If a, b, c are three natural numbres in A.P. such that a+b+c =21, then possible number of odered triplet (a, b, c), is

Options

A. 15
B. 14
C. 13
D. 16

Answer

To solve the problem, we need to find the number of ordered triplets (a, b, c) of natural numbers that are in Arithmetic Progression (A.P.) and satisfy the equation \( a + b + c = 21 \).

### Step-by-Step Solution:

1. **Understanding A.P.**:
Since \( a, b, c \) are in A.P., we know that:
\[ 2b = a + c \]
This means that the middle term \( b \) is the average of \( a \) and \( c \).

2. **Using the given sum**:
We also have the equation:
\[ a + b + c = 21 \]

3. **Substituting for \( a + c \)**:
From the first equation, we can express \( a + c \) in terms of \( b \):
\[ a + c = 2b \]

4. **Substituting into the sum equation**:
Now, substitute \( a + c \) into the sum equation:
\[ a + b + c = 21 \implies 2b + b = 21 \implies 3b = 21 \]

5. **Solving for \( b \)**:
Dividing both sides by 3, we find:
\[ b = 7 \]

6. **Finding \( a + c \)**:
Since \( b = 7 \), we can find \( a + c \):
\[ a + c = 2b = 2 \times 7 = 14 \]

7. **Expressing \( c \) in terms of \( a \)**:
We can express \( c \) as:
\[ c = 14 - a \]

8. **Finding the range for \( a \)**:
Since \( a \) and \( c \) are natural numbers, \( a \) must be at least 1. Therefore:
\[ 1 \leq a < 14 \]
This gives us the possible values for \( a \) as \( 1, 2, 3, \ldots, 13 \).

9. **Counting the ordered triplets**:
For each value of \( a \) from 1 to 13, there is a corresponding value of \( c \) (since \( c = 14 - a \)). Thus, we can have the following ordered triplets:
- For \( a = 1, c = 13 \) → (1, 7, 13)
- For \( a = 2, c = 12 \) → (2, 7, 12)
- For \( a = 3, c = 11 \) → (3, 7, 11)
- ...
- For \( a = 13, c = 1 \) → (13, 7, 1)

Therefore, there are 13 possible ordered triplets.

### Final Answer:
The total number of ordered triplets \( (a, b, c) \) is **13**.