If the sum of the first `100` terms of an `AP` is `-1` and the sum of even terms lying in first `100` terms is `1`, then which of the following is not true ?

Options

A. Common difference of the sequence is `(3)/(50)`
B. First term of the sequence is `(-149)/(50)`
C. `100^(th)` term `=(74)/(25)`
D. None of these

Explanation

To solve the problem step by step, we need to find the values of the first term \( a \) and the common difference \( d \) of the arithmetic progression (AP) based on the given conditions.

### Step 1: Use the formula for the sum of the first \( n \) terms of an AP
The formula for the sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) \]
For the first 100 terms, we have:
\[ S_{100} = \frac{100}{2} \left( 2a + (100 - 1)d \right) = 50(2a + 99d) \]
According to the problem, \( S_{100} = -1 \). Therefore, we can set up the equation:
\[ 50(2a + 99d) = -1 \]
Dividing both sides by 50 gives:
\[ 2a + 99d = -\frac{1}{50} \quad \text{(Equation 1)} \]

### Step 2: Find the sum of the even terms
The even terms in the first 100 terms of the AP are \( a_2, a_4, a_6, \ldots, a_{100} \). The first even term is \( a_2 = a + d \) and the common difference for the even terms is \( 2d \). There are 50 even terms. The sum of these even terms can be calculated as:
\[ S_{even} = \frac{n}{2} \left( 2a_2 + (n - 1) \cdot \text{(common difference)} \right) \]
Substituting \( n = 50 \), \( a_2 = a + d \), and common difference \( 2d \):
\[ S_{even} = \frac{50}{2} \left( 2(a + d) + (50 - 1) \cdot 2d \right) = 25 \left( 2(a + d) + 98d \right) \]
This simplifies to:
\[ S_{even} = 25(2a + 2d + 98d) = 25(2a + 100d) \]
According to the problem, \( S_{even} = 1 \). Therefore, we can set up the equation:
\[ 25(2a + 100d) = 1 \]
Dividing both sides by 25 gives:
\[ 2a + 100d = \frac{1}{25} \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( 2a + 99d = -\frac{1}{50} \) (Equation 1)
2. \( 2a + 100d = \frac{1}{25} \) (Equation 2)

We can subtract Equation 1 from Equation 2:
\[ (2a + 100d) - (2a + 99d) = \frac{1}{25} + \frac{1}{50} \]
This simplifies to:
\[ d = \frac{1}{25} + \frac{1}{50} \]
Finding a common denominator (which is 50):
\[ d = \frac{2}{50} + \frac{1}{50} = \frac{3}{50} \]

### Step 4: Substitute \( d \) back to find \( a \)
Substituting \( d = \frac{3}{50} \) back into Equation 1:
\[ 2a + 99 \cdot \frac{3}{50} = -\frac{1}{50} \]
This simplifies to:
\[ 2a + \frac{297}{50} = -\frac{1}{50} \]
Subtracting \( \frac{297}{50} \) from both sides gives:
\[ 2a = -\frac{1}{50} - \frac{297}{50} = -\frac{298}{50} \]
Thus:
\[ a = -\frac{149}{50} \]

### Step 5: Find the 100th term
The 100th term \( a_{100} \) can be found using the formula:
\[ a_{100} = a + 99d \]
Substituting the values of \( a \) and \( d \):
\[ a_{100} = -\frac{149}{50} + 99 \cdot \frac{3}{50} \]
Calculating this gives:
\[ a_{100} = -\frac{149}{50} + \frac{297}{50} = \frac{148}{50} = \frac{74}{25} \]

### Conclusion
Now we have:
- Common difference \( d = \frac{3}{50} \)
- First term \( a = -\frac{149}{50} \)
- 100th term \( a_{100} = \frac{74}{25} \)