=30` , then the possible number of ordered triplet (x, y, z) is :

Options

A. 18
B. 19
C. 20
D. 21

Answer

To solve the problem, we need to find the number of ordered triplets (x, y, z) of natural numbers that are in arithmetic progression (A.P.) and satisfy the equation \( x + y + z = 30 \).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression (A.P.):**
- If \( x, y, z \) are in A.P., then there exists a common difference \( d \) such that:
\[ y - x = z - y \implies y = \frac{x + z}{2} \]

2. **Expressing x, y, z in terms of d:**
- We can express \( x, y, z \) as:
\[ x = 10 - d, \quad y = 10, \quad z = 10 + d \]
- Here, \( 10 \) is the middle term, and \( d \) is the common difference.

3. **Substituting into the sum equation:**
- Substitute \( x, y, z \) into the equation \( x + y + z = 30 \):
\[ (10 - d) + 10 + (10 + d) = 30 \]
- Simplifying this gives:
\[ 30 = 30 \]
- This confirms that our expressions for \( x, y, z \) satisfy the equation.

4. **Finding the conditions for natural numbers:**
- Since \( x, y, z \) must be natural numbers, we need:
\[ 10 - d > 0 \quad \text{and} \quad 10 + d > 0 \]
- From \( 10 - d > 0 \):
\[ d < 10 \]
- From \( 10 + d > 0 \):
\[ d > -10 \]
- Therefore, \( d \) must satisfy:
\[ -10 < d < 10 \]

5. **Determining the integer values for d:**
- The integer values of \( d \) range from \( -9 \) to \( 9 \) (inclusive).
- This gives us:
\[ d = -9, -8, -7, \ldots, 0, \ldots, 7, 8, 9 \]
- Counting these values:
- From \( -9 \) to \( 9 \) gives \( 9 - (-9) + 1 = 19 \) values.

6. **Conclusion:**
- Thus, the total number of ordered triplets \( (x, y, z) \) is \( 19 \).

### Final Answer:
The possible number of ordered triplets \( (x, y, z) \) is **19**.