Question

What is the 103rd term of the arithmetic sequence left brace a subscript n baseline right brace equals left brace negative 3 fourths comma negative 2 thirds comma negative Start Fraction 7 over 12 End Fraction comma negative 1 half comma ellipsis right brace?

(1 point)
Responses

Start Fraction 47 over 6 End Fraction
Image with alt text: Start Fraction 47 over 6 End Fraction

Start Fraction 31 over 4 End Fraction
Image with alt text: Start Fraction 31 over 4 End Fraction

Start Fraction 113 over 12 End Fraction
Image with alt text: Start Fraction 113 over 12 End Fraction

Start Fraction 37 over 4 End Fraction
Image with alt text: Start Fraction 37 over 4 End Fraction

Answer 1

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:

1. The first term \(a_1 = -\frac{3}{4}\)
2. 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:

\[
\boxed{\frac{31}{4}}
\]