If for an A.P. `S_(31) = 186`,find `t_(16)`.

Answer

To solve the problem, we need to find the 16th term (T16) of the arithmetic progression (A.P.) given that the sum of the first 31 terms (S31) is 186.

### Step-by-Step Solution:

1. **Understanding the formula for the sum of the first n terms of an A.P.:**
The formula for the sum of the first n terms (Sn) of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
where:
- \( n \) is the number of terms,
- \( A \) is the first term,
- \( D \) is the common difference.

2. **Substituting the known values into the formula:**
For our problem, we have \( n = 31 \) and \( S_{31} = 186 \). Plugging these values into the formula gives:
\[ 186 = \frac{31}{2} \times (2A + (31 - 1)D) \]
Simplifying this, we have:
\[ 186 = \frac{31}{2} \times (2A + 30D) \]

3. **Multiplying both sides by 2 to eliminate the fraction:**
\[ 372 = 31 \times (2A + 30D) \]

4. **Dividing both sides by 31:**
\[ 12 = 2A + 30D \]
This can be rearranged to:
\[ 2A + 30D = 12 \]

5. **Finding T16:**
The formula for the nth term (Tn) of an A.P. is given by:
\[ T_n = A + (n - 1)D \]
For T16:
\[ T_{16} = A + (16 - 1)D = A + 15D \]

6. **Expressing T16 in terms of the equation we derived:**
From the equation \( 2A + 30D = 12 \), we can express \( A + 15D \):
- Divide the entire equation by 2:
\[ A + 15D = \frac{12}{2} = 6 \]

7. **Conclusion:**
Thus, we find that:
\[ T_{16} = 6 \]

### Final Answer:
The 16th term \( T_{16} \) is \( 6 \).