The sum of all odd numbers between 1 to 1000 which are divisible by 3 is

Options

A. 83667
B. 56128
C. 90000
D. none

Answer

To find the sum of all odd numbers between 1 and 1000 that are divisible by 3, we can follow these steps:

### Step 1: Identify the odd numbers divisible by 3
The odd numbers that are divisible by 3 start from 3 and continue with a common difference of 6 (since every second odd number is divisible by 3). The sequence of odd numbers divisible by 3 is:
3, 9, 15, 21, 27, ..., up to the largest odd number less than or equal to 1000 that is divisible by 3.

### Step 2: Determine the last term of the sequence
To find the largest odd number less than or equal to 1000 that is divisible by 3, we can check:
- The largest odd number less than or equal to 1000 is 999.
- Now, we check if 999 is divisible by 3: \( 999 \div 3 = 333 \), which is an integer.
Thus, the last term \( a_n = 999 \).

### Step 3: Find the number of terms in the sequence
The first term \( a_1 = 3 \), the last term \( a_n = 999 \), and the common difference \( d = 6 \).

Using the formula for the \( n \)-th term of an arithmetic sequence:
\[ a_n = a_1 + (n - 1) \cdot d \]
Substituting the known values:
\[ 999 = 3 + (n - 1) \cdot 6 \]
\[ 999 - 3 = (n - 1) \cdot 6 \]
\[ 996 = (n - 1) \cdot 6 \]
\[ n - 1 = \frac{996}{6} = 166 \]
\[ n = 166 + 1 = 167 \]

### Step 4: Calculate the sum of the sequence
The sum \( S_n \) of the first \( n \) terms of an arithmetic sequence can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (a_1 + a_n) \]
Substituting the values we have:
\[ S_{167} = \frac{167}{2} \cdot (3 + 999) \]
\[ S_{167} = \frac{167}{2} \cdot 1002 \]
\[ S_{167} = 167 \cdot 501 \]
Calculating this gives:
\[ S_{167} = 83667 \]

### Final Answer
The sum of all odd numbers between 1 and 1000 that are divisible by 3 is **83667**.

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