If the nth term of an arithmetic progression is `2n-1`, then what is the sum upto n terms?

Options

A. `n^(2)`
B. `n^(2)-1`
C. `n^(2)+1`
D. `(1)/(2)n(n+1)`

Explanation

To find the sum of the first \( n \) terms of the arithmetic progression (AP) where the \( n \)th term is given by \( T_n = 2n - 1 \), we can follow these steps:

### Step 1: Identify the first term and common difference
The \( n \)th term of the AP is given by:
\[ T_n = 2n - 1 \]
To find the first term \( T_1 \):
\[ T_1 = 2(1) - 1 = 1 \]
To find the second term \( T_2 \):
\[ T_2 = 2(2) - 1 = 3 \]
To find the third term \( T_3 \):
\[ T_3 = 2(3) - 1 = 5 \]
From these calculations, we can see that the first term \( a = 1 \) and the common difference \( d \) can be calculated as:
\[ d = T_2 - T_1 = 3 - 1 = 2 \]

### Step 2: Write the formula for the sum of the first \( n \) terms
The formula for the sum of the first \( n \) terms \( S_n \) of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting the values of \( a \) and \( d \):
\[ S_n = \frac{n}{2} \times (2 \cdot 1 + (n - 1) \cdot 2) \]

### Step 3: Simplify the expression
Now we simplify the expression:
\[ S_n = \frac{n}{2} \times (2 + 2(n - 1)) \]
\[ S_n = \frac{n}{2} \times (2 + 2n - 2) \]
\[ S_n = \frac{n}{2} \times 2n \]
\[ S_n = n^2 \]

### Final Result
Thus, the sum of the first \( n \) terms of the arithmetic progression is:
\[ S_n = n^2 \]

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