Find the sum of all the numbers between 100 and 200 which are divisible by 7.

Answer

To find the sum of all the numbers between 100 and 200 that are divisible by 7, we can follow these steps:

### Step 1: Identify the first and last terms
The first number greater than 100 that is divisible by 7 is 105 (since \( 7 \times 15 = 105 \)). The last number less than 200 that is divisible by 7 is 196 (since \( 7 \times 28 = 196 \)).

### Step 2: Determine the common difference
The common difference \( D \) between the terms in this arithmetic sequence is 7, as we are looking for numbers that are divisible by 7.

### Step 3: Use the formula for the nth term of an arithmetic sequence
The nth term of an arithmetic sequence can be expressed as:
\[ a_n = a + (n - 1)D \]
Where:
- \( a \) is the first term (105)
- \( D \) is the common difference (7)
- \( a_n \) is the last term (196)

Setting up the equation:
\[ 196 = 105 + (n - 1) \cdot 7 \]

### Step 4: Solve for \( n \)
Rearranging the equation:
\[ 196 - 105 = (n - 1) \cdot 7 \]
\[ 91 = (n - 1) \cdot 7 \]
Dividing both sides by 7:
\[ n - 1 = \frac{91}{7} = 13 \]
Adding 1 to both sides gives:
\[ n = 14 \]

### Step 5: Calculate the sum of the arithmetic series
The sum \( S_n \) of the first \( n \) terms of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
Where:
- \( n \) is the number of terms (14)
- \( a \) is the first term (105)
- \( l \) is the last term (196)

Substituting the values:
\[ S_{14} = \frac{14}{2} \cdot (105 + 196) \]
\[ S_{14} = 7 \cdot 301 \]
\[ S_{14} = 2107 \]

### Final Answer
The sum of all the numbers between 100 and 200 that are divisible by 7 is **2107**.

---