The 5th term of an A.P. is 11 and the 9th term is 7. Find the 16th term.

Answer

To find the 16th term of an arithmetic progression (A.P.) given that the 5th term is 11 and the 9th term is 7, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for the nth term of an A.P.**:
The nth term of an A.P. can be expressed as:
\[ T_n = a + (n - 1) \cdot d \]
where \( T_n \) is the nth term, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

2. **Write the equations for the 5th and 9th terms**:
- For the 5th term:
\[ T_5 = a + (5 - 1) \cdot d = a + 4d \]
Given \( T_5 = 11 \), we can write:
\[ a + 4d = 11 \quad \text{(Equation 1)} \]

- For the 9th term:
\[ T_9 = a + (9 - 1) \cdot d = a + 8d \]
Given \( T_9 = 7 \), we can write:
\[ a + 8d = 7 \quad \text{(Equation 2)} \]

3. **Solve the system of equations**:
We have two equations:
\[ a + 4d = 11 \quad \text{(1)} \]
\[ a + 8d = 7 \quad \text{(2)} \]

To eliminate \( a \), we can subtract Equation 1 from Equation 2:
\[ (a + 8d) - (a + 4d) = 7 - 11 \]
This simplifies to:
\[ 4d = -4 \]
Dividing both sides by 4 gives:
\[ d = -1 \]

4. **Substitute \( d \) back into one of the equations to find \( a \)**:
We can substitute \( d = -1 \) into Equation 1:
\[ a + 4(-1) = 11 \]
Simplifying this gives:
\[ a - 4 = 11 \]
Adding 4 to both sides results in:
\[ a = 15 \]

5. **Find the 16th term**:
Now that we have \( a = 15 \) and \( d = -1 \), we can find the 16th term:
\[ T_{16} = a + (16 - 1) \cdot d \]
Substituting the values:
\[ T_{16} = 15 + 15 \cdot (-1) \]
This simplifies to:
\[ T_{16} = 15 - 15 = 0 \]

### Conclusion:
The 16th term of the given A.P. is \( 0 \).