An A.P. whose first term is unity and in which the sum of the first half of any even number of terms so that of the second half of the same number of terms is in constant ratio, the common difference 'd' is given by :

Options

A. a. 1
B. b. 2
C. c. 3
D. d. 4

Answer

To solve the problem, we need to find the common difference \( d \) of an arithmetic progression (A.P.) where the first term \( a = 1 \) and the sum of the first half of any even number of terms is in constant ratio to the sum of the second half of the same number of terms.

### Step-by-Step Solution:

1. **Understanding the A.P.**:
The first term of the A.P. is given as \( a = 1 \). The common difference is denoted as \( d \). The \( n \)-th term of the A.P. can be expressed as:
\[ a_n = a + (n-1)d = 1 + (n-1)d \]

2. **Sum of the First \( n \) Terms**:
The formula for the sum of the first \( n \) terms \( S_n \) of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Substituting \( a = 1 \):
\[ S_n = \frac{n}{2} \times (2 \cdot 1 + (n-1)d) = \frac{n}{2} \times (2 + (n-1)d) \]

3. **Sum of the First Half and Second Half**:
For an even number of terms \( 2n \):
- The sum of the first half (first \( n \) terms) is:
\[ S_n = \frac{n}{2} \times (2 + (n-1)d) \]
- The sum of the second half (next \( n \) terms) is:
\[ S_{2n} - S_n = S_{2n} - S_n \]
Where \( S_{2n} \) is:
\[ S_{2n} = \frac{2n}{2} \times (2 + (2n-1)d) = n(2 + (2n-1)d) \]
Thus, the sum of the second half becomes:
\[ S_{2n} - S_n = n(2 + (2n-1)d) - \frac{n}{2}(2 + (n-1)d) \]

4. **Setting Up the Ratio**:
The problem states that the ratio of the sum of the first half to the second half is constant:
\[ \frac{S_n}{S_{2n} - S_n} = k \quad \text{(constant)} \]

5. **Solving for \( d \)**:
After substituting the expressions for \( S_n \) and \( S_{2n} - S_n \) into the ratio and simplifying, we will find that the coefficient of \( n \) must equal zero for the ratio to remain constant. This leads to an equation involving \( d \).

6. **Finding the Value of \( d \)**:
By solving the resulting equation, we find that the common difference \( d \) is:
\[ d = 2 \]

### Conclusion:
Thus, the common difference \( d \) of the arithmetic progression is \( 2 \).