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Answer:
To find the first term (\(a\)) and the common difference (\(d\)) of an arithmetic sequence (A.S.) given the 9th term (\(a_9\)) and the 17th term (\(a_{17}\)), we can use the formula for the \(n\)th term of an arithmetic sequence:
\[ a_n = a + (n-1)d \]
Given:
- The 9th term, \(a_9 = 15\)
- The 17th term, \(a_{17} = 27\)
We can set up two equations based on the formula for the \(n\)th term:
1. For the 9th term:
\[ a + 8d = 15 \]
2. For the 17th term:
\[ a + 16d = 27 \]
Now, we have a system of linear equations:
\[ a + 8d = 15 \]
\[ a + 16d = 27 \]
To solve for \(d\), we subtract the first equation from the second:
\[ (a + 16d) - (a + 8d) = 27 - 15 \]
\[ 8d = 12 \]
\[ d = \frac{12}{8} \]
\[ d = \frac{3}{2} \]
Now that we have \(d\), we can substitute it back into the first equation to find \(a\):
\[ a + 8 \left(\frac{3}{2}\right) = 15 \]
\[ a + 12 = 15 \]
\[ a = 15 - 12 \]
\[ a = 3 \]
So, the first term (\(a\)) is 3 and the common difference (\(d\)) is \(\frac{3}{2}\).