To find the 103rd term of the arithmetic sequence, we will first identify the first term and the common difference.
The first term \( a_1 \) is:
\[ a_1 = -\frac{3}{4} \]
Next, we need to determine the common difference \( d \). We can find it by subtracting the first term from the second term:
\[ d = a_2 - a_1 = -\frac{2}{3} - \left(-\frac{3}{4}\right) \]
To subtract these fractions, we first need a common denominator. The least common multiple of 3 and 4 is 12. Thus, we convert both fractions:
\[ -\frac{2}{3} = -\frac{8}{12} \]
\[ -\frac{3}{4} = -\frac{9}{12} \]
Now we can find the common difference \( d \):
\[ d = -\frac{8}{12} + \frac{9}{12} = \frac{1}{12} \]
Now we have:
- The first term \( a_1 = -\frac{3}{4} \)
- The common difference \( d = \frac{1}{12} \)
The formula for the \( n \)-th term of an arithmetic sequence is given by:
\[ a_n = a_1 + (n - 1) \cdot d \]
We want to find the 103rd term (\( n = 103 \)):
\[ a_{103} = -\frac{3}{4} + (103 - 1) \cdot \frac{1}{12} \]
Calculating \( (103 - 1) \cdot \frac{1}{12} \):
\[ 102 \cdot \frac{1}{12} = \frac{102}{12} = \frac{51}{6} \]
Now substitute this back into the formula:
\[ a_{103} = -\frac{3}{4} + \frac{51}{6} \]
To perform the addition, we need to express both fractions with a common denominator. The least common multiple of 4 and 6 is 12.
Convert \( -\frac{3}{4} \):
\[ -\frac{3}{4} = -\frac{9}{12} \]
Convert \( \frac{51}{6} \):
\[ \frac{51}{6} = \frac{102}{12} \]
Now we can add these fractions together:
\[ a_{103} = -\frac{9}{12} + \frac{102}{12} = \frac{102 - 9}{12} = \frac{93}{12} \]
Now we simplify \( \frac{93}{12} \):
\[ \frac{93}{12} = \frac{31}{4} \]
Thus, the 103rd term of the arithmetic sequence is:
\[ \boxed{\frac{31}{4}} \]