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 solve the problem, we need to find the 100th term of an arithmetic progression (AP) given two conditions:

1. The sum of the first 100 terms is -1.
2. The sum of the even terms is 1.

Let's denote the first term of the AP as \( A \) and the common difference as \( D \).

### 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 AP is given by:

\[ S_n = \frac{n}{2} \times (2A + (n-1)D) \]

For \( n = 100 \):

\[ S_{100} = \frac{100}{2} \times (2A + 99D) = 50 \times (2A + 99D) \]

Given that \( S_{100} = -1 \), we can set up the equation:

\[ 50 \times (2A + 99D) = -1 \]

Dividing both sides by 50:

\[ 2A + 99D = -\frac{1}{50} \quad \text{(Equation 1)} \]

### Step 2: Find the sum of the even terms

The even terms of the AP are \( A + D, A + 3D, A + 5D, \ldots \) up to the 100th term. The number of even terms in the first 100 terms is 50.

The first even term is \( A + D \) and the common difference for the even terms is \( 2D \).

Using the sum formula for the first 50 even terms:

\[ S_{50} = \frac{50}{2} \times (2(A + D) + (50 - 1) \times 2D) = 25 \times (2A + 2D + 98D) = 25 \times (2A + 100D) \]

Given that \( S_{50} = 1 \), we can set up the equation:

\[ 25 \times (2A + 100D) = 1 \]

Dividing both sides by 25:

\[ 2A + 100D = \frac{1}{25} \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

Now we have two equations:

1. \( 2A + 99D = -\frac{1}{50} \)
2. \( 2A + 100D = \frac{1}{25} \)

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 (50):

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

### Step 4: Substitute \( D \) back to find \( A \)

Substituting \( D = \frac{3}{50} \) into Equation 1:

\[ 2A + 99 \left(\frac{3}{50}\right) = -\frac{1}{50} \]

Calculating \( 99 \times \frac{3}{50} = \frac{297}{50} \):

\[ 2A + \frac{297}{50} = -\frac{1}{50} \]

Subtracting \( \frac{297}{50} \) from both sides:

\[ 2A = -\frac{1}{50} - \frac{297}{50} = -\frac{298}{50} \]

Thus,

\[ A = -\frac{149}{50} \]

### Step 5: Find the 100th term

The formula for the \( n \)th term of an AP is:

\[ T_n = A + (n-1)D \]

For \( n = 100 \):

\[ T_{100} = A + 99D \]

Substituting the values of \( A \) and \( D \):

\[ T_{100} = -\frac{149}{50} + 99 \left(\frac{3}{50}\right) \]

Calculating \( 99 \times \frac{3}{50} = \frac{297}{50} \):

\[ T_{100} = -\frac{149}{50} + \frac{297}{50} = \frac{148}{50} = \frac{74}{25} \]

### Final Answer

The 100th term of the arithmetic progression is:

\[ \boxed{\frac{74}{25}} \]