Find the 31st term of an A.P. whose 10th term is 38 and 16th term is 74.

Answer

To find the 31st term of an arithmetic progression (A.P.) where the 10th term is 38 and the 16th term is 74, 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:
\[ T_n = A + (n - 1) \cdot d \]
where \( A \) is the first term and \( d \) is the common difference.

### Step 2: Set up equations for the given terms
From the problem, we know:
- The 10th term \( T_{10} = 38 \)
- The 16th term \( T_{16} = 74 \)

Using the formula for the nth term, we can write:
1. For the 10th term:
\[ T_{10} = A + (10 - 1) \cdot d = A + 9d = 38 \quad \text{(Equation 1)} \]
2. For the 16th term:
\[ T_{16} = A + (16 - 1) \cdot d = A + 15d = 74 \quad \text{(Equation 2)} \]

### Step 3: Subtract Equation 1 from Equation 2
Now, we can subtract Equation 1 from Equation 2 to eliminate \( A \):
\[ (A + 15d) - (A + 9d) = 74 - 38 \]
This simplifies to:
\[ 6d = 36 \]

### Step 4: Solve for \( d \)
Now, divide both sides by 6:
\[ d = \frac{36}{6} = 6 \]

### Step 5: Substitute \( d \) back into Equation 1 to find \( A \)
Now that we have \( d \), we can substitute it back into Equation 1 to find \( A \):
\[ A + 9d = 38 \]
Substituting \( d = 6 \):
\[ A + 9 \cdot 6 = 38 \]
\[ A + 54 = 38 \]
Now, solve for \( A \):
\[ A = 38 - 54 = -16 \]

### Step 6: Find the 31st term
Now that we have both \( A \) and \( d \), we can find the 31st term \( T_{31} \):
\[ T_{31} = A + (31 - 1) \cdot d = A + 30d \]
Substituting \( A = -16 \) and \( d = 6 \):
\[ T_{31} = -16 + 30 \cdot 6 \]
\[ T_{31} = -16 + 180 \]
\[ T_{31} = 164 \]

### Final Answer
Thus, the 31st term of the A.P. is:
\[ \boxed{164} \]