If the first term of an A.P. is 100 and sum of its first 6 terms is 5 times the sum of next 6 terms, then find the common difference of the A.P.

Explanation

To solve the problem, we need to find the common difference \( D \) of an arithmetic progression (A.P.) where the first term \( A = 100 \) and the sum of the first 6 terms is 5 times the sum of the next 6 terms.

### Step-by-step Solution:

1. **Identify the given values**:
- First term \( A = 100 \)
- Let the common difference be \( D \).

2. **Write the formula for the sum of the first \( n \) terms of an A.P.**:
The sum of the first \( n \) terms \( S_n \) of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n-1)D) \]

3. **Calculate the sum of the first 6 terms \( S_6 \)**:
Using the formula:
\[ S_6 = \frac{6}{2} \times (2 \times 100 + (6-1)D) = 3 \times (200 + 5D) = 600 + 15D \]

4. **Calculate the sum of the next 6 terms \( S'_{6} \)**:
The sum of the next 6 terms can be expressed as:
\[ S'_{6} = S_{12} - S_{6} \]
First, we need to find \( S_{12} \):
\[ S_{12} = \frac{12}{2} \times (2 \times 100 + (12-1)D) = 6 \times (200 + 11D) = 1200 + 66D \]
Now substituting this into the equation for \( S'_{6} \):
\[ S'_{6} = (1200 + 66D) - (600 + 15D) = 600 + 51D \]

5. **Set up the equation based on the problem statement**:
According to the problem, \( S_6 = 5 \times S'_{6} \):
\[ 600 + 15D = 5 \times (600 + 51D) \]

6. **Expand and simplify the equation**:
\[ 600 + 15D = 3000 + 255D \]
Rearranging gives:
\[ 600 - 3000 = 255D - 15D \]
\[ -2400 = 240D \]

7. **Solve for \( D \)**:
Dividing both sides by 240:
\[ D = \frac{-2400}{240} = -10 \]

### Final Result:
The common difference \( D \) of the A.P. is \( -10 \).