`lt a_i gt` sequence is an A.P. with common difference 1 and `sum_(i=1)^n a_i=192 , sum_(i=1)^(n/2) a_(2i)=120`
then find the value of n , where n is an even integer

Options

A. 48
B. 96
C. 18
D. 36

Answer

To solve the problem, we need to analyze the information given about the arithmetic progression (A.P.) and the sums provided.

### Step-by-Step Solution:

1. **Understanding the A.P.**:
Let the first term of the A.P. be \( a_1 \). Since the common difference is 1, the terms can be expressed as:
\[ a_1, a_1 + 1, a_1 + 2, \ldots, a_1 + (n-1) \]

2. **Sum of the first n terms**:
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (a_1 + a_n) \]
Here, \( a_n = a_1 + (n-1) \). Therefore,
\[ S_n = \frac{n}{2} \times \left(a_1 + (a_1 + (n-1))\right) = \frac{n}{2} \times (2a_1 + n - 1) \]
We know from the problem statement that \( S_n = 192 \). Thus,
\[ \frac{n}{2} \times (2a_1 + n - 1) = 192 \tag{1} \]

3. **Sum of the even indexed terms**:
The even indexed terms are \( a_2, a_4, a_6, \ldots, a_n \) (since \( n \) is even). The number of even indexed terms is \( \frac{n}{2} \). The first even term is \( a_2 = a_1 + 1 \) and the last even term is \( a_n = a_1 + (n-1) \).

The sum of the even indexed terms can be expressed as:
\[ S_{even} = \frac{n}{4} \times \left((a_1 + 1) + (a_1 + (n-1))\right) = \frac{n}{4} \times (2a_1 + n) \]
According to the problem, this sum equals 120:
\[ \frac{n}{4} \times (2a_1 + n) = 120 \tag{2} \]

4. **Solving the equations**:
From equation (1):
\[ n(2a_1 + n - 1) = 384 \tag{3} \]
From equation (2):
\[ n(2a_1 + n) = 480 \tag{4} \]

5. **Subtracting equations (3) and (4)**:
Subtract equation (3) from equation (4):
\[ n(2a_1 + n) - n(2a_1 + n - 1) = 480 - 384 \]
Simplifying gives:
\[ n = 96 \]

6. **Conclusion**:
The value of \( n \) is \( 96 \).