If 6 times the `6^(th)` term of an A.P, is equal to 9 times the `9^(th)` term, show that its `15^(th)` term is zero.

Answer

To solve the problem, we need to show that the 15th term of the arithmetic progression (A.P.) is zero, given that 6 times the 6th term is equal to 9 times the 9th term.

### Step-by-step Solution:

1. **Define the nth term of an A.P.**:
The nth term \( T_n \) of an arithmetic progression can be expressed as:
\[ T_n = a + (n - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Write the expressions for the 6th and 9th terms**:
- The 6th term \( T_6 \) is:
\[ T_6 = a + (6 - 1)d = a + 5d \]
- The 9th term \( T_9 \) is:
\[ T_9 = a + (9 - 1)d = a + 8d \]

3. **Set up the equation based on the problem statement**:
According to the problem, 6 times the 6th term is equal to 9 times the 9th term:
\[ 6 \times T_6 = 9 \times T_9 \]
Substituting the expressions for \( T_6 \) and \( T_9 \):
\[ 6(a + 5d) = 9(a + 8d) \]

4. **Expand both sides**:
Expanding the left side:
\[ 6a + 30d \]
Expanding the right side:
\[ 9a + 72d \]
So, we have:
\[ 6a + 30d = 9a + 72d \]

5. **Rearrange the equation**:
Bringing all terms involving \( a \) to one side and all terms involving \( d \) to the other side:
\[ 6a - 9a = 72d - 30d \]
This simplifies to:
\[ -3a = 42d \]

6. **Divide by -3**:
Dividing both sides by -3 gives:
\[ a = -14d \]

7. **Find the 15th term**:
Now, we can find the 15th term \( T_{15} \):
\[ T_{15} = a + (15 - 1)d = a + 14d \]
Substitute \( a = -14d \):
\[ T_{15} = -14d + 14d = 0 \]

8. **Conclusion**:
Thus, we have shown that the 15th term \( T_{15} \) is equal to 0.