The sum of all odd numbers between 1 and 100 which are divisible by 3, is

Options

A. 83667
B. 90000
C. 83660
D. None of these

Answer

To find the sum of all odd numbers between 1 and 100 that are divisible by 3, we can follow these steps:

### Step 1: Identify the odd numbers between 1 and 100 that are divisible by 3.

The odd numbers divisible by 3 in this range are:
- 3
- 9
- 15
- 21
- 27
- 33
- 39
- 45
- 51
- 57
- 63
- 69
- 75
- 81
- 87
- 93
- 99

### Step 2: Recognize the pattern.

These numbers form an arithmetic progression (AP) where:
- The first term \( a = 3 \)
- The common difference \( d = 6 \)

### Step 3: Determine the number of terms in the sequence.

To find the number of terms \( n \), we can use the formula for the nth term of an AP:
\[ a_n = a + (n - 1) \cdot d \]
Setting \( a_n = 99 \) (the last term), we have:
\[ 99 = 3 + (n - 1) \cdot 6 \]
Subtracting 3 from both sides:
\[ 96 = (n - 1) \cdot 6 \]
Dividing both sides by 6:
\[ n - 1 = 16 \]
Thus:
\[ n = 17 \]

### Step 4: Calculate the sum of the terms.

The sum \( S_n \) of the first \( n \) terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (a + a_n) \]
Substituting the values we found:
\[ S_{17} = \frac{17}{2} \cdot (3 + 99) \]
Calculating inside the parentheses:
\[ 3 + 99 = 102 \]
Now substituting back:
\[ S_{17} = \frac{17}{2} \cdot 102 \]
Calculating further:
\[ S_{17} = 17 \cdot 51 = 867 \]

### Final Answer:
The sum of all odd numbers between 1 and 100 that are divisible by 3 is **867**.

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