If `11^(th)` term of an A.P. Is 38 and 16th term is 73, its first term is :

Options

A. 7
B. 32
C. `-32`
D. `45(1)/(2)`.

Answer

To find the first term of the arithmetic progression (A.P.) given the 11th term and the 16th term, we can follow these steps:

### Step 1: Write the formulas for the terms of the 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 the equations based on the given terms.
From the problem, we know:
- The 11th term \( T_{11} = 38 \)
- The 16th term \( T_{16} = 73 \)

Using the formula for the nth term:
1. For the 11th term:
\[ T_{11} = a + (11-1)d = a + 10d = 38 \]
This gives us our first equation:
\[ a + 10d = 38 \quad \text{(Equation 1)} \]

2. For the 16th term:
\[ T_{16} = a + (16-1)d = a + 15d = 73 \]
This gives us our second equation:
\[ a + 15d = 73 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously.
We have the following system of 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:
\[ 5d = 35 \]
Dividing both sides by 5:
\[ d = 7 \]

### Step 4: Substitute \( d \) back into one of the equations to find \( a \).
Now that we have \( d \), we can substitute it back into Equation 1:
\[ a + 10(7) = 38 \]
This simplifies to:
\[ a + 70 = 38 \]
Subtracting 70 from both sides:
\[ a = 38 - 70 \]
Thus:
\[ a = -32 \]

### Final Answer:
The first term \( a \) of the A.P. is \( -32 \).

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