The sum of all 2 digit odd numbers is

Options

A. ` 3n - 5 `
B. `4n - 3 `
C. `4n + 3`
D. `3n + 5 `

Answer

To find the sum of all 2-digit odd numbers, we can follow these steps:

### Step 1: Identify the first and last 2-digit odd numbers
The first 2-digit odd number is 11, and the last 2-digit odd number is 99.

### Step 2: Determine the common difference
The common difference (d) between consecutive odd numbers is 2.

### Step 3: Use the formula for the nth term of an arithmetic sequence
The nth term of an arithmetic sequence can be calculated using the formula:
\[ a_n = a + (n - 1) \cdot d \]
where:
- \( a \) is the first term (11),
- \( d \) is the common difference (2),
- \( a_n \) is the nth term (99).

### Step 4: Set up the equation
We can set up the equation:
\[ 99 = 11 + (n - 1) \cdot 2 \]

### Step 5: Solve for n
Rearranging the equation:
1. Subtract 11 from both sides:
\[ 99 - 11 = (n - 1) \cdot 2 \]
\[ 88 = (n - 1) \cdot 2 \]

2. Divide both sides by 2:
\[ 44 = n - 1 \]

3. Add 1 to both sides:
\[ n = 45 \]

### Step 6: Use the sum formula for an arithmetic series
The sum of the first n terms of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (a + a_n) \]
where:
- \( n \) is the number of terms (45),
- \( a \) is the first term (11),
- \( a_n \) is the last term (99).

### Step 7: Substitute the values into the sum formula
Substituting the values:
\[ S_{45} = \frac{45}{2} \cdot (11 + 99) \]
\[ S_{45} = \frac{45}{2} \cdot 110 \]

### Step 8: Calculate the sum
1. Calculate \( \frac{45}{2} \):
\[ \frac{45}{2} = 22.5 \]

2. Multiply by 110:
\[ S_{45} = 22.5 \cdot 110 = 2475 \]

### Final Answer
The sum of all 2-digit odd numbers is **2475**.

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