Find the `31^(st)` term of an A.P. whose `11^(th)` term is 38 and the `16^(th)` term is 73.

Answer

To find the 31st term of an arithmetic progression (A.P.) where the 11th term is 38 and the 16th term is 73, 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:
- \( T_n \) is the nth term,
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

### Step 2: Set up equations for the given terms.
From the problem, we know:
- The 11th term \( T_{11} = 38 \):
\[ T_{11} = A + (11 - 1)D = A + 10D = 38 \] (Equation 1)

- The 16th term \( T_{16} = 73 \):
\[ T_{16} = A + (16 - 1)D = A + 15D = 73 \] (Equation 2)

### Step 3: Solve the equations simultaneously.
We have two equations:
1. \( A + 10D = 38 \) (Equation 1)
2. \( A + 15D = 73 \) (Equation 2)

To eliminate \( A \), we can subtract Equation 1 from Equation 2:
\[ (A + 15D) - (A + 10D) = 73 - 38 \]
This simplifies to:
\[ 5D = 35 \]
Now, divide both sides by 5:
\[ D = 7 \]

### Step 4: Substitute \( D \) back to find \( A \).
Now that we have \( D = 7 \), we can substitute it back into Equation 1 to find \( A \):
\[ A + 10(7) = 38 \]
\[ A + 70 = 38 \]
Now, subtract 70 from both sides:
\[ A = 38 - 70 \]
\[ A = -32 \]

### Step 5: Find the 31st term.
Now that we have both \( A \) and \( D \), we can find the 31st term \( T_{31} \):
\[ T_{31} = A + (31 - 1)D \]
Substituting the values of \( A \) and \( D \):
\[ T_{31} = -32 + (30)(7) \]
Calculating \( 30 \times 7 \):
\[ T_{31} = -32 + 210 \]
Now, perform the addition:
\[ T_{31} = 178 \]

### Final Answer:
The 31st term of the A.P. is \( 178 \).

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