The series of natural numbers is divided into groups `(1), (2, 3, 4), (3, 4, 5, 6, 7), (4, 5, 6, 7, 8, 9, 10)……….` If sum of elements in the `20^(th)` groups is `l`, then `l` is equal to

Options

A. `1368`
B. `(38)^(2)`
C. `(39)^(2)`
D. `38xx39`

Explanation

To find the sum of the elements in the 20th group of the series of natural numbers, we can follow these steps:

### Step 1: Identify the first element of the 20th group
The first element of the nth group is equal to n. Therefore, for the 20th group, the first element is:
\[ a = 20 \]

### Step 2: Determine the number of terms in the 20th group
From the pattern observed, the number of terms in the nth group can be expressed as:
\[ \text{Number of terms} = 2n - 1 \]
For the 20th group:
\[ \text{Number of terms} = 2 \times 20 - 1 = 39 \]

### Step 3: List the elements in the 20th group
The elements in the 20th group start from 20 and continue for 39 terms. Thus, the elements are:
\[ 20, 21, 22, \ldots, 20 + 39 - 1 = 20 + 38 = 58 \]
So the elements are:
\[ 20, 21, 22, \ldots, 58 \]

### Step 4: Calculate the sum of the elements in the 20th group
The sum of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Where:
- \( n \) = number of terms = 39
- \( a \) = first term = 20
- \( d \) = common difference = 1

Substituting the values:
\[ S_{39} = \frac{39}{2} \times (2 \times 20 + (39 - 1) \times 1) \]
\[ S_{39} = \frac{39}{2} \times (40 + 38) \]
\[ S_{39} = \frac{39}{2} \times 78 \]
\[ S_{39} = 39 \times 39 = 1521 \]

### Conclusion
Thus, the sum of the elements in the 20th group, denoted as \( l \), is:
\[ l = 1521 \]