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]
