The sum of the first twelve terms of an Arithmetic Progression is 168. If the third term is 7, find the values of the common difference and the first term.
Explanation
A.P ; (S_{12} = 168 ; T_{3} = 7)
(S_{n} = frac{n}{2}(2a + (n – 1) d))
(T_{n} = a + (n – 1)d)
(168 = frac{12}{2}(2a + (12 – 1)d implies 168 = 6(2a + 11d))
(168 = 12a + 66d implies 84 = 6a + 33d ….. (1))
(7 = a + 2d …. (2))
From (2), a = 7 – 2d. Put into (1), we have
(84 = 6(7 – 2d) + 33d )
(84 = 42 – 12d + 33d implies 84 – 42 = 21d)
(21d = 42 implies d = 2)
(a = 7 – 2(2) = 7 – 4 = 3)
(a(text{first term}) = 3 ; d(text{common difference}) = 2)