To find the 103rd term of the arithmetic sequence given by:
\[ { a_n } = \left{ -\frac{3}{4}, -\frac{2}{3}, -\frac{7}{12}, -\frac{1}{2}, \ldots \right} \]
we first need to determine the common difference of the sequence. The common difference \(d\) is calculated by subtracting any two consecutive terms:
- The first term \(a_1 = -\frac{3}{4}\)
- The second term \(a_2 = -\frac{2}{3}\)
Finding a common denominator for these fractions to calculate the difference:
- The common denominator for 4 and 3 is 12.
Converting the fractions:
- \(a_1 = -\frac{3}{4} = -\frac{9}{12}\)
- \(a_2 = -\frac{2}{3} = -\frac{8}{12}\)
Now, calculate the common difference:
\[ d = a_2 - a_1 = -\frac{8}{12} - \left(-\frac{9}{12}\right) = -\frac{8}{12} + \frac{9}{12} = \frac{1}{12} \]
Next, we can use the formula for the \(n\)-th term of an arithmetic sequence:
\[ a_n = a_1 + (n-1)d \]
Substituting \(a_1 = -\frac{3}{4}\), \(d = \frac{1}{12}\), and \(n = 103\):
\[ a_{103} = -\frac{3}{4} + (103 - 1) \cdot \frac{1}{12} \]
Simplifying:
\[ = -\frac{3}{4} + 102 \cdot \frac{1}{12} \]
Calculating \(102 \cdot \frac{1}{12}\):
\[ 102 \cdot \frac{1}{12} = \frac{102}{12} = \frac{17}{2} \]
Now substitute back:
\[ a_{103} = -\frac{3}{4} + \frac{17}{2} \]
Finding a common denominator (which is 4):
Convert \(\frac{17}{2}\):
\[ \frac{17}{2} = \frac{34}{4} \]
Now calculate:
\[ a_{103} = -\frac{3}{4} + \frac{34}{4} = \frac{-3 + 34}{4} = \frac{31}{4} \]
Thus, the 103rd term of the arithmetic sequence is:
\[ \boxed{\frac{31}{4}} \]