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Find the 9th term of the arithmetic sequence.

[tex]\[
\begin{array}{l}
5, \frac{11}{3}, \frac{7}{3}, \ldots \\
a_9=\square
\end{array}
\][/tex]


Sagot :

To solve for the 9th term [tex]\(a_9\)[/tex] of the given arithmetic sequence, we follow these steps:

1. Identify the terms: The first three terms of the sequence are:
[tex]\[ a_1 = 5, \quad a_2 = \frac{11}{3}, \quad a_3 = \frac{7}{3} \][/tex]

2. Find the common difference [tex]\(d\)[/tex]: The common difference in an arithmetic sequence is found by subtracting any term from its subsequent term. We have:
[tex]\[ d = a_2 - a_1 = \frac{11}{3} - 5 \][/tex]

To perform this subtraction, we need to express 5 with the same denominator:
[tex]\[ d = \frac{11}{3} - \frac{15}{3} = \frac{11 - 15}{3} = \frac{-4}{3} \][/tex]

Therefore, the common difference [tex]\(d\)[/tex] is:
[tex]\[ d = -\frac{4}{3} \][/tex]

3. Use the formula for the [tex]\(n\)[/tex]-th term: The formula to find the [tex]\(n\)[/tex]-th term [tex]\(a_n\)[/tex] of an arithmetic sequence is:
[tex]\[ a_n = a_1 + (n - 1) d \][/tex]

4. Substitute the values to find [tex]\(a_9\)[/tex]:
Since [tex]\(n = 9\)[/tex], [tex]\(a_1 = 5\)[/tex], and [tex]\(d = -\frac{4}{3}\)[/tex],
[tex]\[ a_9 = 5 + (9 - 1) \left(-\frac{4}{3}\right) \][/tex]
Simplify the expression inside the parentheses:
[tex]\[ a_9 = 5 + 8 \left(-\frac{4}{3}\right) \][/tex]
Multiply [tex]\(8\)[/tex] by [tex]\(-\frac{4}{3}\)[/tex]:
[tex]\[ 8 \left(-\frac{4}{3}\right) = -\frac{32}{3} \][/tex]
Therefore,
[tex]\[ a_9 = 5 - \frac{32}{3} \][/tex]
Convert 5 to a fraction with denominator 3:
[tex]\[ 5 = \frac{15}{3} \][/tex]
Now, subtract the fractions:
[tex]\[ a_9 = \frac{15}{3} - \frac{32}{3} = \frac{15 - 32}{3} = \frac{-17}{3} \][/tex]

Thus, the 9th term of the arithmetic sequence is:
[tex]\[ a_9 = -\frac{17}{3} \approx -5.67 \][/tex]

Therefore, the 9th term [tex]\(a_9\)[/tex] is roughly:
[tex]\[ a_9 = -5.67 \][/tex]