What is the sum of all positive integers lying between 200 and 400 that are multiples of 7?

Options

A. 8729
B. 8700
C. 8428
D. 8278

Answer

To find the sum of all positive integers lying between 200 and 400 that are multiples of 7, we can follow these steps:

### Step 1: Identify the first and last multiples of 7 in the range
- The first multiple of 7 greater than 200 can be found by calculating \( \lceil \frac{200}{7} \rceil \times 7 \).
- \( \frac{200}{7} \approx 28.57 \), so \( \lceil 28.57 \rceil = 29 \).
- Therefore, the first multiple is \( 29 \times 7 = 203 \).

- The last multiple of 7 less than 400 can be found by calculating \( \lfloor \frac{400}{7} \rfloor \times 7 \).
- \( \frac{400}{7} \approx 57.14 \), so \( \lfloor 57.14 \rfloor = 57 \).
- Therefore, the last multiple is \( 57 \times 7 = 399 \).

### Step 2: Determine the number of terms (n) in the sequence
- The multiples of 7 between 203 and 399 form an arithmetic sequence where:
- First term \( a = 203 \)
- Last term \( l = 399 \)
- Common difference \( d = 7 \)

- To find the number of terms \( n \), we use the formula for the nth term of an arithmetic sequence:
\[ l = a + (n-1) \cdot d \]
Plugging in the values:
\[ 399 = 203 + (n-1) \cdot 7 \]
Rearranging gives:
\[ 399 - 203 = (n-1) \cdot 7 \]
\[ 196 = (n-1) \cdot 7 \]
\[ n-1 = \frac{196}{7} = 28 \]
\[ n = 28 + 1 = 29 \]

### Step 3: 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:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
Substituting the known values:
\[ S_{29} = \frac{29}{2} \cdot (203 + 399) \]
\[ S_{29} = \frac{29}{2} \cdot 602 \]
\[ S_{29} = 29 \cdot 301 \]
\[ S_{29} = 8729 \]

### Final Answer
The sum of all positive integers lying between 200 and 400 that are multiples of 7 is **8729**.