To solve for [tex]\(a_{12}\)[/tex], [tex]\(a_{13}\)[/tex], and [tex]\(a_{14}\)[/tex] in the given arithmetic sequence:
1. Identify the given terms and the common difference:
- The first term, [tex]\(a_1\)[/tex], is 3.
- The second term, [tex]\(a_2\)[/tex], is 12.
- The third term, [tex]\(a_3\)[/tex], is 21.
2. Calculate the common difference ([tex]\(d\)[/tex]):
- The common difference [tex]\(d\)[/tex] in an arithmetic sequence can be found by subtracting the first term from the second term. Hence, [tex]\(d = a_2 - a_1 = 12 - 3 = 9\)[/tex].
3. Recall the formula for the [tex]\(n\)[/tex]-th term of an arithmetic sequence:
- The general formula for the [tex]\(n\)[/tex]-th term of an arithmetic sequence is given by:
[tex]\[
a_n = a_1 + (n - 1) \cdot d
\][/tex]
4. Substitute the values to find [tex]\(a_{12}\)[/tex], [tex]\(a_{13}\)[/tex], and [tex]\(a_{14}\)[/tex]:
- For [tex]\(a_{12}\)[/tex]:
[tex]\[
a_{12} = a_1 + (12 - 1) \cdot d = 3 + (12 - 1) \cdot 9 = 3 + 11 \cdot 9 = 3 + 99 = 102
\][/tex]
- For [tex]\(a_{13}\)[/tex]:
[tex]\[
a_{13} = a_1 + (13 - 1) \cdot d = 3 + (13 - 1) \cdot 9 = 3 + 12 \cdot 9 = 3 + 108 = 111
\][/tex]
- For [tex]\(a_{14}\)[/tex]:
[tex]\[
a_{14} = a_1 + (14 - 1) \cdot d = 3 + (14 - 1) \cdot 9 = 3 + 13 \cdot 9 = 3 + 117 = 120
\][/tex]
5. Write the results in the table:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$a_1$ & $a_2$ & $a_3$ & $a_{12}$ & $a_{13}$ & $a_{14}$ \\
\hline
3 & 12 & 21 & 102 & 111 & 120 \\
\hline
\end{tabular}
\][/tex]
Therefore, the values of [tex]\(a_{12}\)[/tex], [tex]\(a_{13}\)[/tex], and [tex]\(a_{14}\)[/tex] are 102, 111, and 120 respectively.
