Answer :

To find the sum of the first 36 terms of the given arithmetic series (13, 19, 25, ...), we can follow these steps:

1. Identify the first term (a) and the common difference (d):
- First term, \( a = 13 \)
- Common difference, \( d = 19 - 13 = 6 \)

2. Determine the number of terms (n):
- Number of terms, \( n = 36 \)

3. Use the formula for the sum of the first \( n \) terms of an arithmetic series:
The formula is:
[tex]\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \][/tex]
Where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

4. Substitute the known values into the formula:
- \( a = 13 \)
- \( d = 6 \)
- \( n = 36 \)

[tex]\[ S_{36} = \frac{36}{2} \times (2 \times 13 + (36-1) \times 6) \][/tex]

5. Simplify the expression inside the parentheses:
- Calculate \( 2 \times 13 \):
[tex]\[ 2 \times 13 = 26 \][/tex]

- Calculate \( (36-1) \times 6 \):
[tex]\[ 35 \times 6 = 210 \][/tex]

- Add the two results together:
[tex]\[ 26 + 210 = 236 \][/tex]

6. Multiply by \(\frac{36}{2}\):
[tex]\[ S_{36} = 18 \times 236 \][/tex]

7. Complete the multiplication:
[tex]\[ 18 \times 236 = 4248 \][/tex]

Therefore, the sum of the first 36 terms of the arithmetic series 13, 19, 25, ..., to the nearest integer, is:
[tex]\[ \boxed{4248} \][/tex]

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