Find the sum given below :
`(i) 3+6+9+ ....+96 " " (ii) 2+4+6+....+ 50`

Answer

To find the sums of the given arithmetic progressions (APs), we will follow these steps for each part of the question.

### Part (i): Find the sum of the series 3 + 6 + 9 + ... + 96

**Step 1: Identify the first term (a), common difference (d), and last term (l).**
- The first term \( a = 3 \)
- The common difference \( d = 6 - 3 = 3 \)
- The last term \( l = 96 \)

**Step 2: Use the formula for the nth term of an AP to find n.**
The nth term of an AP is given by:
\[ A_n = a + (n - 1) \cdot d \]
Setting \( A_n = 96 \):
\[ 96 = 3 + (n - 1) \cdot 3 \]
Subtracting 3 from both sides:
\[ 93 = (n - 1) \cdot 3 \]
Dividing by 3:
\[ 31 = n - 1 \]
Thus,
\[ n = 32 \]

**Step 3: Use the formula for the sum of the first n terms of an AP.**
The sum \( S_n \) is given by:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
Substituting the values we found:
\[ S_{32} = \frac{32}{2} \cdot (3 + 96) = 16 \cdot 99 \]
Calculating:
\[ S_{32} = 1584 \]

### Part (ii): Find the sum of the series 2 + 4 + 6 + ... + 50

**Step 1: Identify the first term (a), common difference (d), and last term (l).**
- The first term \( a = 2 \)
- The common difference \( d = 4 - 2 = 2 \)
- The last term \( l = 50 \)

**Step 2: Use the formula for the nth term of an AP to find n.**
Setting \( A_n = 50 \):
\[ 50 = 2 + (n - 1) \cdot 2 \]
Subtracting 2 from both sides:
\[ 48 = (n - 1) \cdot 2 \]
Dividing by 2:
\[ 24 = n - 1 \]
Thus,
\[ n = 25 \]

**Step 3: Use the formula for the sum of the first n terms of an AP.**
Substituting the values we found:
\[ S_{25} = \frac{25}{2} \cdot (2 + 50) = \frac{25}{2} \cdot 52 \]
Calculating:
\[ S_{25} = 25 \cdot 26 = 650 \]

### Final Answers:
- The sum of the series \( 3 + 6 + 9 + ... + 96 \) is **1584**.
- The sum of the series \( 2 + 4 + 6 + ... + 50 \) is **650**.