To find the sum of the first 36 terms of an arithmetic sequence, we can use the formula for the sum of the first \( n \) terms of an arithmetic sequence:
\[ S_n = \frac{n}{2} \times (a_1 + a_n) \]
where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( n \) is the number of terms,
- \( a_1 \) is the first term,
- \( a_n \) is the \( n \)-th term.
We are given:
- \( a_{36} = 14 \) (the 36th term),
- \( d = \frac{1}{8} \) (the common difference),
- \( n = 36 \).
To use the formula, we need to find \( a_1 \) (the first term). The \( n \)-th term of an arithmetic sequence can be expressed in terms of the first term and the common difference:
\[ a_n = a_1 + (n-1)d \]
Substituting \( n = 36 \):
\[ a_{36} = a_1 + (36-1)d = a_1 + 35 \cdot \frac{1}{8} \]
Now we can set this equal to 14:
\[ 14 = a_1 + \frac{35}{8} \]
To solve for \( a_1 \), first, convert 14 to a fraction with the same denominator:
\[ 14 = \frac{112}{8} \]
Now we can substitute:
\[ \frac{112}{8} = a_1 + \frac{35}{8} \]
Subtract \( \frac{35}{8} \) from both sides:
\[ a_1 = \frac{112}{8} - \frac{35}{8} = \frac{112 - 35}{8} = \frac{77}{8} \]
Now we have \( a_1 \) and we can calculate \( S_{36} \):
Next, we need to calculate \( a_{36} \):
\[ S_{36} = \frac{36}{2} \times \left(a_1 + a_{36}\right) = 18 \times \left(\frac{77}{8} + 14\right) \]
Again, convert 14 to a fraction with a common denominator:
\[ 14 = \frac{112}{8} \]
Now combine:
\[ S_{36} = 18 \times \left(\frac{77}{8} + \frac{112}{8}\right) = 18 \times \frac{189}{8} \]
Calculate \( 18 \times 189 = 3402 \):
\[ S_{36} = \frac{3402}{8} \]
Now simplify \( \frac{3402}{8} \):
\[ 3402 \div 8 = 425.25 \]
Thus, the sum of the first 36 terms of the arithmetic sequence is \( \boxed{425.25} \).