If the `3^(rd)` and the `9^(th)` terms of an A.P. are 4 and -8 respectively ,which term of thi A.P. is zero ?

Answer

To solve the problem, we need to find which term of the arithmetic progression (A.P.) is zero, given that the 3rd term is 4 and the 9th term is -8.

### Step-by-Step Solution:

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

2. **Set up equations for the given terms:**
From the problem, we know:
- The 3rd term \( T_3 = 4 \):
\[ T_3 = a + 2d = 4 \quad \text{(Equation 1)} \]
- The 9th term \( T_9 = -8 \):
\[ T_9 = a + 8d = -8 \quad \text{(Equation 2)} \]

3. **Subtract Equation 1 from Equation 2:**
To eliminate \( a \), we can subtract Equation 1 from Equation 2:
\[ (a + 8d) - (a + 2d) = -8 - 4 \]
Simplifying this gives:
\[ 6d = -12 \]
Therefore, we can solve for \( d \):
\[ d = \frac{-12}{6} = -2 \]

4. **Substitute \( d \) back into Equation 1 to find \( a \):**
Now that we have \( d \), we can substitute it back into Equation 1 to find \( a \):
\[ a + 2(-2) = 4 \]
Simplifying this gives:
\[ a - 4 = 4 \implies a = 8 \]

5. **Find the term that is zero:**
We need to find \( n \) such that \( T_n = 0 \):
\[ T_n = a + (n-1)d = 0 \]
Substituting \( a = 8 \) and \( d = -2 \):
\[ 8 + (n-1)(-2) = 0 \]
Simplifying this:
\[ 8 - 2(n-1) = 0 \implies 8 - 2n + 2 = 0 \implies 10 - 2n = 0 \]
Rearranging gives:
\[ 2n = 10 \implies n = 5 \]

### Conclusion:
The term of the A.P. that is zero is the **5th term**.