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

Options

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

Answer

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

### Step 1: Identify the first term and common difference
The \( n \)th term of an AP can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

Given \( a_n = 2n - 1 \):
- For \( n = 1 \):
\[ a_1 = 2(1) - 1 = 1 \]
- For \( n = 2 \):
\[ a_2 = 2(2) - 1 = 3 \]
- For \( n = 3 \):
\[ a_3 = 2(3) - 1 = 5 \]

From these calculations, we can see that the first term \( a = 1 \) and the common difference \( d = a_2 - a_1 = 3 - 1 = 2 \).

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

### Step 3: Substitute the values of \( a \) and \( d \)
Substituting \( a = 1 \) and \( d = 2 \) into the formula:
\[ S_n = \frac{n}{2} \times (2 \cdot 1 + (n-1) \cdot 2) \]
\[ = \frac{n}{2} \times (2 + 2(n-1)) \]
\[ = \frac{n}{2} \times (2 + 2n - 2) \]
\[ = \frac{n}{2} \times 2n \]
\[ = n^2 \]

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