The formula for the nth term of an arithmetic progression (A.P.) is \(a_n = a + (n − 1) d \). This implies that \(a_4 = a + 3d \), \(a_8 = a + 7d \), \(a_6 = a + 5d \), and \(a_{10} = a + 9d \). We are given that \(a_4 + a_8 = 24 \), which translates to \(a + 3d + a + 7d = 24 \). Simplifying this equation yields \(2a + 10d = 24 \), which further simplifies to \(a + 5d = 12 \) (Equation 1). We are also given that \(a_6 + a_{10} = 44 \), which translates to \(a + 5d + a + 9d = 44 \). Simplifying this equation yields \(2a + 14d = 44 \), which further simplifies to \(a + 7d = 22 \) (Equation 2). Subtracting Equation 1 from Equation 2 gives \(2d = 22 − 12 \), so \(2d = 10 \), and thus \(d = 5 \). Substituting \(d = 5 \) into Equation 1 (\(a + 5d = 12 \)) gives \(a + 5(5) = 12 \), so \(a + 25 = 12 \), which means \(a = −13 \). The second term is \(a_2 = a + d = −13 + 5 = −8 \). The third term is \(a_3 = a_2 + d = −8 + 5 = −3 \). Therefore, the first three terms of this A.P. are \(−13, −8,\) and \(−3\).