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

Answer 1

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:

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