Given sum of the first `n` terms of an A.P is `2n + 3n^(2)`. Another A.P. is formed with the same first term and double of the common difference, the sum of `n` terms

Options

A. `n + 4n^(2)`
B. `n^(2) + 4n`
C. `3n + 2n^(2)`
D. `6n^(2) - n`

Answer

To solve the problem step by step, we will first determine the nth term of the given arithmetic progression (A.P.) and then find the sum of the first n terms of the new A.P. formed with the same first term and double the common difference.

### Step 1: Identify the given sum of the first n terms of the A.P.
The sum of the first n terms \( S_n \) is given as:
\[ S_n = 2n + 3n^2 \]

### Step 2: Find the nth term of the A.P.
The nth term \( T_n \) can be derived from the sum of the first n terms using the formula:
\[ T_n = S_n - S_{n-1} \]
where \( S_{n-1} \) is the sum of the first \( n-1 \) terms.

To find \( S_{n-1} \):
\[ S_{n-1} = 2(n-1) + 3(n-1)^2 \]
Expanding this:
\[ S_{n-1} = 2n - 2 + 3(n^2 - 2n + 1) = 2n - 2 + 3n^2 - 6n + 3 = 3n^2 - 4n + 1 \]

Now, substituting \( S_n \) and \( S_{n-1} \) into the formula for \( T_n \):
\[ T_n = S_n - S_{n-1} = (2n + 3n^2) - (3n^2 - 4n + 1) \]
Simplifying this:
\[ T_n = 2n + 3n^2 - 3n^2 + 4n - 1 = 6n - 1 \]

### Step 3: Identify the first term and common difference of the original A.P.
From the nth term \( T_n = 6n - 1 \), we can express it in the standard form of an A.P.:
\[ T_n = a + (n-1)d \]
Comparing, we find:
- First term \( a = 5 \) (when \( n = 1 \))
- Common difference \( d = 6 \)

### Step 4: Form the new A.P. with the same first term and double the common difference.
The new common difference \( d' \) is:
\[ d' = 2d = 2 \times 6 = 12 \]

### Step 5: Find the sum of the first n terms of the new A.P.
The sum of the first n terms \( S_n' \) of an A.P. is given by:
\[ S_n' = \frac{n}{2} [2a + (n-1)d'] \]
Substituting the values:
\[ S_n' = \frac{n}{2} [2 \times 5 + (n-1) \times 12] \]
Simplifying:
\[ S_n' = \frac{n}{2} [10 + 12n - 12] = \frac{n}{2} [12n - 2] = n(6n - 1) \]

### Final Result:
Thus, the sum of the first n terms of the new A.P. is:
\[ S_n' = 6n^2 - n \]