If the sum of n terms of an arithmetic progression is `n^(2)-2n`, then what is the `n^(th)` term ?

Options

A. A)`3n-n^(2)`
B. B)`2n-3`
C. C)`2n+3`
D. D)`2n-5`

Answer

To find the nth term of the arithmetic progression (AP) given that the sum of n terms \( S_n \) is \( n^2 - 2n \), we can follow these steps:

### Step 1: Understand the formula for the sum of n terms
The sum of the first n terms of an arithmetic progression can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up the equation
Given that \( S_n = n^2 - 2n \), we can equate this to the formula for the sum of n terms:
\[ \frac{n}{2} \times (2a + (n-1)d) = n^2 - 2n \]

### Step 3: Multiply both sides by 2 to eliminate the fraction
\[ n(2a + (n-1)d) = 2(n^2 - 2n) \]
This simplifies to:
\[ n(2a + (n-1)d) = 2n^2 - 4n \]

### Step 4: Divide both sides by n (assuming \( n \neq 0 \))
\[ 2a + (n-1)d = 2n - 4 \]

### Step 5: Rearrange to find the nth term
The nth term \( T_n \) of an AP can be expressed as:
\[ T_n = a + (n-1)d \]
From our previous equation, we can express \( 2a + (n-1)d \):
\[ 2a + (n-1)d = 2n - 4 \]
Now, we can express \( a \) in terms of \( T_n \):
\[ T_n = a + (n-1)d = \frac{(2a + (n-1)d) + (n-1)d}{2} \]
This leads to:
\[ T_n = \frac{(2n - 4) + (n-1)d}{2} \]

### Step 6: Find the value of \( T_n \)
To find \( T_n \), we can express \( a \) and \( d \) in terms of \( n \). From the equation \( 2a + (n-1)d = 2n - 4 \), we can substitute \( a \) and \( d \) back to find \( T_n \):
1. Let’s find \( T_1 \):
- For \( n = 1 \):
\[ S_1 = 1^2 - 2 \cdot 1 = 1 - 2 = -1 \]
Thus, \( T_1 = -1 \).

2. Let’s find \( T_2 \):
- For \( n = 2 \):
\[ S_2 = 2^2 - 2 \cdot 2 = 4 - 4 = 0 \]
Thus, \( T_2 = S_2 - S_1 = 0 - (-1) = 1 \).

3. Let’s find \( T_3 \):
- For \( n = 3 \):
\[ S_3 = 3^2 - 2 \cdot 3 = 9 - 6 = 3 \]
Thus, \( T_3 = S_3 - S_2 = 3 - 0 = 3 \).

### Step 7: Generalize for \( n \)
From the pattern, we can see that:
\[ T_n = 2n - 3 \]

### Final Result
Thus, the nth term \( T_n \) of the arithmetic progression is:
\[ \boxed{2n - 3} \]