The sum of all 2 digit odd numbers is

Options

A. 2375
B. 2475
C. 2560
D. 4920

Answer

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

### Step 1: Identify the first and last terms
The first two-digit odd number is 11, and the last two-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 progression (AP)
The nth term of an AP can be expressed as:
\[ a_n = a + (n - 1) \cdot d \]
where:
- \( a \) is the first term (11),
- \( d \) is the common difference (2),
- \( a_n \) is the nth term (99).

Setting up the equation:
\[ 99 = 11 + (n - 1) \cdot 2 \]

### Step 4: Solve for n
Rearranging the equation:
\[ 99 - 11 = (n - 1) \cdot 2 \]
\[ 88 = (n - 1) \cdot 2 \]
\[ n - 1 = \frac{88}{2} \]
\[ n - 1 = 44 \]
\[ n = 44 + 1 = 45 \]

So, there are 45 terms in this sequence.

### Step 5: Use the formula for the sum of the first n terms of an AP
The sum \( S_n \) of the first n terms of an AP is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]

Substituting the known values:
- \( n = 45 \)
- \( a = 11 \)
- \( d = 2 \)

### Step 6: Calculate the sum
\[ S_{45} = \frac{45}{2} \cdot (2 \cdot 11 + (45 - 1) \cdot 2) \]
\[ S_{45} = \frac{45}{2} \cdot (22 + 44 \cdot 2) \]
\[ S_{45} = \frac{45}{2} \cdot (22 + 88) \]
\[ S_{45} = \frac{45}{2} \cdot 110 \]
\[ S_{45} = 45 \cdot 55 \]
\[ S_{45} = 2475 \]

Thus, the sum of all two-digit odd numbers is **2475**.