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
