Let the first term be \(a\) and the common difference be \(d\).
The sum of the first \(n\) terms of an arithmetic progression is given by:
\(S_n = \frac{n}{2} (2a + (n-1)d)\)
For the first nine terms:
\(S_9 = \frac{9}{2} (2a + 8d) = 117\)
Solving for \(2a + 8d\):
\(9(2a + 8d) = 234\)
\(2a + 8d = 26\) (Equation 1)
The sum of the next four terms (terms 10 to 13) is 91:
\(S_{13} - S_9 = 91\)
\(S_{13} = \frac{13}{2} (2a + 12d)\)
\(\frac{13}{2} (2a + 12d) - \frac{9}{2} (2a + 8d) = 91\)
\(13(2a + 12d) - 9(2a + 8d) = 182\)
\(26a + 156d - 18a - 72d = 182\)
\(8a + 84d = 182\)
\(4a + 42d = 91\) (Equation 2)
Solving Equations 1 and 2 simultaneously:
From Equation 1: \(2a + 8d = 26\)
From Equation 2: \(4a + 42d = 91\)
Multiply Equation 1 by 2:
\(4a + 16d = 52\)
Subtract from Equation 2:
\((4a + 42d) - (4a + 16d) = 91 - 52\)
\(26d = 39\)
\(d = 1.5\)
Substitute \(d = 1.5\) into Equation 1:
\(2a + 8(1.5) = 26\)
\(2a + 12 = 26\)
\(2a = 14\)
\(a = 7\)
Thus, the first term is 7, and the common difference is 1.5.
