If in an A.P. 7th term is 9 and 9th term is 7, then find 16th term.

Answer

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

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

### Step 2: Set up equations for the 7th and 9th terms
From the problem, we know:
- The 7th term \( T_7 = 9 \)
- The 9th term \( T_9 = 7 \)

Using the formula:
1. For the 7th term:
\[ T_7 = a + (7-1)d = a + 6d = 9 \]
This gives us our first equation:
\[ a + 6d = 9 \quad \text{(Equation 1)} \]

2. For the 9th term:
\[ T_9 = a + (9-1)d = a + 8d = 7 \]
This gives us our second equation:
\[ a + 8d = 7 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( a + 6d = 9 \)
2. \( a + 8d = 7 \)

We can subtract Equation 1 from Equation 2 to eliminate \( a \):
\[ (a + 8d) - (a + 6d) = 7 - 9 \]
This simplifies to:
\[ 2d = -2 \]
Dividing both sides by 2:
\[ d = -1 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now that we have \( d \), we can substitute it back into either equation to find \( a \). Let's use Equation 1:
\[ a + 6(-1) = 9 \]
This simplifies to:
\[ a - 6 = 9 \]
Adding 6 to both sides:
\[ a = 15 \]

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

### Final Answer
Thus, the 16th term of the A.P. is:
\[ \boxed{0} \]