Sure! Let's find the partial sum of the arithmetic series [tex]\(1 + 5 + 9 + \cdots + 109\)[/tex].
### Step-by-Step Solution:
1. Identify the first term ([tex]$a$[/tex]) and the last term ([tex]$l$[/tex]):
- The first term [tex]\(a = 1\)[/tex].
- The last term [tex]\(l = 109\)[/tex].
2. Determine the common difference ([tex]$d$[/tex]) between consecutive terms:
- The second term is 5.
- Therefore, the common difference [tex]\(d = 5 - 1 = 4\)[/tex].
3. Find the total number of terms ([tex]$n$[/tex]) in the series:
- The nth term of an arithmetic series can be found using the formula:
[tex]\[
T_n = a + (n - 1)d
\][/tex]
- Set [tex]\(T_n = 109\)[/tex] to find [tex]\(n\)[/tex]:
[tex]\[
109 = 1 + (n - 1) \cdot 4
\][/tex]
Subtract 1 from both sides:
[tex]\[
108 = (n - 1) \cdot 4
\][/tex]
Divide both sides by 4:
[tex]\[
n - 1 = \frac{108}{4} = 27
\][/tex]
Add 1 to both sides:
[tex]\[
n = 27 + 1 = 28
\][/tex]
4. Calculate the sum of the first [tex]$n$[/tex] terms ([tex]$S_n$[/tex]):
- The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series is given by:
[tex]\[
S_n = \frac{n}{2} (a + l)
\][/tex]
- Substitute [tex]\(n = 28\)[/tex], [tex]\(a = 1\)[/tex], and [tex]\(l = 109\)[/tex]:
[tex]\[
S_n = \frac{28}{2} (1 + 109)
\][/tex]
Simplify:
[tex]\[
S_n = 14 \cdot 110
\][/tex]
Multiply:
[tex]\[
S_n = 1540
\][/tex]
Therefore, the partial sum [tex]\(1 + 5 + 9 + \cdots + 109\)[/tex] equals [tex]\(1540\)[/tex].
