The 12th term = a + 11d where a is the first term and d is the common difference a + 11d = 74
The first three terms are a, a + d, a + 2d a + (a + d) + (a + 2d) = 42 3a + 3d = 42 a + d = 42 / 3 = 14 By subtracting this equation from the T12 equation (a + 11d) - (a + d) = 74 - 14 10d = 60 d = 6 (the common difference = 6) a + 6 = 14 a = 14 - 6 = 8 (the first term = 8)
S(n) = n/2 x (2a + (n-1)d) S(10) = 10/2 x (2x8 + (10-1)x6) S(10) = 5 x (16 + 54) = 5 x 70 = 350