The sum of n terms of a series is `3n^(2) + 4n`. Then the series is

Options

A. A.P.
B. G.P.
C. H.P.
D. A.G.S.

Answer

To find the series given that the sum of n terms \( S_n \) is \( 3n^2 + 4n \), we will follow these steps:

### Step 1: Understand the given information
We know that the sum of the first n terms of the series is given by:
\[ S_n = 3n^2 + 4n \]

### Step 2: Find the nth term of the series
The nth term \( T_n \) can be calculated using the formula:
\[ T_n = S_n - S_{n-1} \]
First, we need to find \( S_{n-1} \):
\[ S_{n-1} = 3(n-1)^2 + 4(n-1) \]
Expanding this:
\[ S_{n-1} = 3(n^2 - 2n + 1) + 4(n - 1) = 3n^2 - 6n + 3 + 4n - 4 = 3n^2 - 2n - 1 \]

### Step 3: Calculate \( T_n \)
Now we can find \( T_n \):
\[ T_n = S_n - S_{n-1} = (3n^2 + 4n) - (3n^2 - 2n - 1) \]
Simplifying this:
\[ T_n = 3n^2 + 4n - 3n^2 + 2n + 1 = 6n + 1 \]

### Step 4: Identify the series
The nth term of the series is:
\[ T_n = 6n + 1 \]
To find the first few terms of the series, we can substitute values of \( n \):
- For \( n = 1 \): \( T_1 = 6(1) + 1 = 7 \)
- For \( n = 2 \): \( T_2 = 6(2) + 1 = 13 \)
- For \( n = 3 \): \( T_3 = 6(3) + 1 = 19 \)
- For \( n = 4 \): \( T_4 = 6(4) + 1 = 25 \)

Thus, the series starts as:
\[ 7, 13, 19, 25, \ldots \]

### Step 5: Determine the nature of the series
The first term is 7, and the common difference can be calculated as:
\[ T_2 - T_1 = 13 - 7 = 6 \]
Since the difference between consecutive terms is constant (6), this series is an arithmetic progression (AP).

### Final Answer
The series is:
\[ T_n = 6n + 1 \quad \text{(an arithmetic progression)} \]