Natural numbers are divided into groups as (1). (2,3), (4,5,6), (7,8,9,10) and so on. What is the sum of the numbers in the 11th group ?

Options

A. 605
B. 615
C. 671
D. 693

Answer

To find the sum of the numbers in the 11th group of natural numbers divided into groups as described, we can follow these steps:

### Step 1: Identify the pattern of groups
The groups are formed as follows:
- 1st group: (1)
- 2nd group: (2, 3)
- 3rd group: (4, 5, 6)
- 4th group: (7, 8, 9, 10)
- ...

We can see that the number of elements in each group increases by 1 for each subsequent group. The 1st group has 1 element, the 2nd group has 2 elements, the 3rd group has 3 elements, and so forth.

### Step 2: Determine the starting number of the 11th group
To find the starting number of the 11th group, we need to calculate the total number of elements in the first 10 groups. The number of elements in the first n groups is given by the sum of the first n natural numbers, which is:

\[ \text{Sum} = \frac{n(n + 1)}{2} \]

For n = 10:

\[ \text{Sum} = \frac{10(10 + 1)}{2} = \frac{10 \times 11}{2} = 55 \]

Thus, the first 10 groups contain a total of 55 numbers. Therefore, the first number of the 11th group is:

\[ 56 \]

### Step 3: List the numbers in the 11th group
The 11th group will contain 11 numbers starting from 56. Thus, the numbers in the 11th group are:

\[ 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66 \]

### Step 4: Calculate the sum of the numbers in the 11th group
To find the sum of these numbers, we can use the formula for the sum of an arithmetic series:

\[ S_n = \frac{n}{2} \times (a + l) \]

Where:
- \( n \) is the number of terms (11),
- \( a \) is the first term (56),
- \( l \) is the last term (66).

Substituting the values:

\[ S_{11} = \frac{11}{2} \times (56 + 66) = \frac{11}{2} \times 122 = 11 \times 61 = 671 \]

### Final Answer
The sum of the numbers in the 11th group is **671**.

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