Find the sum of all numbers from 50 to 250 which are divisible by 6 and find `t_(13)`.

Answer

To solve the problem, we will break it down into two parts:

1. Finding the sum of all numbers from 50 to 250 that are divisible by 6.
2. Finding the 13th term (t13) of the arithmetic progression formed by these numbers.

### Step 1: Identify the first and last terms divisible by 6

- **First term (a)**: The smallest number greater than or equal to 50 that is divisible by 6.
- We can find this by calculating \( 50 \div 6 = 8.33 \). The next whole number is 9.
- Therefore, the first term is \( 6 \times 9 = 54 \).

- **Last term (l)**: The largest number less than or equal to 250 that is divisible by 6.
- We can find this by calculating \( 250 \div 6 = 41.67 \). The largest whole number is 41.
- Therefore, the last term is \( 6 \times 41 = 246 \).

### Step 2: Determine the number of terms (n)

- The numbers divisible by 6 from 54 to 246 form an arithmetic progression where:
- First term \( a = 54 \)
- Last term \( l = 246 \)
- Common difference \( d = 6 \)

To find the number of terms \( n \):
- The formula for the nth term of an arithmetic progression is given by:
\[ l = a + (n-1) \cdot d \]
Substituting the values we have:
\[ 246 = 54 + (n-1) \cdot 6 \]
Rearranging gives:
\[ 246 - 54 = (n-1) \cdot 6 \]
\[ 192 = (n-1) \cdot 6 \]
\[ n-1 = \frac{192}{6} = 32 \]
\[ n = 32 + 1 = 33 \]

### Step 3: Calculate the sum of the series (S_n)

The sum \( S_n \) of the first \( n \) terms of an arithmetic progression can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
Substituting the values we have:
\[ S_{33} = \frac{33}{2} \cdot (54 + 246) \]
Calculating inside the parentheses:
\[ 54 + 246 = 300 \]
Now substituting back:
\[ S_{33} = \frac{33}{2} \cdot 300 = 33 \cdot 150 = 4950 \]

### Step 4: Find the 13th term (t13)

The nth term of an arithmetic progression can be calculated using the formula:
\[ t_n = a + (n-1) \cdot d \]
For \( n = 13 \):
\[ t_{13} = 54 + (13-1) \cdot 6 \]
Calculating:
\[ t_{13} = 54 + 12 \cdot 6 = 54 + 72 = 126 \]

### Final Answers

- The sum of all numbers from 50 to 250 that are divisible by 6 is **4950**.
- The 13th term \( t_{13} \) is **126**.