In which group of sequence (1), (2, 3), (4, 5, 6), (7, 8, 9, 10), (11, 12, 13, 14, 15),……….. does 5310 occurs.

Answer

To solve the problem of determining in which group the number 5310 occurs in the given sequence, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Pattern of Groups**:
The groups are as follows:
- Group 1: (1)
- Group 2: (2, 3)
- Group 3: (4, 5, 6)
- Group 4: (7, 8, 9, 10)
- Group 5: (11, 12, 13, 14, 15)
- ...

Each group \( n \) contains \( n \) numbers.

2. **Calculate the Last Term of Each Group**:
The last term of each group can be calculated as follows:
- Group 1: Last term = 1
- Group 2: Last term = 3
- Group 3: Last term = 6
- Group 4: Last term = 10
- Group 5: Last term = 15
- Group 6: Last term = 21
- ...

The last term of the \( n \)-th group can be represented by the formula:
\[ L_n = \frac{n(n + 1)}{2} \]
This formula gives the sum of the first \( n \) natural numbers.

3. **Set Up the Inequality**:
We need to find \( n \) such that:
\[ L_n \geq 5310 \]
This translates to:
\[ \frac{n(n + 1)}{2} \geq 5310 \]
Multiplying both sides by 2 gives:
\[ n(n + 1) \geq 10620 \]

4. **Estimate \( n \)**:
To estimate \( n \), we can solve the quadratic equation:
\[ n^2 + n - 10620 = 0 \]
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 4 \times 10620}}{2} \]
\[ n = \frac{-1 \pm \sqrt{42481}}{2} \]
\[ n = \frac{-1 \pm 206}{2} \]
Taking the positive root:
\[ n = \frac{205}{2} \approx 102.5 \]

5. **Check Integer Values**:
Since \( n \) must be an integer, we check \( n = 102 \) and \( n = 103 \):
- For \( n = 102 \):
\[ L_{102} = \frac{102 \times 103}{2} = 5253 \]
- For \( n = 103 \):
\[ L_{103} = \frac{103 \times 104}{2} = 5356 \]

6. **Determine the Group**:
Since \( 5253 < 5310 < 5356 \), the number 5310 lies in group 103.

### Final Answer:
The number 5310 occurs in **Group 103**.