If 11th term of A.P. is 38 and 16th term is 73, then its first term is :

Options

A. 7
B. 32
C. `-32`
D. `-35`

Answer

To find the first term of the arithmetic progression (A.P.) given that 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:
\[ T_n = a + (n - 1) \cdot d \]
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 11th and 16th terms
Using the formula, we can write the equations for the 11th and 16th terms:
1. For the 11th term (\( T_{11} \)):
\[ T_{11} = a + (11 - 1) \cdot d = a + 10d = 38 \]
2. For the 16th term (\( T_{16} \)):
\[ T_{16} = a + (16 - 1) \cdot d = a + 15d = 73 \]

### Step 3: Write the equations
Now we have the following two equations:
1. \( a + 10d = 38 \) (Equation 1)
2. \( a + 15d = 73 \) (Equation 2)

### Step 4: Subtract Equation 1 from Equation 2
To eliminate \( a \), we can subtract Equation 1 from Equation 2:
\[ (a + 15d) - (a + 10d) = 73 - 38 \]
This simplifies to:
\[ 5d = 35 \]

### Step 5: Solve for the common difference \( d \)
Now, divide both sides by 5:
\[ d = \frac{35}{5} = 7 \]

### Step 6: Substitute \( d \) back into one of the equations to find \( a \)
Now that we have \( d \), we can substitute it back into Equation 1 to find \( a \):
\[ a + 10 \cdot 7 = 38 \]
This simplifies to:
\[ a + 70 = 38 \]
Now, solve for \( a \):
\[ a = 38 - 70 = -32 \]

### Conclusion
The first term \( a \) of the A.P. is:
\[ \boxed{-32} \]