If the sum of the first 100 terms of an arithmetic progression is `-1` and the sum of the even terms is 1, then the `100^("th")` term of the arithmetic progression is

Options

A. `(47)/(25)`
B. `(149)/(50)`
C. `(74)/(25)`
D. `-(149)/(50)`

Answer

To find the 100th term of the arithmetic progression (AP) given the conditions, we can follow these steps:

### 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 (S_n) of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \left(2A + (n-1)D\right) \]
where:
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the number of terms.

Given that \( S_{100} = -1 \), we can substitute \( n = 100 \):
\[ S_{100} = \frac{100}{2} \left(2A + 99D\right) = -1 \]
This simplifies to:
\[ 50(2A + 99D) = -1 \]
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 of the AP are the 2nd, 4th, 6th, ..., 100th terms. There are 50 even terms in total. The sum of the first n even terms can be expressed as:
\[ S_{even} = \frac{n}{2} \left(2A_{even} + (n-1)D_{even}\right) \]
where:
- \( A_{even} = A + D \) (the first even term),
- \( D_{even} = 2D \) (the common difference between even terms).

Since \( S_{even} = 1 \) and \( n = 50 \):
\[ S_{even} = \frac{50}{2} \left(2(A + D) + (50-1)(2D)\right) = 1 \]
This simplifies to:
\[ 25(2A + 2D + 98D) = 1 \]
So,
\[ 25(2A + 100D) = 1 \]
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)

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 (50):
\[ D = \frac{2}{50} + \frac{1}{50} = \frac{3}{50} \]

### Step 4: Substitute D back to find A.
Now substitute \( D = \frac{3}{50} \) back into Equation 1:
\[ 2A + 99\left(\frac{3}{50}\right) = -\frac{1}{50} \]
This simplifies to:
\[ 2A + \frac{297}{50} = -\frac{1}{50} \]
Subtract \( \frac{297}{50} \) from both sides:
\[ 2A = -\frac{1}{50} - \frac{297}{50} \]
\[ 2A = -\frac{298}{50} \]
Dividing by 2 gives:
\[ A = -\frac{149}{50} \]

### Step 5: Find the 100th term.
The 100th term of the AP is given by:
\[ T_{100} = A + 99D \]
Substituting the values of \( A \) and \( D \):
\[ T_{100} = -\frac{149}{50} + 99\left(\frac{3}{50}\right) \]
This simplifies to:
\[ T_{100} = -\frac{149}{50} + \frac{297}{50} \]
\[ T_{100} = \frac{148}{50} = \frac{74}{25} \]

Thus, the 100th term of the arithmetic progression is:
\[ \boxed{\frac{74}{25}} \]